Lesson Objectives

- Demonstrate an understanding of how to factor a trinomial
- Learn how to factor a polynomial using substitution

## How to Factor by Substitution

In some cases, we will be able to factor a more complex polynomial by making a simple substitution. Up to this point, we have only tried to factor a polynomial
of the form:

ax

As we move higher in math, our polynomials will get more complex. Let's look at an example.

Example 1: Factor using substitution

7x

Notice how the highest power (4) is double that of the smaller power (2). This matches the pattern of our usual trinomial, where we have a highest power of (2) and a smaller power of (1). We will let a variable be equal to our variable raised to the smaller power.

let z = x

Using the rules of exponents, we can replace each x

7x

7(x

7z

Let's factor using the AC method:

a = 7, b = -20, c = 12

ac:

7 • 12 = 84

Since b is -20, we want two integers whose product is 84 and whose sum is -20. We can find these as -6 and -14. We will use these integers to rewrite our middle term.

7z

Factor using grouping:

7z(z - 2) - 6(z - 2)

(7z - 6)(z - 2)

We are not done, we need our answer in terms of x. We will substitute an x

7x

Example 2: Factor each using substitution

45x

Again, we can see that the largest exponent (6) is double that of the smallest (3). We can use substitution to factor this polynomial.

Let's first factor out our GCF of 3:

3(15x

Now we can make a substitution. We let a variable be equal to our variable raised to the smaller power.

let z = x

Using the rules of exponents, we can replace each x

3(15x

3(15(x

3(15z

a = 15, b = 77, c = 90

ac:

15 • 90 = 1350

Since b is 77, we want two integers whose product is 1350 and whose sum is 77. We can find these as 27, and 50. We will use these integers to rewrite our middle term.

3(15z

Factor using grouping:

3(5z(3z + 10) + 9(3z + 10))

3(5z + 9)(3z + 10)

We are not done, we need our answer in terms of x. We will substitute an x

45x

Example 3: Factor each using substitution

10(x + 1)

Here the binomial (x + 1) has an exponent of 2 and an exponent of 1. We can use a simple substitution and factor. Let's let z be equal to the binomial (x + 1).

z = (x + 1)

Now we will replace (x + 1) in our polynomial with z and factor:

10(x + 1)

10z

a = 10, b = -7, c = 1

ac:

10 • 1 = 10

Since b is -7, we want two integers whose product is 10 and whose sum is -7. We can find these as -2, and -5. We will use these integers to rewrite our middle term.

10z

Factor using grouping:

5z(2z - 1) - 1(2z - 1)

(5z - 1)(2z - 1)

Replace z with (x + 1):

(5(x + 1) - 1)(2(x + 1) - 1)

Simplify:

(5x + 5 - 1)(2x + 2 - 1)

(5x + 4)(2x + 1)

10(x + 1)

ax

^{2}+ bx + cAs we move higher in math, our polynomials will get more complex. Let's look at an example.

Example 1: Factor using substitution

7x

^{4}- 20x^{2}+ 12Notice how the highest power (4) is double that of the smaller power (2). This matches the pattern of our usual trinomial, where we have a highest power of (2) and a smaller power of (1). We will let a variable be equal to our variable raised to the smaller power.

let z = x

^{2}Using the rules of exponents, we can replace each x

^{2}with a z and factor:7x

^{4}- 20x^{2}+ 127(x

^{2})^{2}- 20x^{2}+ 127z

^{2}- 20z + 12Let's factor using the AC method:

a = 7, b = -20, c = 12

ac:

7 • 12 = 84

Since b is -20, we want two integers whose product is 84 and whose sum is -20. We can find these as -6 and -14. We will use these integers to rewrite our middle term.

7z

^{2}- 14z - 6z + 12Factor using grouping:

7z(z - 2) - 6(z - 2)

(7z - 6)(z - 2)

We are not done, we need our answer in terms of x. We will substitute an x

^{2}for z in our factored form:7x

^{4}- 20x^{2}+ 12 = (7x^{2}- 6)(x^{2}- 2)Example 2: Factor each using substitution

45x

^{6}+ 231x^{3}+ 270Again, we can see that the largest exponent (6) is double that of the smallest (3). We can use substitution to factor this polynomial.

Let's first factor out our GCF of 3:

3(15x

^{6}+ 77x^{3}+ 90)Now we can make a substitution. We let a variable be equal to our variable raised to the smaller power.

let z = x

^{3}Using the rules of exponents, we can replace each x

^{3}with a z and factor:3(15x

^{6}+ 77x^{3}+ 90)3(15(x

^{3})^{2}+ 77x^{3}+ 90)3(15z

^{2}+ 77z + 90)a = 15, b = 77, c = 90

ac:

15 • 90 = 1350

Since b is 77, we want two integers whose product is 1350 and whose sum is 77. We can find these as 27, and 50. We will use these integers to rewrite our middle term.

3(15z

^{2}+ 50z + 27z + 90)Factor using grouping:

3(5z(3z + 10) + 9(3z + 10))

3(5z + 9)(3z + 10)

We are not done, we need our answer in terms of x. We will substitute an x

^{3}for z in our factored form:45x

^{6}+ 231x^{3}+ 270 = 3(5x^{3}+ 9)(3x^{3}+ 10)Example 3: Factor each using substitution

10(x + 1)

^{2}- 7(x + 1) + 1Here the binomial (x + 1) has an exponent of 2 and an exponent of 1. We can use a simple substitution and factor. Let's let z be equal to the binomial (x + 1).

z = (x + 1)

Now we will replace (x + 1) in our polynomial with z and factor:

10(x + 1)

^{2}- 7(x + 1) + 110z

^{2}- 7z + 1a = 10, b = -7, c = 1

ac:

10 • 1 = 10

Since b is -7, we want two integers whose product is 10 and whose sum is -7. We can find these as -2, and -5. We will use these integers to rewrite our middle term.

10z

^{2}- 5z - 2z + 1Factor using grouping:

5z(2z - 1) - 1(2z - 1)

(5z - 1)(2z - 1)

Replace z with (x + 1):

(5(x + 1) - 1)(2(x + 1) - 1)

Simplify:

(5x + 5 - 1)(2x + 2 - 1)

(5x + 4)(2x + 1)

10(x + 1)

^{2}- 7(x + 1) + 1 = (5x + 4)(2x + 1)
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