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6. Polynomials & Factoring
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United States of AmericaFL
Grade 912

6.07 Factoring trinomials where a=1

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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

MA.912.AR.1.1

Identify and interpret parts of an equation or expression that represent a quantity in terms of a mathematical or real-world context, including viewing one or more of its parts as a single entity.

MA.912.AR.1.7

Rewrite a polynomial expression as a product of polynomials over the real number system.

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