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6. Polynomials & Factoring
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United States of AmericaTennessee
Algebra 1

6.07 Factoring trinomials where a=1

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Lesson

Concept summary

To factor a quadratic trinomial in the form ax^{2} + bx + c where a = 1, we aim to find two numbers p and q whose product is c and whose sum is b.

Factoring quadratic trinomials where a = 1

x^{2} + bx + c= \left(x + p\right)\left(x + q\right)

It is usually easiest to find these numbers by first looking at the factors of the constant term c.

Steps in factoring a quadratic trinomial where a = 1:

  1. Identify the factors of the constant term, c.

  2. Find p and q as a pair of factors of c which also have a sum of b.

  3. Write the quadratic expression in the form \left(x + p\right)\left(x + q\right).

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

Note: Some trinomials are not factorable with integer values of p and q. Such polynomials are sometimes called prime polynomials.

Worked examples

Example 1

Consider the trinomial x^{2} + 5x + 6.

a

Find two integer values that have a sum of b and a product of c.

Approach

We find the factors of the constant term, 6, and choose the pair of factors that add up to 5.

Solution

The factors of 6 include 1, 2, 3 and 6. Among these factors, 2 and 3 are the ones that add up to 5.

b

Factor the quadratic trinomial.

Approach

We use the values of p and q found in (a) to write the trinomial in the form \left(x + p\right)\left(x + q\right).

Solution

Since we already identified that 2 and 3 are the values of p and q respectively, we can now write the trinomial in the form \left(x + p\right)\left(x + q\right).

Substituting the values, we get \left(x + 2\right)\left(x + 3\right) as the final answer.

Reflection

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

\displaystyle \left(x + 2\right)\left(x + 3\right)\displaystyle =\displaystyle \left(x\right)\left(x + 2\right) + \left(3\right)\left(x + 2\right)Distribute the multiplication of x + 2
\displaystyle =\displaystyle x^{2} + 2x + 3x + 6Distribute the multiplication of x and 3
\displaystyle =\displaystyle x^{2} + 5x + 6Combine like terms

Example 2

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

Approach

First, we find the factors of the constant term - 24 then choose the pair of factors, p and q, that add up to 10. After finding p and q, we factor the trinomial by writing it in the form \left(x + p\right)\left(x + q\right).

Solution

  • Since the constant term is a negative number, we know that one factor will be positive and the other will be negative. The factors of -24 include \pm1, \pm2, \pm3, \pm4, \pm6, \pm8, \pm12, and \pm24. Among these factors, 12 and -2 , are the pairs that add up to 10. Therefore 12 and -2 are the values of p and q respectively.

    Note: While 6 and 4 have a sum of 10, they don't have a product of -24 as they are both positive numbers.

  • Since we already identified that 12 and -2 are the values of p and q respectively, we can now write the trinomial in the form \left(x + p\right)\left(x + q\right). Substituting the values, we get \left(x + 12\right)\left(x - 2\right) as the final answer.

Reflection

We can check the answer by multiplying the factored form \left(x + 12\right)\left(x - 2\right).

\displaystyle \left(x + 12\right)\left(x - 2\right)\displaystyle =\displaystyle \left(x\right)\left(x + 12\right) + \left(- 2\right)\left(x + 12\right)Distribute the multiplication of x + 12
\displaystyle =\displaystyle x^{2} + 12x - 2x - 24Distribute the multiplication of x and -2
\displaystyle =\displaystyle x^{2} + 10x - 24Combine like terms

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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