SAT 2016
CHAPTER 9
THE SAT MATH: ADVANCED MATHEMATICS
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 ax2 + 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:
x2 – a2 = (x + a)(x – a)
Perfect Square Trinomials:
x2 + 2ax + a2 = (x + a)(x + a) = (x + a)2
x2 – 2ax + a2 = (x – a)(x – a) = (x – a)2
Which of the following is a factor of x2 + 8x + 16?
A) x – 4
B) x – 8
C) x + 4
D) x + 8
(Easy) Notice that this quadratic fits the pattern x2 + 2ax + a2 and therefore can be factored using the second formula above: x2 + 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 6x2 + 7x + 2?
A) 3x – 2
B) 3x + 2
C) 3x – 1
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: 6x2 + 7x + 2 = 6x2 + (3x + 4x) + 2
Step 4: Use the associative law of addition to group the first two terms together and the last two terms together: 6x2 + (3x + 4x) + 2 = (6x2 + 3x) + (4x + 2)
Step 5: Factor out the greatest common factor from each pair. (6x2 + 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 6x2 + 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 6x2 + 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, 6x2, 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) = 6x2 – 3x – 4x + 2 = 6x2 – 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:
(3x4 + 5x3 – 2x + 2) – (x4 – 5x3 + 2x2 + 6)
Distribute to eliminate parentheses:
3x4 + 5x3 – 2x + 2 – x4 + 5x3 – 2x2 – 6
Combine like terms:
(3x4 – x4) + (5x3 + 5x3) – 2x2 – 2x + (2 – 6)
Simplify:
2x4 + 10x3 – 2x2 – 2x – 4
Which of the following is equivalent to 2x(x + 1) – x2(x + 1) for all values of x?
A) x2 + x
B) x3 – x2 + 2x
C) –x3 + x2 + 2x
D) –x3 + x2 + 2x + 1
(Easy) Original expression:
2x(x + 1) – x2(x + 1)
Distribute:
2x2 + 2x – x3 – x2
Combine like terms:
–x3 + x2 + 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)x2
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)x2 + (ad)x + (bc)x + bd
Which of the following is equivalent to (2x – 7)(3x + 1) for all values of x?
A) 6x2 – 7
B) 6x2 + 5x – 7
C) 6x2 – 21x – 7
D) 6x2 – 19x – 7
(Easy) Original expression:
(2x – 7)(3x + 1)
FOIL:
(2x)(3x) + (2x)(1) +
(–7)(3x) + (–7)(1)
Simplify:
6x2 + 2x – 21x – 7
Combine like terms:
6x2 – 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:
(2x2 – x + 2) × (x3 + x – 1)
Distribute:
(2x2)(x3) + (2x2)(x) – (2x2)(1) –
(x)(x3) – (x)(x) + (x)(1) + (2)(x3) + (2)(x) – (2)(1)
Simplify:
2x5 + 2x3 – 2x2 – x4 – x2 + x + 2x3 + 2x – 2
Combine like terms:
2x5 – x4 + 4x3 – 3x2 + 3x – 2
Lesson 5: Solving quadratic equations
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 – x2 = –2x?
A) –4
B) –3
C) –2
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 ax2 + bx + c = 0 has no real solutions if b2 – 4ac < 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 3x2 = 4x + 2?
A) 
B) 
C) 
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:
3x2 = 4x + 2
Subtract 4x and 2 to set right side to 0:
3x2 – 4x – 2 = 0
Use Quadratic Formula:
Therefore, the correct answer is (D).
If x > 0 and x2 – 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 x2 – 5x – 6 = 0:
If a quadratic equation has the form x2 + 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 2x2 – 7x + k = 0 is x = 5, what is the other possible value of x?
A) 
B) 
C) 
D) 
(Medium-hard) We can start by substituting x = 5 into the original equation and solving:
Original equation:
2x2 – 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:
2x2 – 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:
2x2 – 7x + k = 0
Divide by 2:
Since the quadratic is now in the form x2 + 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) = ax2 + bx + c, has the following important features:
• It is a parabola with a vertical axis of symmetry at 
.
• The y-intercept is c, since f(0) = a(0)2 + b(0) + c = c.
• If it crosses the x-axis, it does so at the points 
and 
.
• If a is positive, the parabola is “open up,” and if a is negative, it is “open down.”
• If the quadratic is in the form y = a(x – h)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?
A) g(4)
B) g(5)
C) g(6)
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?
A) f(x) = –2(x + 3)(x + 1)
B) f(x) = 3(x + 2)2
C) f(x) = –4(x – 2)(x – 3)
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) =ax2 +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?
A) 
B) 
C) 
D) 
(Easy) The graph of y = ax2 + 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 x2 + 10 =
2
If (a – 3)(a + k) = a2 + 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 = 3x2 – kx – 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(x – b)(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) = 2x2 – 5x – 12?
8
If x = –5 is one of the solutions of the equation 0 = x2 – ax – 12, what is the other solution?
9
Which of the following is equivalent to 2a(a – 5) + 3a2(a + 1) for all values of a?
A) 6a4 – 24a3 – 6
B) 5a5 + 3a2 – 10a
C) 3a3 + 5a2 – 10a
D) 3a3 + 2a2 – 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?
A) y = –x2 – 6x – 9
B) y = –x2 + 6x – 9
C) y = x2 + 6x + 9
D) y = x2 – 6x + 9
11
If 2x2 + 8x = 42 and x < 0, what is the value of x2?
A) 4
B) 9
C) 49
D) 64
12
When the function y = h(x) = ax2 + 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?
A) h(5)
B) h(6)
C) h(8)
D) h(9)
Exercise Set 2 (Calculator)
13
If x > 0 and 2x2 – 4x = 30, what is the value of x?
14
If x2 + 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 = 2x2 + kx – 3 for all values of x, what is the value of k?
18
If b2 + 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)?
A) f(2)
B) f(4)
C) f(7)
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?
A) y = –2(x – 1)(x + 5)
B) y = –2(x + 3)2
C) y = –2(x – 5)2
D) y = –2(x – 1)(x – 5)
21
Which of the following equations has no real solutions?
A) x2 – 3x + 2 = 0
B) x2 – 3x – 2 = 0
C) x2 + 2x – 3 = 0
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?
A) –7
B) –6
C) 7
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?
A) f(–2)
B) f(3.5)
C) f(12)
D) f(13.5)
EXERCISE SET 2 ANSWER KEY
No Calculator
1. 14
(x – 2)(x + 2) = 0
FOIL:
x2 – 4 = 0
Add 14:
x2 + 10 = 14
2. 6
(a – 3)(a + k) = a2 + 3a – 18
FOIL:
a2 + (k – 3)a – 3k = a2 + 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)2 – k(3) – 12
Simplify:
0 = 27 – 3k – 12
Simplify:
0 = 15 – 3k
Add 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 = 2x2 – 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 = x2 – 2.5x – 12
and recall that any quadratic in the form x2 + 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:
x2 – ax – 12 = (x + 5)(x – b)
FOIL:
x2 – ax – 12 = x2 + (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) + 3a2(a + 1)
Distribute:
2a2 – 10a + 3a3 + 3a2
Collect like terms:
3a3 + 5a2 – 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
2x2 + 8x = 42
Divide by 2:
x2 + 4x = 21
Subtract 21:
x2 + 4x – 21 = 0
Factor:
(x + 7)(x – 3) = 0
Therefore, x = –7 or 3, but since x < 0, x = –7 and therefore, x2 = (–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
2x2 – 4x = 30
Divide by 2:
x2 – 2x = 15
Subtract 15:
x2 – 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:
x2 + bx + 9 = (x – a)(x – a)
FOIL:
x2 + bx + 9 = x2 – 2ax + a2
Therefore, b = –2a and a2 = 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 = 2x2 + kx – 3
FOIL:
2x2 + 5x – 3 + 2x = 2x2 + kx – 3
Simplify:
2x2 + 7x – 3 = 2x2 + kx – 3
Subtract 2x2 and add 3:
7x = kx
Divide by x:
7 = k
18. 14
b2 + 20b = 96
Subtract 96:
b2 + 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:
b2 + 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 b2 – 4ac < 0. The only choice for which b2 – 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 of x for which the polynomial equals 0) at x = a, it must have a factor of (x – a).
• Conversely, if a polynomial has a factor of (x – a), it must have a zero at x = a.
The function f is defined by the equation f(x) = x3 – ax2 – bx + 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?
A) 4
B) 5
C) 10
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 (x –p) as factors.
f(x) = x3 – ax2 – bx + 20 = (x + 2)(x – 2)(x – p)
FOIL (x + 2)(x – 2):
= (x2 – 4)(x – p)
FOIL (x2 – 4)(x – p):
= x3 – px2 – 4x + 4p
Since x3 – px2 – 4x + 4p must be equivalent to x3 – ax2 – bx + 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?
A) x < –2 or x > 2
B) –2 < x < 5
C) –2 < x < 2 or x > 5
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).
Lesson 8: Systems involving quadratics
The figure above shows the graph of a system of two equations in the xy-plane. How many solutions does this system have?
A) Zero
B) One
C) Two
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 = x2 + 3x
Given the system above, which of the following could be the value of y?
A) 1 or –6
B) 0 or –5
C) 0 or 10
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 = x2 + 3x:
x2 + 3x + 2x = 6
Subtract 6:
x2 + 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
x2 + y2 = 4
y = x2
How many distinct ordered pairs (x, y) satisfy the three-equation system above?
A) Zero
B) One
C) Two
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:
x2 + (1)2 = 4
1 = x2
Use x2 = 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 x3 – 7x2 + 16x – 12 = (x – a)(x – b)(x – c) for all values of x, what is the value of abc?
2
If x3 – 7x2 + 16x – 12 = (x – a)(x – b)(x – c) for all values of x, what is the value of a + b + c?
3
1. If x3 – 7x2 + 16x – 12 = (x – a)(x – b)(x – c) for all values of x, what is the value of ab + bc + ac?
4
If x2 – ax + 12 has a zero at x = 3, what is the value of a?
5
If x2 – ax + 12 has a zero at x = 3, at what other value of x does it have a zero?
6
y = 4x2 + 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
x2 + y2 = 9
Which of the following equations, if graphed in the xy-plane, would intersect the graph of the equation above in exactly one point?
A) y = –4
B) y = –3
C) y = –1
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?
A) g(0.5)
B) g(1.5)
C) g(2.5)
D) g(3.5)
9
If 2x2 + ax + b has zeros at x = 5 and x = –1, what is the value of a + b?
A) –18
B) –9
C) –2
D) –1
10
If the graph of the equation y = ax4 + bx in the xy-plane passes through the points (2, 12) and (–2, 4), what is the value of a + b?
A) 0.5
B) 1.5
C) 2.0
D) 2.5
11
If the function y = 3(x2 + 1)(x3 – 1)(x + 2) is graphed in the xy-plane, in how many distinct points will it intersect the x-axis?
A) Two
B) Three
C) Four
D) Five
Exercise Set 3 (Calculator)
12
If x2 + y = 10x and y = 25, what is the value of x?
13
If 2x3 – 5x – a has a zero at x = 4, what is the value of a?
14
If x > 0 and x4 – 9x3 – 22x2 = 0, what is the value of x?
15
If d is a positive constant and the graph in the xy-plane of y = (x2)(x2 + x – 72)(x – d) has only one positive zero, what is the value of d?
16
y = 2x2 + 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
4a2 – 5b = 16
3a2 – 5b = 7
Given the system of equations above, what is the value of a2b2?
18
For how many distinct positive integer values of n is (n –1)(n – 9)(n – 17) less than 0?
A) Six
B) Seven
C) Eight
D) Nine
19
x2 + 2y2 = 44
y2 = 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?
A) –8
B) –6
C) 6
D) 8
20
m2 + 2n = 10
2m2 + 2n = 14
Given the system of equations above, which of the following could be the value of m + n?
A) –7
B) –2
C) 1
D) 2
21
For how many distinct values of x does (x2 – 4)(x – 4)2(x2 + 4) equal 0?
A) Three
B) Four
C) Five
D) Six
22
The function f(x) is defined by the equation f(x) = a(x + 2)(x – a)(x – 8) where a is a constant. If f(2.5) is negative, which of the following could be the value of a?
A) –2
B) 0
C) 2
D) 4
EXERCISE SET 3 ANSWER KEY
No Calculator
1. 12 When the expression (x – a)(x – b)(x – c) is fully distributed and simplified, it yields the expression x3 – (a + b + c)x2 + (ab + bc + ac)x – abc. If this is equivalent to x3 – 7x2 + 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 x2 – ax + 12 = 0 when x = 3, then
(3)2 – 3a + 12 = 0
Simplify:
21 – 3a = 0
Add 3a:
21 = 3a
Divide by 3:
7 = a
5. 4 As we saw in question 4, a = 7.
x2 – 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 = 4x2 + 2
Subtract 16, add x:
0 = 4x2 + 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
2x2 + 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:
a – b = 2
Add equations:
6a = –48
Divide by 6:
a = –8
Substitute a = –8:
–8 – b = 2
Add 8:
–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
Add two equations:
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 (x2 + 1) cannot be zero for any value of x, (x3 – 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:
x2 + 25 = 10x
Subtract 10x:
x2 – 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
Add a:
108 = a
14. 11
x4 – 9x3 – 22x2 = 0
Divide by x2:
x2 – 9x – 22 = 0
Factor:
(x – 11)(x + 2) = 0
Use Zero Product Property:
x = 11 or –2
15. 8
y = (x2)(x2 + x – 72)(x – d)
Factor:
y = (x2)(x + 9)(x – 8)(x – d)
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 = 2x2 + 18
Substitute y = ax:
ax = 2x2 + 18
Subtract ax:
0 = 2x2 – ax + 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:
b2 = 9
2b = a/2
The only positive solution to this system is b = 3 and a = 12.
17. 144
4a2 – 5b = 16
3a2 – 5b = 7
Subtract equations:
a2 = 9
Substitute a2 = 9:
3(9) – 5b = 7
Subtract 27:
–5b = –20
Divide by –5:
b = 4
Therefore, a2b2 = 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
x2 + 2y2 = 44
Substitute y2 = x – 2:
x2 + 2(x – 2) = 44
Distribute:
x2 + 2x – 4 = 44
Subtract 44:
x2 + 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 y2 cannot equal –8. Therefore, the correct answer is (C) 6, and the points of intersection are (6, 2) and (6, –2).
20. C
2m2 + 2n = 14
m2 + 2n = 10
Subtract equations:
m2 = 4
Take square root:
m = ±2
Substitute m2 = 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. (x2 – 4) equals 0 if x is 2 or –2, (x – 4) equals 0 if x is 4, and (x2 + 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.