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6.08 Factoring trinomials where a is not 1

Lesson

Concept summary

To factor a quadratic trinomial in the form ax^{2} + bx + c where a \neq 1, we first look for any common factors that can be taken out of the whole expression. If this leaves behind a polynomial with a leading coefficient of 1, then we can proceed as normal.

If, even after taking out any common factors, the leading coefficient is not equal to 1, we then want to look for two integers whose sum is b and whose product is ac. We can use then these integers, r and s below, to rewrite the trinomial with four terms and use the method of factoring by grouping.

Factoring quadratic trinomials where a \neq 1

ax^{2} + bx + c= ax^{2} + rx + sx + c

Steps in factoring a quadratic trinomial where a \neq 1:

  1. Factor out any GCF.

    (If a is negative, we can also take 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 that are taken out at the start, so that each step results in an equivalent expression.

Worked examples

Example 1

Factor 2 x^{2} + 11x + 5.

Approach

Since there are no common factors for all three terms, we proceed with finding the value of two integers that multiply to ac = 2 \cdot 5 = 10 and add up to b = 11. After finding these integers, we use them to rewrite the middle term 11x as a sum of two terms, and then factor the trinomial by grouping.

Solution

The factors of 10 are 1, 2, 5, and 10. Among these factors, 10 and 1 are the pair that add up to 11.

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

\displaystyle 2x^2 + 11x + 5\displaystyle =\displaystyle 2x^2 + x + 10x + 5Rewrite polynomial with four terms
\displaystyle =\displaystyle x\left(2x + 1\right) + 5\left(2x + 1\right)Factor each pair
\displaystyle =\displaystyle \left(2x + 1\right)\left(x + 5\right)Take out common factor of \left(2x + 1\right)

There are no more factors to be taken out, so the fully factored form of the polynomial is \left(2x+1\right)\left(x+5\right).

Reflection

We can check the answer by multiplying the factored form \left(2x+1\right)\left(x+5\right).

\displaystyle \left(2x+1\right)\left(x+5\right)\displaystyle =\displaystyle x\left(2x+1\right)+5\left(2x+1\right)Distribute the multiplication by \left(2x + 1\right)
\displaystyle =\displaystyle 2x^{2} + x + 10x + 5Distribute the remaining multiplication
\displaystyle =\displaystyle 2x^{2} + 11x + 5Combine like terms

Also note that we could have also rewritten the polynomial as 2x^2 + 10x + x + 5. This would have resulted in a different middle step in factoring by grouping, but the same end result.

Example 2

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

Approach

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.

Solution

The factors of 45 are \pm 1, \pm 3, \pm 5, \pm 9, \pm 15, and \pm 45. Note that since the middle term of the trinomial is negative, we need to consider negative factors as well as positive ones. Of these factors, -15 and -3 are the pair that add 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)Take out 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).

Reflection

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

\displaystyle \left(5x-3\right)\left(x-3\right)\displaystyle =\displaystyle x\left(5x-3\right)-3\left(5x-3\right)Distribute the multiplication by \left(5x-3\right)
\displaystyle =\displaystyle 5x^{2} - 3x - 15x + 9Distribute the remaining multiplication
\displaystyle =\displaystyle 5x^{2} - 18x + 9Combine like terms

Example 3

Factor 15 x^{2} - 27x - 6.

Approach

This polynomial has a common factor between all three terms. We can start by taking out this factor, then proceeding to look for integers that we can use to rewrite the polynomial and use factoring by grouping.

First, we find any common factors for all three terms and factor it out. Then, from the simpler trinomial, we find the values of r and s, that multiply to ac and add up to b. After finding r and s, we write the trinomial in the form ax^{2} + rx + sx + c and factor it by grouping.

Solution

The greatest common factor of the three terms is 3. Taking out this factor, we have15x^2 - 27x - 6 = 3\left(5 x^2 - 9x - 2\right)

From the trinomial 5 x^2 - 9x - 2, we want to find two numbers that have a product of ac = \left(5\right)\left(-2\right) = -10 and a sum of b = -9. The factors of -10 include \pm 1, \pm 2, \pm 5, and \pm 10. Of these factors, -10 and 1 are the pair that add up to -9.

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

\displaystyle 15x^2 - 27x - 6\displaystyle =\displaystyle 3\left(5 x^2 - 9x - 2\right)Previous work
\displaystyle =\displaystyle 3\left(5 x^2 + x - 10x - 2\right)Rewrite polynomial with four terms
\displaystyle =\displaystyle 3\left(x\left(5x + 1\right) - 2\left(5x + 1\right)\right)Factor each pair to leave behind a common binomial
\displaystyle =\displaystyle 3\left(5x + 1\right)\left(x - 2\right)Take out common factor of \left(5x + 1\right)

There are no more factors to be taken out, so the fully factored form of the polynomial is 3\left(5x + 1\right)\left(x - 2\right).

Reflection

We can check the answer by multiplying the factored form 3\left(5x+1\right)\left(x-2\right).

\displaystyle 3\left(5x+1\right)\left(x-2\right)\displaystyle =\displaystyle 3\left(x\left(5x+1\right)-2\left(5x+1\right)\right)Distribute the multiplication by \left(5x + 1\right)
\displaystyle =\displaystyle 3\left(5x^{2} + x -10x - 2\right)Distribute the multiplication by x and -2
\displaystyle =\displaystyle 15x^{2} -27x - 6Combine like terms and distribute the multiplication by 3

Outcomes

A1.A.APR.A.1

Add, subtract, and multiply polynomials. Use these operations to demonstrate that polynomials form a closed system that adhere to the same properties of operations as the integers.

A1.MP1

Make sense of problems and persevere in solving them.

A1.MP5

Use appropriate tools strategically.

A1.MP7

Look for and make use of structure.

A1.MP8

Look for and express regularity in repeated reasoning.

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