﻿ ﻿Factoring Polynomial Expressions - Polynomial Expressions - High School Algebra I Unlocked (2016)

## High School Algebra I Unlocked (2016)

### Lesson 3.4. Factoring Polynomial Expressions

Factoring polynomial expressions is a big deal in the world of algebra, and you will end up factoring hundreds of polynomial expressions before you finish your algebra career. Earlier in this book, we discussed prime factorization, a process for factoring numbers. While the process for factoring polynomials is not identical to that used to factor numbers, both processes employ the same concepts.

Look back to Lesson
2.1 for a refresher on
prime factorization,
including example
problems!

Factoring allows us to transform a single polynomial expression into multiple polynomial expressions. In essence, factoring is the reverse of distribution. However, rather than multiplying a term through an expression, factoring involves dividing out a term and placing it in front of a parenthetical expression.

When factoring, you want to identify a common factor in all of the terms of your polynomial. Consider the following question.

While there are a couple pairs of parentheses in this question, following the correct order of operations and distributing properly is the key to finding the answer. When working with multiple parentheticals, start from the inside and work outwards. The first set of parentheses contains the expression y + z; since this expression is already simplified, move outwards. Now you need to tackle the expression x − (y + z). Distribute the negative sign through the parentheses to find that x − (y + z) = xyz. Finally, distribute −w through the expression xyz to find that −w(xyz) = −wxw(−y) −w(−z) = −wx + wy + wz. Thus, the expression −w[x − (y + z)] is −wx + wy + wz, or (C).

Remember distributive
multiplication? We
discussed this topic and
other math properties
in Lesson 1.2.

EXAMPLE

Factor 3x + 9.

What is the greatest common factor between these terms? Both 3x and 9 have the value of 3 in common, so you can factor a 3 out of each term:

3x + 9 = 3(x) + 3(3)

Then, according to the rule of distributive multiplication, which states that ab + ac = a(b + c), we can conclude the following:

3(x) + 3(3) = 3(x + 3)

Therefore, the expression 3x + 9 is factored as 3(x + 3).

You can also factor polynomials by capitalizing on the rules of exponents, as shown in the following example.

EXAMPLE

Factor 12x3 + 8x2 + 4x.

First, find the common factor between the terms. Here the terms have a 4x in common, so 4x will be the factor in front of the parentheses:

12x3 + 8x2 + 4x = 4x( )

Next, take the first term in the original expression, 12x3, and divide it by the common factor, 4x. Using the rules of exponents, you will find that = 3x2, which becomes the first term within the parentheses:

12x3 + 8x2 + 4x = 4x(3x2 )

Then take the second term in the original expression, 8x2, and divide it by the common factor, 4x. Again, use the rules of exponents to find that = 2x, which becomes the second term within the parentheses:

12x3 + 8x2 + 4x = 4x(3x2 + 2x )

Here is how you may see FOIL on the SAT.

(2x − 3)(x + 4) =

A) 2x2 + 5x − 12

B) 2x2 + x − 12

C) 2x2 + 11x + 12

D) 3x + 1

Finally, divide the third term in the original expression, 4x, by the common factor of 4x. You will find that , which is your third term in the parentheses:

12x3 + 8x2 + 4x = 4x(3x2 + 2x + 1)

That’s it! You’ve successfully factored the polynomial 12x3 + 8x2 + 4x into 4x(3x2 + 2x + 1). If you want to be sure you’ve factored correctly, distribute the 4x across each terms in the polynomial (3x2 + 2x + 1) to see if the result is the original, unfactored expression.

4x(3x2 + 2x + 1) = 4x(3x2) + 4x(2x) + 4x(1)

= 12x3 + 8x2 + 4x

If the result of multiplying your factored expression is equal to the original expression, congratulations! You have factored correctly.

You may also encounter some funky-looking factoring questions, like this next one.

EXAMPLE

Factor 3(cd) + f(cd).

This expression looks different from the previous ones, right? Never fear, you can still factor here! Each group of terms has a common factor, (cd), so factor it out of the whole expression. When you factor out (cd) from the first term, you have a value of 3 left over:

3(cd) + f(cd) = (cd)(3 )

When you factor out (cd) from the second term, you have a value of f left over:

3(cd) + f(cd) = (cd)(3 + f)

There you go—you’re done. The final factored expression is (cd)(3 + f).

We mentioned earlier that you’ll be factoring a lot in algebra. In particular, you will be factoring a lot of quadratic equations in the form ax2 + bx + c; if a = 1, then ax2 + bx + c = (x + m)(x + n), where m and n are factors of c, and m + n =b.

Keep in mind the following rules when factoring quadratics:

• If the value of c is positive, both m and n are either positive or negative.

◦ If the value of b is positive, both m and n will be positive.

◦ If the value of b is negative, both m and n will be negative.

• If the value of c is negative, m and n will have different signs.

◦ If the value of b is positive, the larger factor will be positive.

◦ If the value of b is negative, the larger factor will be negative.

Let’s see how this works.

Use FOIL to simplify the expression (2x − 3)(x + 4). Sometimes it can be easier to make a quick chart for each of the FOIL terms:

 Step Terms Result First (2x)(x) 2x2 Outside (2x)(4) 8x Inside (−3)(x) −3x Last (−3)(4) −12

Once you have all of the resulting terms, add them together to find that (2x − 3)(x + 4) = 2x2 + 8x − 3x − 12. Now combine like terms to find that 2x2 + 8x − 3x − 12 = 2x2 + 5x − 12. Thus, the correct answer is (A).

EXAMPLE

Factor x2 + 6x + 8.

When factoring simple quadratics, start by setting up your initial terms. When the first term is x2, you can always set up your initial parentheses as follows:

x2 + 6x + 8 = (x )(x )

Now you need to find factors of the c-term, 8, that add up to the b-term, 6. Since both b and c are positive, both factors will be positive. Make a chart like the following:

 Factors of c-term (8) Sum of the Factors Value of b-term (6) 1, 8 1 + 8 = 9 6 2, 4 2 + 4 = 6 6

In the chart, the only factors of the c-term, 8, that add up to the b-term, 6, are 2 and 4. Thus, you can fill in the rest of the parentheses as follows:

x2 + 6x + 8 = (x + 2)(x + 4)

EXAMPLE

In the expression x2 + kx + 12, k is a negative integer. What are all possible values of k ?

This question is different from previous quadratics we’ve tackled; however, it is always helpful to refer back to the standard form of a quadratic and the standard factored form of a quadratic: x2 + bx + c = (x + m)(x + n).

First, you need to factor the expression x2 + kx + 12. Since the first term is x2, you can set up your initial parentheses as follows:

x2 + kx + 12 = (x )(x )

Moreover, since the value of b, k, is negative, and the value of c, 12, is positive, both factors will be negative.

x2 + kx + 12 = (x − )(x − )

Now you need to find the integer factors of 12, as well as the sum of those factors. These factors are shown in the following chart.

 Factors of c-term (12) Sum of the Factors Value of b-term (k) 1, 12 −1 − 12 = −13 −13 2, 6 −2 − 6 = −8 −8 3, 4 −3 − 4 = −7 −7

Since you know that k is a negative integer, all possible values of k are −13, −8, and −7.

There are three types quadratic expressions that appear repeatedly in algebra. You should memorize their binomial formulas, or the way in which you factor them, so that you can work them quickly and efficiently.

Binomial Formulas

x2y2 = (xy)(x + y)

x2 − 2xy + y2 = (xy)(xy) = (xy)2

x2 + 2xy + y2 = (x + y)(x + y) = (x + y)2

Let’s see how these binomial formulas can be implemented.

EXAMPLE

Factor 4x2z2.

Whenever you see a quadratic equation that has two variables, determine whether you can factor it according to one of the three binomial formulas. In this scenario, the expression 4x2z2 adheres to the binomial formula x2y2 = (xy)(x + y). Therefore, the first term in each pair of parentheses is equal to the principal square root, or positive square root, of 4x2; here, the principal square root is = 2x.

4x2z2 = (2x )(2x )

When you have only two terms in a quadratic in this form, the signs within the parentheses will alternate; that is, one will be positive and one will be negative.

4x2z2 = (2x + )(2x − )

Finally, the last term in each pair of parentheses is equal to the principal square root of z2; here, the principal square root is = z. Therefore, the binomial 4x2z2 is factored as

= (2x + z)(2xz)

Here is how you may see factoring on the ACT.

Which of the following expressions is equivalent to a2 − 8a + 16 ?

A. (a − 4)(a − 4)

B. (a − 4)(a + 4)

C. (a − 2)(a − 8)

D. (a − 2)(a + 8)

E. (2a − 4)(2a − 4)

EXAMPLE

Factor 4x2 − 8xy + 4y2.

You’re given a quadratic that has a coefficient other than one in front of the first term. Before you try to factor the quadratic into two binomials, determine whether the terms have a common factor. In this case, all three terms have the value of 4 in common, which can be factored out as follows:

4x2 − 8xy + 4y2 =

4(x2 − 2xy + y2)

Now that you’ve factored out the 4 from the entire expression, you are left with x2 − 2xy + y2 within the parentheses. You should recognize that this is one of the binomial formulas we mentioned previously. Factor the quadratic as follows:

4(x2 − 2xy + y2) =

4[(xy)(xy)]=

4(xy)2

Thus, the quadratic 4x2 − 8xy + 4y2 is factored as 4(xy)2.

You may not always be able to factor quadratic expressions using binomial formulas, prime factorization, or the principal square root; indeed, not all quadratics can be factored in a straightforward manner. When you face a tricky quadratic, use an alternative method called completing the square.

 Standard Quadratic Form Completing the Square Form ax2 + bx + c a(x + d)2 + e d = and e = c −

Let’s see completing the square in action.

EXAMPLE

Factor 4x2 + 12x + 14 by completing the square.

The quadratic you’re given here is 4x2 + 12x + 14, and the standard form of a quadratic expression is ax2 + bx + c. Therefore, a = 4, b = 12, and c = 14. In order to complete the square and factor the quadratic in the form a(x + d)2 + e, we need to find the values of d and e.

Using the formulas given previously, we can find the value of d:

d =

d =

d =

This question requires you to factor the expression a2 − 8a + 16. You can approach this question in multiple ways. First, you may recognize that a2 − 8a + 16 = a2 − 8a + 42 aligns with one of the binomial formulas: x2 − 2xy + y2 = (xy)(xy) = (xy)2; in this case, x = a and y = 4. Therefore, you can factor the quadratic as follows:

a2 − 8a + 16 = (a − 4)(a − 4)

However, you also know that the standard form of a quadratic and the standard factored form of a quadratic is x2 + bx + c = (x + m)(x + n). Thus, an alternative approach is to set up the parentheses like this:

a2 − 8a + 16 = (a )(a )

You also know that since the value of b, −8, is negative, and the value of c, 16, is positive, both factors will be negative.

a2 − 8a + 16 = (a − )(a − )

So, you need to find factors of the c-term, 16, that add up to the b-term, −8.

 Factors of c-term (16) Sum of the Factors Value of b-term (k) 1, 16 −1 − 16 = −17 −17 2, 8 −2 − 8 = −10 −10 4, 4 −4 − 4 = −8 −8

Here, the only factors that work are 4 and 4, and the quadratic is factored as

a2 − 8a + 16 = (a − 4)(a − 4)

Regardless of the method you use to factor, the result will be (A).

Next, find the value of e, substituting the values of a, b, and c:

e = c

e = 14 −

e = 14 −

e = 14 − 9

e = 5

Finally, substitute in the values a = 4, d = 3/2, and e = 5 into the quadratic form associated with completing the square:

a(x + d)2 + e =

4(x + 3/2)2 + 5

That’s all there is to it! By completing the square, you factored the quadratic 4x2 + 12x + 14 into 4(x + 3/2)2 + 5.

Great work! You are now a pro at working with polynomials, which will be of great value as you continue your algebra journey. Now, before we start working with polynomial equations in the next chapter, practice your polynomial skills by completing the practice questions on the next page.

DRILL

CHAPTER 3 PRACTICE QUESTIONS

Directions: Complete the following problems as specified by each question. For extra practice, try using an alternative method to solve the problem or check your work.

1. Simplify the following expression:
(a2b + 3ab2 + ab3) + (a2b − 2a2b2 + ab2ab3)

2. Simplify the following expression:
(x−1 + x−2 − 3x2 + 2x) − (4x + 2 − x−1 − 2x−2)

3. Simplify the following expression:
(2y4 − 5x)(3y3 + 6x)

4. Factor the expression 4x3 − 400x.

5. Factor the expression 4x3 − 8x2 + 2x − 4.

6. Simplify the expression (6a4 + 5b3)2.

7. Factor the expression 16b8a4.

8. Simplify the expression (2a2 + a)(3a−1a−2).

9. Factor the expression x2 − 32x − 16 by completing the square.

10. Factor the expression 4x2 + 13x − 8 by completing the square.

SOLUTIONS TO CHAPTER 3 PRACTICE QUESTIONS

1. 2a2b − 2a2b2 + 4ab2

This question asks you to simplify the expression (a2b + 3ab2 + ab3) + (a2b − 2a2b2 + ab2ab3). Follow the order of operations and combine like terms to find that

(a2b + 3ab2 + ab3) + (a2b − 2a2b2 + ab2ab3) =

a2b + a2b + (−2a2b2) + 3ab2+ ab2 + ab3 + (−ab3) =

2a2b − 2a2b2 + 4ab2

2. −3x2 − 2x − 2 + 2x−1 + 3x−2

This question requires you to simplify the expression (x−1 + x−2 − 3x2 + 2x) − (4x + 2 − x−1 − 2x−2). While the negative exponents may seem intimidating, treat this polynomial expression like you would any other. Follow the order of operations and combine like terms to find that

(x−1 + x−2 − 3x2 + 2x) − (4x + 2 − x−1 − 2x−2) =

−3x2 + 2x − 4x − 2 + x−1 − (−x−1) + x−2 − (−2x−2) =

−3x2 − 2x − 2 + 2x−1 + 3x−2

3. 6y7 + 12xy4 − 15xy3 − 30x2

Here you need to use the power of FOIL to expand the expression (2y4 − 5x)(3y3 + 6x). Use FOIL to find that

(2y4 − 5x)(3y3 + 6x) = (2y4)(3y3 ) + (2y4)(6x) + (−5x)(3y3) + (−5x)(6x)

Then multiply the terms together, combining similar terms, to find that

(2y4)(3y3) + (2y4)(6x) + (−5x)(3y3) + (−5x)(6x) = 6y7 + 12xy4 − 15xy3 − 30x2

Thus, the expression is (2y4 − 5x)(3y3 + 6x) = 6y7 + 12xy4 − 15xy3 − 30x2.

4. 4x(x + 10)(x − 10)

This question requires you to factor the expression 4x3 − 400x. Start by factoring out the greatest common factor of the two terms, 4x, to find that 4x3 − 400x = 4x(x2 − 100). Next, you can factor the expression within the parentheses, x2 − 100, using the binomial formula x2y2 = (xy)(x + y). Therefore, 4x(x2 − 100) = 4x(x + 10)(x − 10).

5. 2(x2 + 1)(x − 2)

To factor the expression 4x3 − 8x2 + 2x − 4, you need to find a common factor among the terms. Start by factoring out the greatest common factor from both the first two terms and the last two terms to find that 4x3 − 8x2 + 2x − 4 = 4x2(x − 2) + 2(x − 2). Now you have two terms, both of which have the common factor (x − 2). Therefore, you can factor the expression to find that 4x2(x − 2) + 2(x − 2) = (4x2 + 2)(x − 2). But wait! The first factor can be factored even further to find that (4x2 + 2)(x − 2) = 2(2x2 + 1)(x − 2).

6. 36a8 + 60a4b3 + 25b6

You need to expand the expression (6a4 + 5b3)2, which can be expanded according to the binomial formula (x + y)2 = (x + y)(x + y) = x2 + 2xy + y2. Therefore, (6a4 + 5b3)2 = (6a4)2 + 2(6a4)(5b3) + (5b3)2 = 36a8 + 60a4b3 + 25b6. Alternatively, use FOIL to get (6a4+ 5b3)2 = (6a4 + 5b3)(6a4 + 5b3) = 36a8 + 30a4b3 + 30a4b3 + 25b6. Then combine like terms and simplify to find that 36a8 + 30a4b3 + 30a4b3 + 25b6 = 36a8 + 60a4b3 + 25b6.

7. (2b2a)(2b2 + a)(4b4 + a2)

To factor the expression 16b8a4, use the binomial formula x2y2 = (xy)(x + y). Thus, 16b8a4 can be factored as (4b4a2)(4b4 + a2). You can then further factor the first term, (4b4a2), according to the binomial formula x2y2 = (xy)(x + y). Therefore, (4b4a2)(4b4 + a2) = (2b2a)(2b2 + a)(4b4 + a2).

8. 6a + 1 − a−1

Use FOIL to expand the expression (2a2 + a)(3a−1a−2), to find that

(2a2 + a)(3a−1a−2) = (2a2)(3a−1) + (2a2)(−a−2) + (a)(3a−1) + (a)(−a−2)

Then multiply the terms together, combining similar terms, to find that

(2a2)(3a−1) + (2a2)(−a−2) + (a)(3a−1) + (a)(−a−2) = 6a − 2 + 3 − a−1

= 6a + 1 − a−1

Thus, the expression is (2a2 + a)(3 a−1a−2) = 6a + 1 − a−1.

9. (x −16)2 − 272

Here, you’re asked to factor the expression x2 − 32x − 16 by completing the square, factoring the quadratic in the form a(x + d)2 + e. Start by finding the values of d and e with the formulas d = and e = c. Since the standard form of a quadratic expression is ax2 + bx + c, a = 1, b = −32, and c = −16. First, find the value of d; d = and d = − , so d = −16. Next, find the value of e; e = c, e = −16 − , e = −16 − , e = −16 − 256, so e = −272. Finally, substitute in the values a = 1, d = −16, and e = −272 into the quadratic form associated with completing the square: a(x + d)2 + e = (x −16)2 − 272.

10.

Here you need to factor the expression 4x2 + 13x − 8 by completing the square, factoring the quadratic in the form a(x + d)2 + e. Thus, you need to find the values of d and e, using the formulas d = and e = c. Since the standard form of a quadratic expression is ax2 + bx + c, a = 4, b = 13, and c = −8. First, find the value of d: d = , so d = , and d = . Next, find the value of e; e = c, e = −8 −, e = −8 − , e = −, and e = −. Finally, substitute the values a = 4, d = , and e = − into the quadratic form associated with completing the square: a(x + d)2 + e = 4(x + )2.

REFLECT

Congratulations on completing Chapter 3!

Here’s what we just covered.

•Define and explain polynomial expressions

1 2 3 4 5

•Combine polynomial expressions through addition and subtraction

1 2 3 4 5

•Multiply polynomial expressions using the distributive property and FOIL

1 2 3 4 5

•Factor polynomial expressions by reversing the FOIL process or by completing the square

1 2 3 4 5

If you rated any of these topics lower than you’d like, consider reviewing the corresponding lesson before moving on, especially if you found yourself unable to correctly answer one of the related end-of-chapter questions.

CHAPTER 3 KEY POINTS

A polynomial is a finite expression of the sum of more than two terms that contain different powers of the same variable(s). A polynomial can have constants, variables, and positive or non-negative exponents. A polynomial cannot have variables in the denominator of a fraction, a variable under a radical sign, or negative or fractional exponents.

A polynomial with one term is referred to as a monomial, a polynomial with two terms is referred to as a binomial, and a polynomial with three terms is referred to as a trinomial.

The degree of a polynomial having only one variable is equal to the largest exponent associated with that variable.

To express polynomials in standard form is to put the terms in order of degree; the standard form of a quadratic is ax2 + bx + c.

Like terms are terms whose variables and exponents are identical.

A quadratic is a second-degree polynomial, a cubic is a third-degree polynomial, and a quartic is a fourth-degree polynomial.

The basic process for adding or subtracting polynomials involves grouping like terms and simplifying the expression.

When multiplying polynomials together, you can either distribute or use FOIL.

Quadratic expressions in the form ax2 + bx + c can be factored as (x + m)(x + n), where m and n are factors of c and m + n = b.

The principal square root of a term is the positive root.

Memorize the way in which you factor the three most popular binomial formulas:

x2 + 2xy + y2 = (x + y)(x + y)

x2y2 = (xy)(x + y)

x2 − 2xyy2 = (xy)(xy)

Completing the square factors a quadratic in the form ax2 + bx + c into a(x + d)2 + e, where d = and e = c.

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