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6. Polynomials and Factoring
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Algebra I

6.06 Factoring trinomials

Lesson
Practice

Introduction

We learned how to factor by grouping in lesson  6.05 Factoring by grouping  . We will factor trinomial expressions in this lesson using the same factoring application.

Factoring trinomials

Trinomials can be rewritten as polynomials with four terms and factored by grouping.

Exploration

Consider the polynomial expressions factored by grouping below:

Twelve expressions arranged into 3 columns with 4 rows in each column. First column: 3 x squared plus 7 x plus 2, 3 x squared plus 6 x plus 1 x plus 2, 3 x left parenthesis x plus 2 right parenthesis plus 1 left parenthesis x plus 2 right parenthesis, left parenthesis x plus 2 right parenthesis left parenthesis 3 x plus 1 right parenthesis; Second column: 2 x squared plus 11 x plus 12, 2 x squared plus 8 x plus 3 x plus 12, 2 x left parenthesis x plus 4 right parenthesis plus 3 left parenthesis x plus 4 right parenthesis, left parenthesis x plus 4 right parenthesis left parenthesis 2 x plus 3 right parenthesis; Third column: x squared plus 5 x plus 6, x squared plus 3 x plus 2 x plus 6, x left parenthesis x plus 3 right parenthesis plus 2 left parenthesis x plus 3 right parenthesis, left parenthesis x plus 3 right parenthesis left parenthesis x plus 2 right parenthesis.
  1. What patterns do you notice between the original expression and the terms used to rewrite the linear term?
  2. Choose one of the linear terms and rewrite the term in a different way than shown, then determine whether the polynomial can still be factored by grouping.

When using the grouping method to factor a trinomial, the coefficients of the terms used to rewrite the linear term have a sum equivalent to the linear coefficient from the original polynomial and a product equivalent to the product of the trinomial's leading coefficient and constant.

Steps in factoring a quadratic trinomial of the form ax^{2} + bx + c:

  1. Factor out any GCF.

    (If a is negative, we can also divide out a factor of -1 before continuing.)

  2. Find two numbers, r and s, that multiply to ac and add to b.

  3. Rewrite the trinomial with four terms in the form ax^{2} + rx + sx + c.

  4. Factor by grouping.

  5. Check whether the answer will not factor further and verify the factored form by multiplication.

Remember to include any common factors divided out at the start, so each step results in an equivalent expression.

Examples

Example 1

Factor x^{2} + 10 x - 24.

Worked Solution
Create a strategy

Since there are no common factors for all three terms, we proceed with finding the value of two integers that multiply to ac = (1)(-24) = -24 and add up to b = 10. After finding these integers, we use them to rewrite the middle term 10x as a sum of two terms and then factor the trinomial by grouping.

Apply the idea

The factors of -24 are 1 and -24, -1 and 24, 2 and -12, -2 and 12, 3 and -8, -3 and 8, 4 and -6, -4 and 6. Among these factors, -2 and 12 are the pair that add up to 10.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle x^{2} + 10 x - 24\displaystyle =\displaystyle x^{2} + 12x - 2x - 24Rewrite polynomial with four terms
\displaystyle =\displaystyle x(x+12) -2(x+12)Factor each pair
\displaystyle =\displaystyle (x+12)(x-2)Divide out common factor of (x+12)

There are no more common factors to be divided out, so the fully factored form of the polynomial is (x+12)(x-2).

Reflect and check

We can perform a midway check that we are factoring by grouping appropriately when we factor out a GCF from each set of binomials in the step x(x+12) -2(x+12).

If we factor out a GCF at this step and the binomial factors are not equivalent, we may have split the linear term from x^{2} + 10 x - 24 incorrectly or factored out a GCF incorrectly. This is an important place to stop and check that we are factoring appropriately.

Also note that we could have also rewritten the polynomial as x^2-2x+12x-24. This would have resulted in a different middle step in factoring by grouping but the same end result.

Example 2

Factor 3 x^{2} - 27.

Worked Solution
Create a strategy

We can factor a GCF of 3 out of the polynomial and write the polynomial as 3(x^2-9). Since the linear term is missing from the polynomial, we can write the polynomial as 3(x^2+0x-9). There are no common factors. We will find the value of two integers that multiply to ac=(1)(-9)=-9 and add up to b=0. After finding these integers, we use them to rewrite the middle term 0x as a sum of two terms. Then factor the trinomial by grouping.

Apply the idea

The factors of -9 are 1 and -9, -1 and 9, 3 and -3. Among these factors, 3 and -3 are the pair that add up to 0.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle 3(x^{2} + 0x - 9)\displaystyle =\displaystyle 3(x^{2} + 3x - 3x - 9)Rewrite polynomial with four terms
\displaystyle =\displaystyle 3[x(x+3)-3(x+3)]Factor each pair
\displaystyle =\displaystyle 3(x+3)(x-3)Divide out common factor of (x+3)

There are no more common factors to be divided out, so the fully factored form of the polynomial is 3(x+3)(x-3).

Reflect and check

Recall that the special product of a difference of squares \left(a+b\right)\left(a-b\right) = a^{2} - b^{2}. Notice that the factored form of the binomial x^2-9=(x+3)(x-3).

Example 3

Factor 5 x^{2} - 18x + 9.

Worked Solution
Create a strategy

Since there are no common factors for all three terms, we proceed with finding the value of two integers that multiply to ac = 5 \cdot 9 = 45 and add up to b = -18. After finding these integers, we use them to rewrite the middle term -18x as a sum of two terms and then factor the trinomial by grouping.

Apply the idea

The factor pairs of 45 are 1 and 45, -1 and -45, 3 and 15, -3 and -15, 5 and 9, -5 and -9.Note that since the middle term of the trinomial is negative, we need to consider negative and positive factors. Of these factors, -15 and -3 are the pair that adds up to -18.

We can use this to rewrite the trinomial and factor by grouping as follows:

\displaystyle 5x^2 - 18x + 9\displaystyle =\displaystyle 5x^2 - 15x - 3x + 9Rewrite polynomial with four terms
\displaystyle =\displaystyle 5x\left(x - 3\right) - 3\left(x - 3\right)Factor each pair to leave behind a common binomial
\displaystyle =\displaystyle \left(x - 3\right)\left(5x - 3\right)Divide out the common factor of \left(x - 3\right)

There are no more factors to be taken out, so the fully factored form of the polynomial is \left(x - 3\right)\left(5x - 3\right).

Reflect and check

We can check the answer by multiplying the factored form \left(5x-3\right)\left(x-3\right).

A rectangle divided into 2 rows of 2 rectangles. From top to bottom, the left column of rectangles is labeled 5 x squared and negative 15 x, and the right column of rectangles is labeled negative 3 x and 9. The outside of the rectangle is labeled 5 x above the left column, 3 above the right column, and a minus sign above the line between the left and right column. The side is labeled x next to the top row and negative 3 next to the bottom row.

The polynomial 5x^2-3x-15x+9 simplifies to 5x^2-18x+9.

Idea summary

Steps in factoring a quadratic trinomial:

  1. Factor out any GCF.

    (If a is negative, we can also divide out a factor of -1 before continuing.)

  2. Find two numbers, r and s, that multiply to ac and add to b.

  3. Rewrite the trinomial with four terms, in the form ax^{2} + rx + sx + c.

  4. Factor by grouping.

  5. Check whether the answer will not factor further and verify the factored form by multiplication.

Remember to include any common factors divided out at the start, so each step results in an equivalent expression.

Outcomes

A.2c

Perform operations on polynomials, including factoring completely first- and second-degree binomials and trinomials in one variable

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