﻿ ﻿THE SAT MATH: ADVANCED MATHEMATICS - McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)

## McGraw-Hill Education SAT 2017 Edition (Mcgraw Hill's Sat) (2016)

### THE SAT MATH: ADVANCED MATHEMATICS

1. Understanding Functions
2. Working with Quadratics and Other Polynomials
3. Working with Exponentials and Radicals
4. Working with Rational Expressions

Why are the Advanced Mathematics topics important on the SAT Math test?

About 27% (16 out of 58 points) of the SAT Math questions are Advanced Mathematics questions. Questions in this category are about

understanding of the structure of expressions and the ability to analyze, manipulate, and rewrite these expressions. This includes an understanding of the key parts of expressions, such as terms, factors, and coefficients, and the ability to interpret complicated expressions made up of these components .

It will also assess your skill in

rewriting expressions, identifying equivalent forms of expressions, and understanding the purpose of different forms .

The specific topics include

• solving, graphing, and analyzing quadratic equations
• solving equations with radicals that may include extraneous solutions
• solving systems including linear and quadratic equations
• creating exponential or quadratic functions from their properties
• calculating with and simplifying rational expressions
• analyzing radicals and exponentials with rational exponents
• creating equivalent forms of expressions to reveal their properties
• working with compositions and transformations of functions
• analyzing higher-order polynomial functions, particularly in terms of their factors and zeros

How is it used?

Fluency in these topics in advanced math is essential to success in postsecondary mathematics, science, engineering, and technology. Since these subjects constitute a portion of any liberal arts curriculum, and a substantial portion of any STEM (science, technology, engineering, or mathematics) program, colleges consider these to be essential college preparatory skills for potential STEM majors.

Sound intimidating? It”s not.

If you take the time to master the four core skills presented in these 14 lessons, you will gain the knowledge and practice you need to master SAT Advanced Math questions.

Skill 1: Understanding Functions

Lesson 1: What is a function?

A function is just a “recipe” for turning any “input” number into another number, called the “output” number. The input number is usually called x , and the output number is f (x ) or y . For instance, the function f (x ) = 3x 2 + 2 is a three-step recipe for turning any input number, x , into another number, f (x ), by the following steps: (1) square x , (2) multiply this result by 3, and (3) add 2 to this result. The final result is called f (x ) or y .

If f (2x ) = x + 2 for all values of x , which of the following equals f (x )?

1. A)
2. B)
3. C)
4. D) 2x– 2

(Medium ) Let”s use the “function-as-recipe” idea. The equation tells us that f is a function that turns an input of 2x into x + 2. What steps would we need to take to accomplish this?

Therefore, f is a two-step function that takes an input, divides it by 2, and then adds 2. Therefore, f (x ) equals the result when an input of x is put through the same steps, which yields .

Another way to think about this problem is to pick a value for x , like x = 1. Substituting this into the given equation gives us f (2(1)) = 1 + 2, or f (2) = 3. Therefore, the correct function must take an input of 2 and turn it into 3. If we substitute x = 2 into all of the choices, we get (A) f (2) = 2, (B) f (2) = 3, (C) f (2) = 0, and (D) f (2) = –1. Clearly, the only function that gives the correct output is (B).

The graph above shows the depth of water in a right cylindrical tank as a function of time as the tank drains. Which of the following represents the graph of the situation in which the tank starts with twice as much water as the original tank had, and the water drains at three times the original rate?

1. A)
2. B)
3. C)
4. D)

(Medium ) Although no increments are shown on the axes (so, for instance, the tick marks on the time axis could indicate minutes, or hours, or days, or any other time unit, and the tick marks on the depth axis could represent meters, or centimeters, or any other depth unit), we do know that the point at which the axes cross is the origin , or the point (0, 0). The given graph shows that the tank starts at 2 depth units and drains completely after 6 time units. In other words, the tank drains at 1/3 of a depth unit per time unit. (Remember from Chapter 8 that the slope equals the unit rate of change .) In the new graph, then, the tank should start at a depth of 2 × 2 = 4 depth units, and it should drain at 3 × 1/3 = 1 depth unit per time unit. In other words, it should take 4 time units for the tanks to drain completely. The only graph that shows this correctly is (A).

Lesson 2: Functions as graphs, equations, or tables

Make sure you”re fluent in expressing functions in three ways: as graphs in the xy -plane, equations in functional notation, or tables of ordered pairs. Also, make sure you can go from one format to another. Every input-output (xy ) pair can be represented in any of these three ways . For instance, if the function g turns an input value of –2 into an output value of 4, we can translate this in three ways:

• The graph ofy = g (x ) in the xy -plane contains the point (–2, 4).
• g(–2) = 4
• In a table of ordered pairs for the function,x = –2 is paired with y = 4.

The graphs of functions f and g are shown above for –3 ≤ x ≤ 3. Which of the following describes the set of all x for which f (x ) ≤ g (x )?

1. A) x≥ –3
2. B) –3 ≤x≤ –1 or 2 ≤ x ≤ 3
3. C) –1 ≤x≤ 2
4. D) 3 ≤x≤ 5

(Easy ) The key to this problem is understanding what the statement f (x ) ≤ g (x ) means. Since f (x ) and g (x ) are the y -values of the respective functions, f (x ) is less than or equal to g (x ) wherever the graphs cross or the graph of g (x ) is above the graph of f (x ). The two graphs cross at the points (–1, 4) and (2, 3), and g (x ) is above f (x ) at every point in between, so the correct answer is (C) –1 ≤ x ≤ 2.

Given the table of values for functions g and h above, for what value of x must g (h (x )) = 6?

1. A) 2
2. B) 5
3. C) 6
4. D) 12

(Medium-hard ) The notation g (h (x )) = 6 means that when the input number, x , is put into the function h , and this result is then placed into function g , the result is 6. Working backward, we should ask: what input to g would yield an output of 6? According to the table, only an input of 3 into g would yield an output of 6. This means that h (x ) = 3. So what input into h would yield an output of 3? Consulting the table again, we can see that g (5) = 3, and therefore x = 5 and the correct answer is (B).

Lesson 3: Compositions and transformations of functions

The notation f (g (x )) indicates the composition of two functions, g and f . The number x is put into the function g and this result is put into the function f and the result is called f (g (x )).

If f (x ) = x + 2 and f (g (1)) = 6, which of the following could be g (x )?

1. A) g(x) = 3x
2. B) g(x) = x + 3
3. C) g(x) = x – 3
4. D) g(x) = 2x + 1

(Medium-hard ) The notation f (g (1)) = 6 indicates that the number 1 is placed into function g , then the result is placed into function f , and the result is an output of 6.

Given equation:

f (g (1)) = 6

Use the given definition of f :

g (1) + 2 = 6

Subtract 2:

g (1) = 4

In other words, g is function that gives an output of 4 when its input is 1. The only function among the choices that has this property is (B) g (x ) = x + 3.

If f (x ) = x 2 + 1 and g (f (x )) = 2x 2 + 4 for all values of x , which of the following expresses g (x )?

1. A) g(x) = 2x + 1
2. B) g(x) = 2x + 2
3. C) g(x) = 2x + 3
4. D) g(x) = 2x 2 + 1

(Medium-hard ) As with the previous question, it helps to use the law of substitution to simplify the problem. By the definition of f , g (f (x )) = g (x 2 + 1) = 2x 2 + 4. Therefore, the function g turns an input of x 2 + 1 into an output of 2x 2 + 4. What series of steps would accomplish this?

Starting expression:

x 2 + 1

Multiply by 2:

2x 2 + 2

2x 2 + 4

Therefore, g is a two-step function that takes an input, multiplies it by 2, and adds 2, which is the function in choice (B).

When the function y = g (x ) is graphed in the xy -plane, it has a minimum value at the point (1, –2). What is the maximum value of the function h (x ) = –3g (x ) – 1?

1. A) 4
2. B) 5
3. C) 6
4. D) 7

(Medium ) The graph of y = h (x ) = –3g (x ) – 1 is the graph of g after it has been stretched vertically by a factor of 3, reflected over the x -axis, and then shifted down 1 unit. This would transform the minimum value point of (1, –2) to a maximum value point on the new graph at (1, –3(–2) – 1) or (1, 5), so the correct answer is (B).

Function Transformations

If the function y = f (x ) is graphed in the xy -plane (as in the example above), then the following represent transformations of function f .

The graph of y = f (x + k ), where k is a positive number, is the graph of y = f (x ) shifted left k units.

The graph of y = f (xk ), where k is a positive number, is the graph of y = f (x ) shifted right k units.

The graph of y = f (x ) + k , where k is a positive number, is the graph of y = f (x ) shifted up k units .

The graph of y = kf (x ) is the graph of y = f (x ) stretched vertically by a factor of k (if k > 1) or shrunk vertically by a factor of k (if k < 1).

The graph of y = –f (x ) is the graph of y = f (x ) reflected over the x -axis.

Exercise Set 1 (Calculator)

1

If f (x ) = x 2 + x + k , where k is a constant, and f (2) = 10, what is the value of f (–2)?

2

The minimum value of the function y = h (x ) corresponds to the point (–3, 2) on the xy -plane. What is the maximum value of g (x ) = 6 – h (x + 2)?

3

The function g is defined by the equation g (x ) = ax + b , where a and b are constants. If g (1) = 7 and g (3) = 6, what is the value of g (–5)?

4

Let the function h be defined by the equation h (x ) = f (g (x )) where f (x ) = x 2 – 1 and g (x ) = x + 5. What is the value of h (2)?

Questions 5–9 refer to the table below.

5

According to the table above, f (3) =

6

According to the table above, f (k (6)) =

7

According to the table above, k (k (6)) =

8

According to the table above, if k (f (x )) = 5, then what is the value of x ?

9

Which of the following is true for all values of x indicated in the earlier table?

1. A) f(k(x )) – k (f (x )) = 0
2. B) f(k(x )) + k (f (x )) = x
3. C) f(k(x )) – k (f (x )) = x
4. D) f(k(x )) + k (f (x )) = 0

10

If g (x – 1) = x 2 + 1, which of the following is equal to g (x )?

1. A) x2+ 2
2. B) x2+ 2x
3. C) x2+ 2x + 1
4. D) x2+ 2x + 2

11

If and f (x ) = (x – 1)2 , then which of the following is equal to f (h (x )) for all x ?

1. A)
2. B)
3. C)
4. D)

Exercise Set 1 (No Calculator)

Questions 12–19 are based on the graph below.

12

What is the value of g (–1)?

13

What is the value of g (f (3))?

14

What is the value of f (g (3))?

15

If g (f (x )) = –1, what is the value of x + 10?

16

If f (k ) + g (k ) = 0, what is the value of k ?

17

If f (a ) = g (a ), where a < 0, and f (b ) = g (b ), where b > 0, what is the value of a + b ?

18

Let h (x ) = f (x ) × g (x ). What is the maximum value of h (x ) if –3 ≤ x ≤ 3?

19

Which of the following graphs represents the function y = f (x ) + g (x )?

1. A)
2. B)
3. C)
4. D)

Calculator

1 . 6 f (2) = 22 + 2 + k = 10, so 6 + k = 10 and k = 4. Therefore, f (–2) = (–2)2 + (–2) + 4 = 6.

2 . 4 The graph of the function g (x ) = 6 – h (x + 2) is the graph of h after (1) a shift 2 units to the left, (2) a reflection over the x -axis, and (3) a shift 6 units up. If we perform these transformations on the point (–3, 2), we get the point (–5, 4), and so the maximum value of g is 4 when x = –5.

3 . 10

g (3) = a (3) + b = 6

g (1) = a (1) + b = 7

Subtract the equations:

2a = –1

Divide by 2:

a = –0.5

Substitute to find b :

–0.5 + b = 7

= 7.5

Therefore

g (x ) = –0.5x + 7.5

g (–5) = –0.5(–5) + 7.5 = 10

4 . 48

h (2) = f (g (2)) = f (2 + 5) = f (7) = (7)2 – 1 = 48

5 . 5

f (3) = 5

6 . 6

f (k (6)) = f (4) = 6

7 . 2

k (k (6)) = k (4) = 2

8 . 5 According to the table, the only input into k that yields an output of 5 is 1. Therefore, f (x ) must be 1, and the only input into f that yields an output of 1 is x = 5.

9 . A Examination of the table reveals that, for all given values of x , f (g (x )) = x and g (f (x )) = x . (This means that f and k are inverse functions , that is, they “undo” each other.) This implies that f (k (x )) – k (f (x )) = xx = 0.

10 . D One way to approach this question is to pick a new variable, z , such that z = x – 1 and therefore x = z + 1.

Original equation:

g (x – 1) = x 2 + 1

Substitute z = x – 1:

g (z ) = (z + 1)2 + 1

FOIL:

g (z ) = z 2 + 2z + 1 + 1

Simplify:

g (z ) = z 2 + 2z + 2

Therefore

g (x ) = x 2 + 2x + 2

11 . D

No Calculator

12 . 2 The graph of g contains the point (–1, 2), therefore g (–1) = 2.

13 . 3 The graph of f contains the point (3, 1); therefore, f (3) = 1, and so g (f (3)) = g (1). Since the graph of g contains the point (1, 3), g (1) = 3.

14 . 2 The graph of g contains the point (3, –1); therefore, g (3) = –1, and so f (g (3)) = f (–1). Since the graph of f contains the point (–1, 2), f (–1) = 2.

15 . 8 The only input to function g that yields an output of –1 is 3. Therefore, if g (f (x )) = –1, f (x ) must equal 3. The only input to f that yields an output of 3 is –2, therefore x = –2 and x + 10 = 8.

16 . 3 The only input for which f and g give outputs that are opposites is 3, because f (3) = 1 and g (x ) = –1.

17 . 1 The two points at which the graphs of g and f cross are (–1, 2) and (2, 1). Therefore, a = –1 and b = 2 and so a + b = 1.

18 . 4 h (x ) = f (x ) × g (x ) has a maximum value when x = –1, where f (1) × g (1) = 2 × 2 = 4.

19 . A To graph y = f (x ) + g (x ), we must simply “plot points” by choosing values of x and finding the corresponding y -values. For instance, if x = –3, y = f (3) + g (3) = 4 + 0 = 4, so the new graph must contain the point (–3, 4). Continuing in this manner for x = –2, x = –1, and so on yields the graph in (A).

Skill 2: Working with Quadratics and Other Polynomials

Lesson 4: Adding, multiplying, and factoring polynomials

A quadratic expression is a second-degree polynomial, that is, an expression of the form ax 2 + bx + c . The SAT Math test may ask you to analyze quadratic expressions and equations, as well as higher-order polynomials.

To factor a simple quadratic expression , first see if it fits any of the basic factoring formulas below.

Difference of Squares:

x 2a 2 = (x + a )(xa )

Perfect Square Trinomials:

x 2 + 2ax + a 2 = (x + a )(x + a ) = (x + a )2

x 2 – 2ax + a 2 = (xa )(xa ) = (xa )2

Which of the following is a factor of x 2 + 8x + 16?

1. A) x– 4
2. B) x– 8
3. C) x+ 4
4. D) x+ 8

(Easy ) Notice that this quadratic fits the pattern x 2 + 2ax + a 2 and therefore can be factored using the second formula above: x 2 + 8x + 16 = (x + 4)(x + 4). Therefore, the correct answer is (C).

To factor a more complex quadratic expression , use the Product-Sum Method illustrated below.

Which of the following is a factor of 6x 2 + 7x + 2?

1. A) 3x– 2
2. B) 3x+ 2
3. C) 3x– 1
4. D) 3x+ 1

(Medium ) First notice that this is a quadratic expression in which a = 6, b = 7, and c = 2. Now we can factor this expression using the Product-Sum Method .

Step 1: Call ac the product number (6 × 2 = 12), and b the sum number (7).

Step 2: Find the two numbers with a product equal to the product number and a sum equal to the sum number . What two numbers have a product of 12 and a sum of 7? A little guessing and checking should reveal that the numbers are 3 and 4.

Step 3: Rewrite the original quadratic, but expand the middle term in terms of the sum you just found: 6x 2 + 7x + 2 = 6x 2 + (3x + 4x ) + 2

Step 4: Use the associative law of addition to group the first two terms together and the last two terms together: 6x 2 + (3x + 4x ) + 2 = (6x 2 + 3x ) + (4x + 2)

Step 5: Factor out the greatest common factor from each pair. (6x 2 + 3x ) + (4x + 2) = 3x (2x + 1) + 2(2x + 1) If we do this correctly, the binomial factors will be the same.

Step 6: Factor out the common binomial factor. (3x + 2)(2x + 1)

Step 7: FOIL this result to confirm that it is equivalent to the original quadratic.

Therefore, the correct answer is (B).

Alternately, we could “test” each choice as a potential factor of 6x 2 + 7x + 2 until we find one that works. For instance, we can test choice (A) by trying to find another binomial factor that when multiplied by (3x – 2) gives a product of 6x 2 + 7x + 2. The best guess would be (2x – 1), because the product of the two first terms (3x × 2x ) gives us the correct first term, 6x 2 , and the product of the two last terms (–2 × –1) gives us the correct last term, 2. Now we FOIL the two binomials completely to see if we get the correct middle term: (3x – 2)(2x – 1) = 6x 2 – 3x – 4x + 2 = 6x 2 – 7x + 2, which has an incorrect middle term (–7x instead of 7x ). The fact that this is the opposite sign of the correct middle term suggests that we need only change the binomial from subtraction to addition, which gives us an answer of (B) 3x + 2.

To add or subtract polynomials , simply combine like terms.

Expression to be simplified:

(3x 4 + 5x 3 – 2x + 2) – (x 4 – 5x 3 + 2x 2 + 6)

Distribute to eliminate parentheses:

3x 4 + 5x 3 – 2x + 2 – x 4 + 5x 3 – 2x 2 – 6

Combine like terms:

(3x 4x 4 ) + (5x 3 + 5x 3 ) – 2x 2 – 2x + (2 – 6)

Simplify:

2x 4 + 10x 3 – 2x 2 – 2x – 4

Which of the following is equivalent to 2x (x + 1) – x 2 (x + 1) for all values of x ?

1. A) x2+ x
2. B) x3x 2 + 2x
3. C) –x3+ x 2 + 2x
4. D) –x3+ x 2 + 2x + 1

(Easy ) Original expression:

2x (x + 1) – x 2 (x + 1)

Distribute:

2x 2 + 2xx 3x 2

Combine like terms:

x 3 + x 2 + 2x

Therefore, the correct answer is (C).

When multiplying binomials , remember to FOIL .

Expression to be multiplied:

(ax + b )(cx + d )

F (product of the two “first” terms):

ax × cx = (ac )x 2

O (product of the two “outside” terms):

ax × d = (ad )x

I (product of the two “inside” terms):

b × cx = (bc )x

L (product of the two “last” terms):

b × d = bd

F + O + I + L:

(ac )x 2 + (ad )x + (bc )x + bd

Which of the following is equivalent to (2x – 7)(3x + 1) for all values of x ?

1. A) 6x2– 7
2. B) 6x2+ 5x – 7
3. C) 6x2– 21x – 7
4. D) 6x2– 19x – 7

(Easy ) Original expression:

(2x – 7)(3x + 1)

FOIL:

(2x )(3x ) + (2x )(1) +

(–7)(3x ) + (–7)(1)

Simplify:

6x 2 + 2x – 21x – 7

Combine like terms:

6x 2 – 19x – 7

Therefore, the correct answer is (D).

To multiply two polynomials , remember to distribute each term in the first polynomial to each term in the second polynomial, then simplify. FOILing is just a special example of this kind of distribution.

Expression to be simplified:

(2x 2x + 2) × (x 3 + x – 1)

Distribute:

(2x 2 )(x 3 ) + (2x 2 )(x ) – (2x 2 )(1) –

(x )(x 3 ) – (x )(x ) + (x )(1) + (2)(x 3 ) + (2)(x ) – (2)(1)

Simplify:

2x 5 + 2x 3 – 2x 2x 4x 2 + x + 2x 3 + 2x – 2

Combine like terms:

2x 5x 4 + 4x 3 – 3x 2 + 3x – 2

To solve tougher quadratic equations , first use the Laws of Equality to set one side of the equation to 0, then factor and use the Zero Product Property.

Zero Product Property: If the product of any set of numbers is 0, then at least one of those numbers must be 0.

Which of the following is a solution to the equation 8 – x 2 = –2x ?

1. A) –4
2. B) –3
3. C) –2
4. D) –1

We could just plug in each number in the choices to the equation until we find one that works. But it”s good to know the general method for finding both solutions. In this case, the fact that the numbers in the choices are all integers suggests that this quadratic is factorable.

To solve tougher quadratic equations, first use the Laws of Equality to set one side of the equation to 0, then factor and use the Zero Product Property .

Quadratic Formula: If ax2 + bx + c = 0,

then

The equation ax 2 + bx + c = 0 has no real solutions if b 24 ac < 0 . This is because the square root of a negative number is not a real number.

Which of the following is a solution to the equation 3x 2 = 4x + 2?

1. A)
2. B)
3. C)
4. D)

(Medium ) Although we could just plug the numbers in the choices back into the equation to see which one works, it”s a bit of a pain to do that with such obnoxious numbers. The ugliness of these numbers also tells us that this quadratic is not easily factorable. Therefore, it”s probably best to use the Quadratic Formula.

Equation to be solved:

3x 2 = 4x + 2

Subtract 4x and 2 to set right side to 0:

3x 2 – 4x – 2 = 0

Therefore, the correct answer is (D).

If x > 0 and x 2 – 5x = 6, what is the value of x ?

Since the problem states that x > 0, the correct answer is 6.

Alternately, we could have used the Quadratic Formula on the equation x 2 – 5x – 6 = 0:

If a quadratic equation has the form x 2 + bx + c = 0, the zeros of the quadratic must have a sum of –b and a product of c .

This is because if a = 1, the quadratic formula gives solutions of and .

Sum of zeros :

Product of zeros:

If one of the solutions to the equation 2x 2 – 7x + k = 0 is x = 5, what is the other possible value of x ?

1. A)
2. B)
3. C)
4. D)

(Medium-hard ) We can start by substituting x = 5 into the original equation and solving:

Original equation:

2x 2 – 7x + k = 0

Substitute x = 5:

2(5)2 – 7(5) + k = 0

Simplify:

15 + k = 0

Subtract 15:

k = –15

Rewrite original equation with k = –15:

2x 2 – 7x – 15 = 0

Factor with Product-Sum Method:

(2x + 3)(x – 5) = 0

Use Zero Product Property:

x = –3/2 or 5

Therefore, the correct answer is (A).

Alternately, we can save a bit of time and effort by using the theorem above.

Original equation:

2x 2 – 7x + k = 0

Divide by 2:

Since the quadratic is now in the form x 2 + bx + c = 0, we know that the sum of the solutions must be 7/2, or 3.5. Therefore, if one of the solutions is 5, the other must be 3.5 – 5 = –1.5, or –3/2.

Lesson 6: Analyzing the graphs of quadratic functions

The graph of any quadratic function in the xy -plane, that is, a function of the form y = f (x ) = ax 2 + bx + c , has the following important features:

• It is a parabola with a vertical axis of symmetry at .
• They -intercept is c , since f (0) = a (0)2 + b (0) + c = c .
• If it crosses thex -axis, it does so at the points and .
• Ifa is positive, the parabola is “open up,” and if a is negative, it is “open down.”
• If the quadratic is in the formy = a (xh )2 + k , then the vertex of the parabola is (h , k ).

The graph of the quadratic function y = g (x ) in the xy -plane is a parabola with vertex at (3, –2). If this graph also passes through the origin, which of the following must equal 0?

1. A) g(4)
2. B) g(5)
3. C) g(6)
4. D) g(7)

(Medium ) It”s helpful to draw a sketch of this parabola so that we can see its shape.

For this question, the axis of symmetry is key. Since the parabola has a vertex of (3, –2), its axis of symmetry is x = 3. The zeros of the parabola (the points where y = 0, or where the graph crosses the x -axis) must be symmetric to this line. Since the origin is 3 units to the left of this axis, the other zero must be three units to the right of the axis, or at the point (6, 0). This means that g (6) must equal 0, and the correct answer is (C).

Notice that we don”t need to do anything complicated, like find the specific quadratic equation (which would be a pain in the neck).

When the quadratic function f is graphed in the xy -plane, its graph has a positive y -intercept and two distinct negative x -intercepts. Which of the following could be f ?

1. A) f(x) = –2(x + 3)(x + 1)
2. B) f(x) = 3(x + 2)2
3. C) f(x) = –4(x – 2)(x – 3)
4. D) f(x) = (x + 1)(x + 3)

(Easy ) Since the functions are all given in factored form, it is easy to see where their zeros lie by using the Zero Product Property. The function in (A) has zeros (x -intercepts) at x = –3 and x = –1, which are both negative, but its y -intercept is f (0) = –2(3)(1) = –6, which is of course not positive. The only choice that gives two distinct x -intercepts and a positive value for f (x ) is choice (D) f (x ) = (x + 1)(x + 3), which has x -intercepts at x = –1 and x = –3, and a y -intercept at y = 3.

The quadratic function h is defined by the equation h (x ) = ax 2 +bx + c , where a is a negative constant and c is a positive constant. Which of the following could be the graph of h in the xy -plane?

1. A)
2. B)
3. C)
4. D)

(Easy ) The graph of y = ax 2 + bx + c is an “open down” parabola if a is negative, and has a y -intercept of c . The only “open down” parabola with a positive y -intercept is choice (B).

Exercise Set 2 (No Calculator)

1

If (x – 2)(x + 2) = 0, then x 2 + 10 =

2

If (a – 3)(a + k ) = a 2 + 3a – 18 for all values of a , what is the value of k ?

3

When the quadratic function y = 10(x + 4)(x + 6) is graphed in the xy -plane, the result is a parabola with vertex at (a , b ). What is the value of ab ?

4

If the function y = 3x 2kx – 12 has a zero at x = 3, what is the value of k ?

5

If the graph of a quadratic function in the xy -plane is a parabola that intersects the x -axis at x = –1.2 and x = 4.8, what is the x -coordinate of its vertex?

6

If the graph of y = a (xb )(x – 4) has a vertex at (5, –3), what is the value of ab ?

7

What is the sum of the zeros of the function h (x ) = 2x 2 – 5x – 12?

8

If x = –5 is one of the solutions of the equation 0 = x 2ax – 12, what is the other solution?

9

Which of the following is equivalent to 2a (a – 5) + 3a 2 (a + 1) for all values of a ?

1. A) 6a4– 24a 3 – 6
2. B) 5a5+ 3a 2 – 10a
3. C) 3a3+ 5a 2 – 10a
4. D) 3a3+ 2a 2 – 10a – 6

10

Which of the following functions, when graphed in the xy -plane, has exactly one negative x -intercept and one negative y -intercept?

1. A) y= –x2 – 6x – 9
2. B) y= –x2 + 6x – 9
3. C) y=x 2 + 6x + 9
4. D) y=x 2 – 6x + 9

11

If 2x 2 + 8x = 42 and x < 0, what is the value of x 2 ?

1. A) 4
2. B) 9
3. C) 49
4. D) 64

12

When the function y = h (x ) = ax 2 + bx + c is graphed in the xy -plane, the result is a parabola with vertex at (4, 7). If h (2) = 0, which of the following must also equal 0?

1. A) h(5)
2. B) h(6)
3. C) h(8)
4. D) h(9)

Exercise Set 2 (Calculator)

13

If x > 0 and 2x 2 – 4x = 30, what is the value of x ?

14

If x 2 + bx + 9 = 0 has only one solution, and b > 0, what is the value of b ?

15

When y = 5(x – 3.2)(x – 4.6) is graphed in the xy -plane, what is the value of the y -intercept?

16

When y = 5(x – 3.2)(x – 4.6) is graphed in the xy -plane, what is the x -coordinate of the vertex?

17

If (2x – 1)(x + 3) + 2x = 2x 2 + kx – 3 for all values of x , what is the value of k ?

18

If b 2 + 20b = 96 and b > 0, what is the value of b + 10?

19

The graph of y = f (x ) in the xy -plane is a parabola with vertex at (3, 7). Which of the following must be equal to f (–1)?

1. A) f(2)
2. B) f(4)
3. C) f(7)
4. D) f(15)

20

Which of the following functions, when graphed in the xy -plane, has two positive x -intercepts and a negative y -intercept?

1. A) y= –2(x– 1)(x + 5)
2. B) y= –2(x+ 3)2
3. C) y= –2(x– 5)2
4. D) y= –2(x– 1)(x – 5)

21

Which of the following equations has no real solutions?

1. A) x2– 3x + 2 = 0
2. B) x2– 3x – 2 = 0
3. C) x2+ 2x – 3 = 0
4. D) x2+ 2x + 3 = 0

22

The graph of the function y = a (x + 6)(x + 8) has an axis of symmetry at x = k . What is the value of k ?

1. A) –7
2. B) –6
3. C) 7
4. D) 8

23

The graph of the quadratic function y = f (x ) in the xy -plane is a parabola with vertex at (6, –1). Which of the following must have the same value as the y -intercept of this graph?

1. A) f(–2)
2. B) f(3.5)
3. C) f(12)
4. D) f(13.5)

No Calculator

1 . 14

(x – 2)(x + 2) = 0

FOIL:

x 2 – 4 = 0

x 2 + 10 = 14

2 . 6

(a – 3)(a + k ) = a 2 + 3a – 18

FOIL:

a 2 + (k – 3)a – 3k = a 2 + 3a – 18

Equate coefficients:

k – 3 = 3; –3k = –18

Therefore k = 6.

3 . 50 By the Factor Theorem, the parabola has x -intercepts at x = –4 and x = –6. The x -coordinate of the vertex is the average of these zeros, or –5. To get the y -coordinate of the vertex, we just plug x = –5 back into the equation: y = 10(–5 + 4)(–5 + 6) = 10(–1)(1) = –10. Therefore a = –5 and b = –10 and so ab = 50.

4 . 5

When x = 3, y = 0: 0 = 3(3)2k (3) – 12

Simplify:

0 = 27 – 3k – 12

Simplify:

0 = 15 – 3k

3k = 15

Divide by 3:

k = 5

5 . 1.8 The x -coordinate of the vertex is the average of the x -intercepts (if they exist): (–1.2 + 4.8)/2 = 3.6/2 = 1.8.

6 . 18 The x -coordinate of the vertex is the average of the x -intercepts (if they exist):

5 = (b + 4)/2

Multiply by 2:

10 = b + 4

Subtract 4:

6 = b

Substitute x = 5 and y = –3 into equation to find the value of a :

–3 = a (5 – 6)(5 – 4) = –a

Multiply by –1:

3 = a

Therefore, ab = (3)(6) = 18

7 . 2.5

0 = 2x 2 – 5x – 12

Factor:

0 = (2x + 3)(x – 4)

Therefore, the zeros are x = –3/2 and x = 4, which have a sum of 2.5. Alternately, you can divide the original equation by 2:

0 = x 2 – 2.5x – 12

and recall that any quadratic in the form x 2 + bx + c = 0 must have zeros that have a sum of –b and a product of c . Therefore, without having to calculate the zeros, we can see that they have a sum of –(–2.5) = 2.5.

8 . 2.4 We know that one of the zeros is x = –5, and we want to find the other, x = b . We can use the Factor Theorem:

x 2ax – 12 = (x + 5)(xb )

FOIL:

x 2ax – 12 = x 2 + (5 – b )x – 5b

Since the constant terms must be equal, 12 = 5b and therefore, b = 12/5 = 2.4.

9 . C

2a (a – 5) + 3a 2 (a + 1)

Distribute:

2a 2 – 10a + 3a 3 + 3a 2

Collect like terms:

3a 3 + 5a 2 – 10a

10 . A Substitute x = 0 to find the y -intercept of each graph. Only (A) and (B) yield negative y -intercepts, so (C) and (D) can be eliminated. Factoring the function in (A) yields y = –(x + 3), which has only a single x -intercept at x= –3.

11 . C

2x 2 + 8x = 42

Divide by 2:

x 2 + 4x = 21

Subtract 21:

x 2 + 4x – 21 = 0

Factor:

(x + 7)(x – 3) = 0

Therefore, x = –7 or 3, but since x < 0, x = –7 and therefore, x 2 = (–7)2 = 49.

12 . B Draw a quick sketch of the parabola. Since it has a vertex at (4, 7), it must have an axis of symmetry of x = 4. The two zeros of the function must be symmetric to the line x = 4, and since the zero x = 2 is two units to the left of the axis, the other must by 2 units to the right, at x = 6.

Calculator

13 . 5

2x 2 – 4x = 30

Divide by 2:

x 2 – 2x = 15

Subtract 15:

x 2 – 2x – 15 = 0

Factor:

(x – 5)(x + 3) = 0

Therefore, x = 5 or –3. But since x > 0, x = 5.

14 . 6 Let”s call the one solution a . If it is the only solution, the two factors must be the same:

x 2 + bx + 9 = (xa )(xa )

FOIL:

x 2 + bx + 9 = x 2 – 2ax + a 2

Therefore, b = –2a and a 2 = 9. This means that x = 3 or –3 and so b = –2(3) = –6 or –2(–3) = 6. Since b must be positive, b = 6.

15 . 73.6 The y -intercept is simply the value of the function when x = 0: y = 5(0 – 3.2)(0 – 4.6) = 73.6.

16 . 3.9 The x -coordinate of the vertex is simply the average of the zeros: (3.2 + 4.6)/2 = 3.9.

17 . 7

(2x – 1)(x + 3) + 2x = 2x 2 + kx – 3

FOIL:

2x 2 + 5x – 3 + 2x = 2x 2 + kx – 3

Simplify:

2x 2 + 7x – 3 = 2x 2 + kx – 3

Subtract 2x 2 and add 3:

7x = kx

Divide by x :

7 = k

18 . 14

b 2 + 20b = 96

Subtract 96:

b 2 + 20b – 96 = 0

Factor:

(b – 4)(b + 24) = 0

Therefore, b = 4 or –24, but if b > 0, then b must equal 4, and therefore, b + 10 = 14. Alternately, you might notice that adding 100 to both sides of the original equation gives a “perfect square trinomial” on the left

side:

b 2 + 20b + 100 = 196

Factor:

(b + 10)2 = 196

Take square root:

b + 10 = ±14

If b > 0:

b + 10 = 14

19 . C Since the vertex of the parabola is at (3, 7), the axis of symmetry is x = 3. Since x = –1 is 4 units to the left of this axis, and x = 7 is 4 units to the right of this axis, f (–1) must equal f (7).

20 . D y = –2(x – 1)(x – 5) has x -intercepts at x = 1 and x = 5 and a y -intercept of y = –10. (Notice that the function in (C) has only one positive x -intercept at x = 5.)

21 . D This one is tough. Since this question allows a calculator, you could solve this by graphing or with the Quadratic Formula. Remember that a quadratic equation has no real solution if b 2 – 4ac < 0. The only choice for which b 2 – 4ac is negative is (D). Alternately, if you graph the left side of each equation as a function in the xy -plane (which I only advise if you have a good graphing calculator), you will see that the function in (D) never crosses the x -axis, implying that it cannot equal 0.

22 . A This quadratic has zeros at x = –6 and x = –8, so its axis of symmetry is at the midpoint of the zeros, at x = –7.

23 . C If the vertex of the parabola is at (6, –1), its axis of symmetry must be x = 6. The y -intercept of the function is f (0), which is the value of y when x = 0. Since this point is 6 units to the left of the axis of symmetry, its reflection over the axis of symmetry is 6 units to the rights of the axis, at f (12).

Lesson 7: Analyzing polynomial equations

The Factor Theorem

• If a polynomial expression has a zero (a value ofx for which the polynomial equals 0) at x = a , it must have a factor of (xa ).
• Conversely, if a polynomial has a factor of (xa ), it must have a zero at x = a .

The function f is defined by the equation f (x ) = x 3ax 2bx + 20 where a and b are constants. In the xy -plane, the graph of y = f (x ) intersects the x -axis at the points (–2, 0), (2, 0), and (p , 0). What is the value of p ?

1. A) 4
2. B) 5
3. C) 10
4. D) 20

(Medium-hard ) Since x = –2 and x = 2 and x = p are zeros of the function (that is, they are inputs that yield an output of 0), the polynomial must have (x + 2), (x – 2), and (xp ) as factors.

f (x ) = x 3ax 2bx + 20 = (x + 2)(x – 2)(xp )

FOIL (x + 2)(x – 2):

= (x 2 – 4)(xp )

FOIL (x 2 – 4)(xp ):

= x 3px 2 – 4x + 4p

Since x 3px 2 – 4x + 4p must be equivalent to x 3ax 2bx + 20, all of the corresponding coefficients must be equal. That is, –p = –a , –4 = –b , and 4p = 20. Therefore, p = 5, a = 5, and b = 4, and the correct answer is (B).

Which range of values defines all of the values of x for which the function f in the previous question is positive?

1. A) x< –2 orx > 2
2. B) –2 <x< 5
3. C) –2 <x< 2 or x > 5
4. D) 2 <x< 5

When analyzing a polynomial function, you may find it very helpful to draw its graph in the xy = plane. Sometimes the x -and y -intercepts are all you need to get a good picture by hand. You should also know how to use the graphing function on your calculator, when it is permitted.

(Hard ) This question is easier to solve if we have a graph of the function. Since we know that the equation of the function is y = (x + 2)(x – 2)(x – 5), we know that it has x -intercepts at x = –2, x = 2, and x = 5, and a y -intercept at y= (0 + 2)(0 – 2)(0 – 5) = 20. Therefore, the graph looks like this:

On this graph, the points where f is positive are the points above the x -axis. This corresponds to the points where x is between –2 and 2, and where x is greater than 5. Therefore, the correct answer is (C).

The figure above shows the graph of a system of two equations in the xy -plane. How many solutions does this system have?

1. A) Zero
2. B) One
3. C) Two
4. D) Three

(Easy ) Finding the solutions to a system of equations means finding the ordered pairs that satisfy all of the equations simultaneously. (If you need to review how to solve systems, see Chapter 7 .) If the equations are graphed, the solutions correspond to any points where all of the graphs meet. In this case, the two graphs cross in two distinct points, so the system has two solutions and the answer is (C).

y + 2x = 6

y = x 2 + 3x

Given the system above, which of the following could be the value of y ?

1. A) 1 or –6
2. B) 0 or –5
3. C) 0 or 10
4. D) 4 or 18

(Medium ) Perhaps the simplest way to solve this system is with the process of substitution, which we applied to linear systems in Chapter 7 , Lesson 12.

First equation:

y + 2x = 6

Substitute y = x 2 + 3x :

x 2 + 3x + 2x = 6

Subtract 6:

x 2 + 5x – 6 = 0

Factor with Product-Sum Method:

(x + 6)(x – 1) = 0

Apply Zero-Product Property:

x = –6 or 1

But be careful. You may be tempted to choose (A) 1 or –6, but the question asks for the value of y , not x . To find the corresponding values of y , we must plug our x -values back into one of the equations: y = (–6)2 + 3(–6) = 18 or y= (1)2 + 3(1) = 4; therefore, the correct answer is (D).

y = 1

x 2 + y 2 = 4

y = x 2

How many distinct ordered pairs (x , y ) satisfy the three-equation system above?

1. A) Zero
2. B) One
3. C) Two
4. D) Three

(Medium ) To find the solutions of a system means to find the ordered pairs (x , y ) that satisfy all of the equations simultaneously. Although graphing this system is not too hard, it is probably simpler to solve this system algebraically.

Substitute the first equation, y = 1, into the other two:

x 2 + (1)2 = 4

1 = x 2

Use x 2 = 1 to substitute into other equation:

(1) + (1)2 = 4

Simplify:

2 = 4

Since this yields an equation that can never be true, regardless of the values of the unknowns, there is no real solution to this system, and the correct answer is (A).

If you graph this system, it will show a horizontal line, a circle, and a parabola. You will see that no point exists where all three graphs meet, indicating that the system has no solution.

Exercise Set 3 (No Calculator)

1

If x 3 – 7x 2 + 16x – 12 = (xa )(xb )(xc ) for all values of x , what is the value of abc ?

2

If x 3 – 7x 2 + 16x – 12 = (xa )(xb )(xc ) for all values of x , what is the value of a + b + c ?

3

1. Ifx3 – 7x 2 + 16x – 12 = (xa )(xb )(xc ) for all values of x , what is the value of ab + bc + ac ?

4

If x 2ax + 12 has a zero at x = 3, what is the value of a ?

5

If x 2ax + 12 has a zero at x = 3, at what other value of x does it have a zero?

6

y = 4x 2 + 2

x + y = 16

When the two equations in the system above are graphed in the xy -plane, they intersect in the point (a , b ). If a > 0, what is the value of a ?

7

x 2 + y 2 = 9

Which of the following equations, if graphed in the xy -plane, would intersect the graph of the equation above in exactly one point?

1. A) y= –4
2. B) y= –3
3. C) y= –1
4. E) y= 0

8

If g (x ) = a (x + 1)(x – 2)(x – 3) where a is a negative constant, which of the following is greatest?

1. A) g(0.5)
2. B) g(1.5)
3. C) g(2.5)
4. D) g(3.5)

9

If 2x 2 + ax + b has zeros at x = 5 and x = –1, what is the value of a + b ?

1. A) –18
2. B) –9
3. C) –2
4. D) –1

10

If the graph of the equation y = ax 4 + bx in the xy -plane passes through the points (2, 12) and (–2, 4), what is the value of a + b ?

1. A) 0.5
2. B) 1.5
3. C) 2.0
4. D) 2.5

11

If the function y = 3(x 2 + 1)(x 3 – 1)(x + 2) is graphed in the xy -plane, in how many distinct points will it intersect the x -axis?

1. A) Two
2. B) Three
3. C) Four
4. D) Five

Exercise Set 3 (Calculator)

12

If x 2 + y = 10x and y = 25, what is the value of x ?

13

If 2x 3 – 5xa has a zero at x = 4, what is the value of a ?

14

If x > 0 and x 4 – 9x 3 – 22x 2 = 0, what is the value of x ?

15

If d is a positive constant and the graph in the xy -plane of y = (x 2 )(x 2 + x – 72)(xd ) has only one positive zero, what is the value of d ?

16

y = 2x 2 + 18

y = ax

In the system above, a is a positive constant. When the two equations are graphed in the xy -plane, they intersect in exactly one point. What is the value of a ?

17

4a 2 – 5b = 16

3a 2 – 5b = 7

Given the system of equations above, what is the value of a 2 b 2 ?

18

For how many distinct positive integer values of n is (n –1)(n – 9)(n – 17) less than 0?

1. A) Six
2. B) Seven
3. C) Eight
4. D) Nine

19

x 2 + 2y 2 = 44

y 2 = x – 2

When the two equations above are graphed in the xy -plane, they intersect in the point (h, k ). What is the value of h ?

1. A) –8
2. B) –6
3. C) 6
4. D) 8

20

m 2 + 2n = 10

2m 2 + 2n = 14

Given the system of equations above, which of the following could be the value of m + n ?

1. A) –7
2. B) –2
3. C) 1
4. D) 2

21

For how many distinct values of x does (x 2 – 4)(x – 4)2 (x 2 + 4) equal 0?

1. A) Three
2. B) Four
3. C) Five
4. D) Six

22

The function f (x ) is defined by the equation f (x ) = a (x + 2)(xa )(x – 8) where a is a constant. If f (2.5) is negative, which of the following could be the value of a ?

1. A) –2
2. B) 0
3. C) 2
4. D) 4

No Calculator

1 . 12 When the expression (xa )(xb )(xc ) is fully distributed and simplified, it yields the expression x 3 – (a + b + c )x 2 + (ab + bc + ac )xabc . If this is equivalent to x 3 – 7x 2 + 16x – 12 for all values of x , then all of the corresponding coefficients must be equal.

2 . 7 See question 1.

3 . 16 See question 1.

4 . 7 If x 2ax + 12 = 0 when x = 3, then

(3)2 – 3a + 12 = 0

Simplify:

21 – 3a = 0

21 = 3a

Divide by 3:

7 = a

5 . 4 As we saw in question 4, a = 7.

x 2 – 7x + 12

Factor:

(x – 3)(x – 4)

Therefore, the zeros are 3 and 4.

6 . 7/4 or 1.75

x + y = 16

Subtract x :

y = 16 – x

Substitute:

16 – x = 4x 2 + 2

0 = 4x 2 + x – 14

Factor:

0 = (4x – 7)(x + 2)

Therefore, x = –2 or 7/4, but if x must be positive, it equals 7/4.

7 . B The graph of the given equation is a circle centered at the origin with a radius of 3. Therefore, the horizontal line at y = –3 just intersects it at (0, –3). You can also substitute y = –3 into the original equation and verify that it gives exactly one solution.

8 . C Just notice the sign of each factor for each input:

g (0.5) = (–)(+)(–)(–) = negative

g (1.5) = (–)(+)(–)(–) = negative

g (2.5) = (–)(+)(+)(–) = positive

g (3.5) = (–)(+)(+)(+) = negative

Since (C) is the only option that yields a positive value, it is the greatest.

9 . A

2x 2 + ax + b

If x = 5 is a zero:

2(5)2 + 5a + b = 0

Subtract 50:

5a + b = –50

If x = –1 is a zero:

2(–1)2 + a (–1) + b = 0

Subtract 2:

a + b = –2

Multiply by –1:

ab = 2

6a = –48

Divide by 6:

a = –8

Substitute a = –8:

–8 – b = 2

b = 10

Multiply by –1:

b = –10

Therefore, a + b = –8 + –10 = –18.

10 . D

Substitute (2, 12):

12 = a (2)4 + b (2)

Simplify:

16a + 2b = 12

Substitute (–2, 4):

4 = a (–2)4 + b (–2)

Simplify:

16a – 2b = 4

32a = 16

Divide by 32:

a = ½

Substitute:

16(1/2) + 2b = 12

Subtract 8:

2b = 4

Divide by 2:

b = 2

Therefore, a + b = 2.5.

11 . A Use the Zero Product Property. The factor (x 2 + 1) cannot be zero for any value of x , (x 3 – 1) is zero when x = 1, and (x + 2) is zero when x = –2. Therefore, there are only two distinct points in which this graph touches the x -axis.

Calculator

12 . 5 Substitute y = 25:

x 2 + 25 = 10x

Subtract 10x :

x 2 – 10x + 25 = 0

Factor:

(x – 5)(x – 5) = 0

Use Zero Product Property:

x = 5

13 . 108 If x = 4 is a zero:

2(4)3 – 5(4) – a = 0

Simplify:

108 – a = 0

108 = a

14 . 11

x 4 – 9x 3 – 22x 2 = 0

Divide by x 2 :

x 2 – 9x – 22 = 0

Factor:

(x – 11)(x + 2) = 0

Use Zero Product Property:

x = 11 or –2

15 . 8

y = (x 2 )(x 2 + x – 72)(xd )

Factor:

y = (x 2 )(x + 9)(x – 8)(xd )

By the Zero Property, the zeros are x = 0, –9, 8, or d . Since d is positive, but there can only be one positive zero, d = 8.

16 . 12

y = 2x 2 + 18

Substitute y = ax :

ax = 2x 2 + 18

Subtract ax :

0 = 2x 2ax + 18

Divide by 2:

If the graphs intersect in only one point, the system must have only one solution, so this quadratic must be a “perfect square trinomial” as discussed in Lesson 4.

Equate coefficients:

b 2 = 9

2b = a /2

The only positive solution to this system is b = 3 and a = 12.

17 . 144

4a 2 – 5b = 16

3a 2 – 5b = 7

Subtract equations:

a 2 = 9

Substitute a 2 = 9:

3(9) – 5b = 7

Subtract 27:

–5b = –20

Divide by –5:

b = 4

Therefore, a 2 b 2 = 9(4)2 =144.

18 . B In order for the product of three numbers to be negative, either all three numbers must be negative or exactly one must be negative and the others positive. Since n must be a positive integer, n – 1 cannot be negative, and so there must be two positive factors and one negative. The only integers that yield this result are the integers from 10 to 16, inclusive, which is a total of seven integers.

19 . C

x 2 + 2y 2 = 44

Substitute y 2 = x – 2:

x 2 + 2(x – 2) = 44

Distribute:

x 2 + 2x – 4 = 44

Subtract 44:

x 2 + 2x – 48 = 0

Factor:

(x – 6)(x + 8) = 0

This seems to imply that the x -coordinate of the point of intersection could be either 6 or –8, both of which are choices. Can they both be correct? No: if we substitute x = –8 into either equation, we get no solution, because y 2cannot equal –8. Therefore, the correct answer is (C) 6, and the points of intersection are (6, 2) and (6, –2).

20 . C

2m 2 + 2n = 14

m 2 + 2n = 10

Subtract equations:

m 2 = 4

Take square root:

m = ±2

Substitute m 2 = 4:

4 + 2n = 10

Subtract 4:

2n = 6

Divide by 2:

n = 3

Therefore, m + n = –2 + 3 = 1 or 2 + 3 = 5.

21 . A Use the Zero Product Property. (x 2 – 4) equals 0 if x is 2 or –2, (x – 4) equals 0 if x is 4, and (x 2 + 4) cannot equal 0. Therefore, there are exactly three distinct zeros.

22 . C

f (2.5) = a (2.5 + 2)(2.5 – a )(2.5 – 8)

Simplify:

(–24.75)(a )(2.5 – a )

This product can only be negative if a and (2.5 – a ) have the same sign, which is only true for (C) a = 2.

Skill 3: Working with Exponentials and Radicals

Lesson 9: The Laws of Exponentials

When working with exponentials you must understand the Laws of Exponentials.

Law #1: If n is a positive integer, then x n means the result when 1 is multiplied by x repeatedly n times.

e.g., 35 = 1 × 3 × 3 × 3 × 3 × 3 = 243

You might think that it”s unnecessary to include the 1 in this product, but including it will help clarify what zero, negative, and fractional exponents mean. For instance, think about the following sequence:

243, 81, 27, 9, 3, ___, ___, ___

What are the missing three terms in this sequence? With a little trial and error, you will see that the rule for getting each term is “divide the previous term by 3,” and therefore the missing terms are 1, 1/3, and 1/9. But notice, also, that these terms are just the descending integer powers of 3:

And so on. If you explore this pattern, and patterns for the powers of other numbers, you will notice that some other laws clearly emerge.

Law #2: As long as x does not equal 0, x 0 = 1.

You can think of x 0 as meaning “1 multiplied by x zero times, or not at all.” Therefore, the result is 1.

Law #3: If n is a positive integer, then x n means the result when 1 is divided by x repeatedly n times.

In other words, .

Law #4: x m × x n = x m + n (When multiplying exponentials with equal bases , add the exponents .)

Law #5: x n × y n =( xy ) n (When multiplying exponentials with equal exponents , multiply the bases .)

This law follows from the Commutative and Associative Laws of Addition.

Law #6: (When dividing exponentials with equal bases , subtract the exponents .)

Law #7: (When dividing exponentials with equal exponents , divide the bases .)

Law #8: ( x m ) n = x mn

Law #9:

Proof: This follows directly from Law #8. If we raise to the n th power, by Law #8 we must get x 1 or x . The number that we must raise to the n th power in order to get x is, by definition, the “n th root of x .”

Law #10: If x >1 and x a = x b , then a = b.

Which of the following expressions is equivalent to

1. A)
2. B) 3n2
3. C) 3n6
4. D) 27n2

Therefore, the correct answer is (D).

Which of the following expressions is equivalent to for all values of n ?

1. A)
2. B) 3
3. C) 3n
4. D) 92n

Therefore, the correct answer is (B).

Alternately, we can plug in various values for n and find that the expression gives a value of n no matter what.

Lesson 10: The Laws of Radicals

The radical symbol is used to indicate roots, which are the inverse of exponentials. For instance, because 2 3 = 8, we can say that 2 is the “third root” or “cube root” of 8 .

Law #9 of exponentials shows us that radicals (or “roots”) can be expressed as exponentials. For instance, . Therefore, we can use the Laws of Exponentials to simplify radical expressions.

Law #1: (This is just the “reflected” ­version of Law of Exponentials #9.)

Law #2: (This follows directly from Law of Exponentials #5.)

Law #3: (This follows directly from Law of Exponentials #7.)

Working with square roots is much easier if you memorize the first 10 or so “perfect square integers”:

22 = 4 , 32 = 9 , 42 = 16 , 52 = 25 , 62 = 36 , 72 = 49 ,

82 = 64 , 92 = 81 , 102 = 100 , 112 = 121 , 122 = 144

• If the radicand has a perfect square factor, the radical can be simplified by factoring.
• If a fraction has a radical in the denominator, eliminate it by multiplying numerator and denominator by the radical.
• To estimate the value of square roots, notice which two consecutive perfect squares the radicand lies between.

e.g., and therefore

Which of the following is equivalent to (No calculator)

1. A)
2. B) 7
3. C) 14
4. D) 19

(Medium ) Notice that each answer choice is much simpler than the original expression. This suggests that the original expression can be simplified. Let”s begin by looking at the radical expressions. If you know your perfect squares you will see that neither radicand (the expression inside the radical) is a perfect square, but one of the radicands—18—is a multiple of a perfect square: 18 = 2 × 9.

Original expression:

Substitute 18 = 9 × 2:

Apply Law #2:

Therefore, the correct answer is (C).

If x 2 = 4, y 2 = 9, and (x – 2)(y + 3) ≠ 0, what is the value of x + y ?

1. A) –5
2. B) –1
3. C) 1
4. D) 5

Every positive number has two distinct square roots . For instance, both 5 and –5 are the square root of 25, because (5)2 = 25 and (–5)2 = 25.

However, the symbol means the principal , or non-negative square root, so and not –5.

(Easy ) If x 2 = 4, then x = ±2, and if y 2 = 9, then y = ±2. But if (x – 2)(y + 3) ≠ 0, the x cannot equal 2 and y cannot equal –3. Therefore, x = –2 and y = 3, and x + y = 1, so the correct answer is (C).

Lesson 11: Solving radical and exponential equations

If , what is the value of x ?

1. A)
2. B)
3. C)
4. D)

(Hard )

Multiply by (x + 2):

Distribute:

Subtract :

Divide by :

Therefore, the correct answer is (D).

If , what is the value of k?

1. A) –3
2. B)
3. C)
4. D)

If , what is the value of y 3 ?

1. A)
2. B)
3. C)
4. D)

Exercise Set 4 (No Calculator)

1

If 2a 2 + 3a – 5a 2 = 9, what is the value of aa 2 ?

2

If (200)(4,000) = 8 × 10 m , what is the value of m ?

3

If w = –1030 , what is the value of

4

If 2 x = 10, what is the value of 5(22 x ) + 2 x ?

5

If (x + 2)(x + 4)(x + 6) = 0, what is the greatest possible value of

6

If , where a and b are integers, what is the value of a +b ?

7

If , what is the value of

8

If 9 x = 25, what is the value of 3 x –1 ?

1. A)
2. B)
3. C)
4. D) 24

9

If and a and b are positive numbers, what is the value of

1. A)
2. B)
3. C) 2
4. D) 4

10

Which of the following is equivalent to for all positive values of n ?

1. A) 2
2. B) 2n
3. C) 2n– 1
4. D) 22n

11

Which of the following is equivalent to 3 m + 3 m + 3 m for all positive values of m ?

1. A) 3m+1
2. B) 32m
3. C) 33m
4. D) 33m+1

12

If x is a positive number and 5 x = y , which of the ­following expresses 5y 2 in terms of x ?

1. A) 52x
2. B) 52x+1
3. C) 53x
4. D) 252x

Exercise Set 4 (Calculator)

13

If and n > 0, what is the value of n ?

14

What is the smallest integer value of m such that

15

If , what is the value of k ?

16

If (x m )3 (x m + 1 )2 = x 37 for all values of x , what is the value of m ?

17

If , what is the value of n ?

18

If , what is the value of n?

19

What is one possible value for x such that

20

Which of the following is equivalent to for all positive values of x ?

1. A)
2. B)
3. C)
4. D)

21

The square root of a certain positive number is twice the number itself. What is the number?

1. A)
2. B)
3. C)
4. D)

22

Which of the following is equivalent to for all positive values of m and n ?

1. A)
2. B)
3. C)
4. D)

23

In the figure above, if n > 1, which of the following expresses x in terms of n ?

1. A)
2. B)
3. C)
4. D)

No Calculator

1 . 3

2a 2 + 3a – 5a 2 = 9

Simplify:

3a – 3a 2 = 9

Divide by 3:

aa 2 = 3

2 . 5 (200)(4,000) = 800,000 = 8 × 105

3 . 1/8 or .125

Exponential Law #5:

Cancel common factors:

4 . 510

5(22 x ) + 2 x

Exponential Law #8:

5(2 x )2 + 2 x

Substitute 2 x = 10:

5(10)2 + 10

Simplify:

5(10)2 + 10 = 510

5 . 64 If (x + 2)(x + 4)(x + 6) = 0, then x = –2, –4, or –6. Therefore 2 x could equal 22 , 24 , or 26 . The greatest of these is 26 = 64.

6 . 80

FOIL:

Simplify:

Simplify:

Therefore a = 48 and b = 32 and a + b = 80.

7 . 8

Cross-multiply:

Simplify:

ab = 9 – 5 = 4

Therefore ab 3/2 = 43/2 = 8

8 . 5/3 or 1.66 or 1.67

9 x = 25

Substitute 9 = 32 :

(32 ) x = 25

Exponential Law #8:

32 x = 25

Take square root:

3 x = 5

Divide by 3:

Exponential Law #6:

9 . B

Simplify:

Simplify:

10 . C

Cancel common factor:

Exponential Law #6:

2 n − 1

11 . A

3 m + 3 m + 3 m

Combine like terms:

3(3 m )

Exponential Law #4:

3 m +1

12 . B

5y 2

Substitute y = 5 x :

5(5 x )2

Exponential Law #8:

5(52 x )

Exponential Law #4:

52 x +1

Calculator

13 . 64

n 2 = (644 )1/2

Exponential Law #8:

n 2 = 642

14 . 5

Scientific Notation:

1 × 10m < 2.5 × 10–5

Substitution and checking makes it clear that m = 5 is the smallest integer that satisfies the inequality.

15 . 2.5

Exponential Law #6:

Simplify:

Express as exponentials:

Exponential Law #4:

3 k +1 = 33.5

Exponential Law #10:

k + 1 = 3.5

Subtract 1:

k = 2.5

16 . 7

(x m )3 (x m +1 )2 = x 37

Exponential Law #8:

(x 3 m )(x 2m +2 ) = x 37

Exponential Law #4:

x 5m +2 = x 37

Exponential Law #10:

5m + 2 = 37

Subtract 2:

5m = 35

Divide by 5:

m = 7

17 . 6

Factor:

Divide by :

Simplify:

18 – 12 = 6 = n

18 . 6

Substitute 8 = 23 :

Exponential Law #8:

Exponential Law #10:

Multiply by 12:

6 = n

19 . 1 < x ≤ 1.56

Middle inequality:

Square both sides:

Divide by x :

(Since x > 0, we do not “swap” the inequality.)

Multiply by 25/16:

Last inequality:

Square both sides:

x < x 2

Divide by x :

1 < x

Therefore, x must be both greater than 1 and less than or equal to 1.56.

20 . B

Simplify:

Simplify:

Cancel common factor:

21 . B Translate:

Square both sides:

x = 4x 2

Divide by x :

22 . D

Factor terms:

Cancel common factors:

Combine like terms:

23 . B Pythagorean Theorem:

Simplify:

1 + x 2 = n

Subtract 1:

x 2 = n – 1

Take square root:

Skill 4: Working with Rational Expressions

Lesson 12: Interpreting and computing with rational expressions

Which of the following is equivalent to for all x greater than 0?

1. A)
2. B)
3. C)
4. D)

When adding, subtracting, multiplying, or dividing rational expressions, just follow the rules for working with fractions.

• When adding or subtracting fractions, first get a common denominator, then combine numerators.
• When multiplying fractions, just multiply straight across.
• To divide by a fraction, just multiply by its reciprocal.

(Medium ) To simplify this difference of fractions, we must find a common denominator.

So the correct answer is (C).

If is equivalent to for all x , which of the following is equivalent to B ?

1. A) 3x– 1
2. B) 3x+ 1
3. C) 9x2
4. D) 9x2– 1

(Hard ) It helps to notice that the given rational expression is “improper,” but that the transformed expression is not. Recall that an “improper fraction,” like 5/3, is one in which the numerator is larger than the denominator. Such fractions can also be expressed as “mixed numbers,” which include an integer and a “proper fraction:” 5/3 = 1 ⅔. Similarly, an “improper rational expression” is one in which the degree of the numerator is greater than the degree of the denominator. In the expression , the numerator has a degree of 2 and the denominator has a degree of 1. Just as with improper fractions, we can convert this to a “mixed” expression by just doing the division:

which means that equals . Therefore, the correct answer is (A).

Let x represent the time, in hours, it takes pump A to fill a standard tank, and let y represent the time, in hours, it takes pump A and pump B, working together, to fill the same standard tank. If the equation above represents this situation, then b must represent

1. A) the time, in hours, it takes pump B, working alone, to fill the standard tank
2. B) the portion of the standard tank that pump B fills when the pumps work together to fill the entire standard tank
3. C) the rate, in standard tanks per hour, of pump B
4. D) the difference between the rates, in standard tanks per hour, of pump B and pump A

Rational expressions are often used to express rates . (Remember: rate , rational , and ratio all derive from the same Latin root.) When working with rational expressions that represent real quantities , it often helps to think in terms of the rate-units that they represent .

For instance, if t represents the amount of time, in hours, it takes someone to paint n rooms, then t/n represents the number of “hours per room” and n/t represents the number of “rooms per hour.”

(Medium-hard ) You may find it helpful to review Chapter 8 , Lesson 5, “Rates and unit rates” before tackling this problem. We are told that x represents the number of “hours per tank” for pump A, that is, the number of hours it takes pump A to fill one standard tank. Therefore, its reciprocal, 1/x , must represent the number of “tanks per hour” for pump A, that is, the number of tanks (or fraction of a tank) that pump A can fill in one hour. Likewise, since yrepresents the number of “hours per tank” when the two pumps work together, 1/y must represent the number of “tanks per hour” that the two pumps can fill when working together.

The essential fact in this situation is that “the rate (in tanks per hour) at which the two pumps work together must equal the sum of the rates (in tanks per hour) of the two pumps working separately.” (For instance, if pump A can fill 2 tanks per hour and pump B can fill 3 tanks per hour, then working together they can fill 5 tanks per hour.)

Since the given equation essentially says, “the rate of pump A plus b = the rate of pump A and pump B working together,” b must represent the rate (in tanks per hour) of pump B. Therefore, the correct answer is (C).

Lesson 13: Simplifying rational expressions

If x = 3a and a ≠ 2, which of the following is equivalent to

1. A)
2. B)
3. C)
4. D)

Since rational expressions are just fractions (although perhaps complicated ones), we simplify them exactly the same way we simplify any fraction , that is, by cancelling common factors in the numerator and denominator(which is equivalent to dividing by 1), or multiplying numerator and denominator by a convenient factor (which is equivalent to multiplying by 1).

• Factoring and cancelling common factors:
• Multiplying by a common factor:

(Medium ) This question is asking us to translate an expression in x into an expression in a , which requires making a substitution. However, it is a bit simpler if we don”t substitute right away, but instead simplify the given expression:

Factor:

Cancel common factor:

Substitute x = 3a :

Divide numerator and denominator by 3:

Therefore the answer is (A). Bonus question: Why did the question have to mention that a ≠ 2?

If for all x > 3, where a and b are constants, what is the value of ab?

1. A)
2. B)
3. C)
4. D)

(Medium-hard ) The expression on the left side of the equation is obnoxious and in desperate need of simplification:

Factor:

Cancel common factor (okay since x > 3):

Divide numerator and denominator by 5:

This last step, which may seem strange, is important because it shows us how the two sides of the equation “match up.” If this equation is to be true for “all x > 3” then a must equal 2/5 and b must equal 3. Therefore, ab = (2/5)(3) = 6/5, and the correct answer is (C). Bonus question: Why did the question mention that x > 3?

Lesson 14: Solving rational equations

If x > 0 and , what is the value of x ? [No calculator]

1. A)
2. B)
3. C)
4. D)

When solving an equation that includes fractions or rational expressions, you may find it helpful to simplify the equation by multiplying both sides by the “common denominator” (that is, the common multiple of the denominators).

Multiply by 5x :

Distribute:

Simplify:

x 2 + 5 = 10x

Notice that, in this case, the equation simplifies to a quadratic, which is relatively easy to work with.

(Hard ) Let”s apply this strategy to our equation:

Multiply by

Distribute:

Simplify:

(x + 1) – (x – 1) = 2x 2 – 2

Simplify:

2 = 2x 2 – 2

4 = 2x 2

Divide by 2:

2 = x 2

Take the square root:

But since the equation states that x > 0, the correct answer is (A).

The function f is defined by the equation f (x ) = x 2 – 3x – 18 and the function h is defined by the equation . For what value of x does h (x ) = 6?

1. A) –6
2. B) –3
3. C) 0
4. D) 9

(Hard ) The first thing we should try to do is simplify the expression for h (x ).

Substitute f (x ) = x 2 – 3x – 18:

Factor using Product-Sum Method:

Cancel common factor:

Solve for x if h (x ) = 6:

Multiply by 2:

12 = x + 3

Subtract 3:

9 = x

Therefore, the correct answer is (D).

Exercise Set 5 (No Calculator)

1

If , what is the value of y ?

2

If and x > 0, what is the value of x ?

3

If , what is the value of x 2 ?

4

If , what is the value of z ?

5

Let g(x ) = x 2 – 9x + 18 and , where a is a constant. If h (4) = , what is the value of a ?

6

If for all values of x greater than 1, what is the value of a + b ?

7

Which of the following is equivalent to for all x greater than 1?

1. A)
2. B)
3. C)
4. D)

8

For how many distinct integer values of n is

1. A) Zero
2. B) One
3. C) Two
4. D) Three

9

If and a > 1, which of the following is equivalent to

1. A)
2. B)
3. C)
4. D)

Exercise Set 5 (Calculator)

10

If , what is the value of x 2 – 10x ?

11

For how many positive integer values of k is

12

If g (x ) = x 2 – 9x + 18 and , what is the value of h (9)?

13

If , what is the value of

14

If , what is the value of c 2 ?

15

If for all values of x , what is the value of a ?

16

Which of the following is equivalent to for all positive values of b ?

1. A)
2. B)
3. C)
4. D)

17

Given the system above, what is the value of a + b ?

1. A)
2. B)
3. C)
4. D)

18

If one proofreader takes n hours to edit 30 pages and another takes m hours to edit 50 pages, and together they can edit x pages per hour, which of the following equations must be true?

1. A)
2. B)
3. C)
4. D)

No Calculator

1 . 6/5 or 1.2

Multiply by 45:

15 – 9 = 5y

(45 is the least common multiple of the denominators.)

Simplify:

6 = 5y

Divide by 5:

6/5 = y

2 . 7

Multiply by 24(x + 1)(x – 1):

24x (x – 1)+ 24(x + 1) = 25(x + 1)(x – 1)

Distribute:

24x 2 – 24x + 24x + 24 = 25x 2 – 25

Gather like terms:

0 = x 2 – 49

49 = x 2

Take square root:

±7 = x

Since x must be positive, x = 7.

3 . 13/2 or 6.5

Multiply by 5(x – 2)(x + 2):

5(x + 2) – 5(x – 2) = 8(x – 2)(x + 2)

Distribute:

5x + 10 – 5x + 10 = 8x 2 – 32

Subtract 20 and simplify:

0 = 8x 2 – 52

52 = 8x 2

Divide by 8:

52/8 = 13/2 = x 2

Remember, the question asks for the value of x 2 , not x , so don”t worry about taking the square root.

4 . 6/17 or .353

Multiply by 6z :

12z – 6 = –5z

17z = 6

Divide by 17:

z = 6/17

5 . 28

Use definition of g :

Simplify:

Cross-multiply:

4 –a = –24

28 = a

6 . 5

Combine fractions:

Simplify:

Since must equal for all values of x , a = 3 and b = 2, so a + b = 5.

7 . D

Since (1 − x ) = −(x − 1):

8 . C

Recall from Chapter 7 , Lesson 9, on solving inequalities, that we need to consider two conditions. First, if n + 2 is positive (that is, n > –2), we can multiply on both sides without “flipping” the inequality:

n + 5 > 2n + 4

Subtract n and 4:

1 > n

So n must be between –2 and 1, and the integer values of –1 and 0 are both solutions. Next, we consider the possibility n + 2 is negative (that is, n < –2), and therefore multiplying both sides by n + 2 requires “flipping” the inequality:

n + 5 < 2n + 4

Subtract n and 4:

1 < n

But there are no numbers that are both less than –2 and greater than 1, so this yields no new solutions.

9 . C

Factor:

Cancel common factors:

Substitute x = 4a :

Cancel common factor:

Calculator

10 . 15

Multiply by 5x :

x 2 – 15 = 10x

x 2 – 10x = 15

Notice that you should not worry about solving for x !

11 . 2

Use common base:

10– k > 10–3

Exponential Law #10:

k > –3

Multiply by –1:

k < 3

Therefore, the two positive integer solutions are 1 and 2.

12 . 3/14 or .214

Use definition of g:

Simplify:

13 . 81/2 or 40.5

Combine fractions:

Simplify:

Multiply by 9/2:

14 . 5

Convert to ×:

Multiply:

Multiply by c 2 – 1:

2c 2 = 10

Divide by 2:

c 2 = 5

15 . 2 Notice that the right-hand side of the equation is the “proper” form of the “improper” fraction on the left, and that a is the remainder when the division of the polynomials is completed:

16 . D

Common denominator:

Combine:

17 . C

Multiply by a :

Divide by 10:

2 = 10a

1/5 = a

Subtract equations:

Multiply by –b :

2 = 6b

Divide by 6:

1/3 = b

Therefore, a + b = 1/5 + 1/3 = 8/15.

18 . A The number of pages they can edit together in an hour must equal the sum of the number of pages they can edit separately. The number of pages the first proofreader can edit per hour is 30/n , and the number of pages the second proofreader can edit per hour is 50/m . Since they can edit x pages per hour together, .

NOTE: You can avoid the most common mistakes with this problem by paying attention to the units of each term. The units of two sides, as well as the unit of each term in a sum or difference, must “match.” Notice that the unit for all of the terms is pages/hour.

﻿