7.2 Factor quadratic trinomials with leading coefficient 1

Elementary algebra Page 1 / 3

By the end of this section, you will be able to:

Factor trinomials of the form $x^{2} + b x + c$
Factor trinomials of the form $x^{2} + b x y + c y^{2}$

Before you get started, take this readiness quiz.

Multiply: $(x + 4) (x + 5) .$
If you missed this problem, review [link] .
Simplify: ⓐ $−9 + (−6)$ ⓑ $−9 + 6 .$
If you missed this problem, review [link] .
Simplify: ⓐ $−9 (6)$ ⓑ $−9 (−6) .$
If you missed this problem, review [link] .
Simplify: ⓐ $| −5 |$ ⓑ $| 3 | .$
If you missed this problem, review [link] .

Factor trinomials of the form x ² + bx + c

You have already learned how to multiply binomials using FOIL. Now you’ll need to “undo” this multiplication—to start with the product and end up with the factors. Let’s look at an example of multiplying binomials to refresh your memory.

This figure shows the steps of multiplying the factors (x + 2) times (x + 3). The multiplying is completed using FOIL to demonstrate. The first term is x squared and is below F. The second term is 3 x below “O”. The third term is 2 x below “I”. The fourth term is 6 below L. The simplified product is then given as x 2 plus 5 x + 6.

To factor the trinomial means to start with the product, $x^{2} + 5 x + 6$ , and end with the factors, $(x + 2) (x + 3)$ . You need to think about where each of the terms in the trinomial came from.

The first term came from multiplying the first term in each binomial. So to get $x^{2}$ in the product, each binomial must start with an x .

\begin{matrix} x^{2} + 5 x + 6 \\ (x) (x) \end{matrix}

The last term in the trinomial came from multiplying the last term in each binomial. So the last terms must multiply to 6.

What two numbers multiply to 6?

The factors of 6 could be 1 and 6, or 2 and 3. How do you know which pair to use?

Consider the middle term . It came from adding the outer and inner terms.

So the numbers that must have a product of 6 will need a sum of 5. We’ll test both possibilities and summarize the results in [link] —the table will be very helpful when you work with numbers that can be factored in many different ways.

Factors of $6$	Sum of factors
$1, 6$	$1 + 6 = 7$
$2, 3$	$2 + 3 = 5$

We see that 2 and 3 are the numbers that multiply to 6 and add to 5. So we have the factors of $x^{2} + 5 x + 6$ . They are $(x + 2) (x + 3)$ .

\begin{matrix} x^{2} + 5 x + 6 & product \\ (x + 2) (x + 3) & factors \end{matrix}

You should check this by multiplying.

Looking back, we started with $x^{2} + 5 x + 6$ , which is of the form $x^{2} + b x + c$ , where $b = 5$ and $c = 6$ . We factored it into two binomials of the form $(x + m) and (x + n)$ .

\begin{matrix} x^{2} + 5 x + 6 & x^{2} + b x + c \\ (x + 2) (x + 3) & (x + m) (x + n) \end{matrix}

To get the correct factors, we found two numbers m and n whose product is c and sum is b .

How to factor trinomials of the form $x^{2} + b x + c$

Factor: $x^{2} + 7 x + 12$ .

Solution

This table gives the steps for factoring x squared + 7 x + 12. The first row states the first step “write the factors as two binomials with first terms x”. In the second column of the first row it states, “write two sets of parentheses and put x as the first term”. In the third column, it has the expression x squared + 7 x +12. Below the expression are two sets of parentheses with x as the first term.

The second row states the second step “find two numbers m and n that multiply to c, m times n = c and add to b, m + n = b”. In the second column of the second row are the factors of 12 and their sums. 1,12 with sum 1 + 12 = 13. 2, 6 with sum 2 + 6 =8. 3, 4 with sum 3 + 4 = 7.

The third row states “use m and n as the last terms of the factors”. The second column states “use 3 and 4 as the last terms of the binomials”. The third column in this row has the product (x + 3)(x + 4).

In the fourth row the statement is “check by multiplying the factors”. The product of (x + 3)(x +4) is shown to be x 2 + 7 x + 12.

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Factor: $x^{2} + 6 x + 8$ .

$(x + 2) (x + 4)$

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Factor: $y^{2} + 8 y + 15$ .

$(y + 3) (y + 5)$

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Let’s summarize the steps we used to find the factors.

Factor trinomials of the form $x^{2} + b x + c$ .

Write the factors as two binomials with first terms x : $(x) (x)$ .
Find two numbers m and n that
Multiply to c , $m \cdot n = c$
Add to b , $m + n = b$
Use m and n as the last terms of the factors: $(x + m) (x + n)$ .
Check by multiplying the factors.

Factor: $u^{2} + 11 u + 24$ .

Solution

Notice that the variable is u , so the factors will have first terms u .

$\begin{matrix} u^{2} + 11 u + 24 \\ Write the factors as two binomials with first terms u . & (u) (u) \end{matrix}$

Find two numbers that: multiply to 24 and add to 11.

Factors of $24$	Sum of factors
$1, 24$	$1 + 24 = 25$
$2, 12$	$2 + 12 = 14$
$3, 8$	$3 + 8 = 11 *$
$4, 6$	$4 + 6 = 10$

$\begin{matrix} Use 3 and 8 as the last terms of the binomials. & (u + 3) (u + 8) \\ Check. \\ (u + 3) (u + 8) \\ u^{2} + 3 u + 8 u + 24 \\ u^{2} + 11 u + 24 ✓ \end{matrix}$

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Factor: $q^{2} + 10 q + 24$ .

$(q + 4) (q + 6)$

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Factor: $t^{2} + 14 t + 24$ .

$(t + 2) (t + 12)$

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Factor: $y^{2} + 17 y + 60$ .

Solution

$\begin{matrix} y^{2} + 17 y + 60 \\ Write the factors as two binomials with first terms y . & (y) (y) \end{matrix}$

Find two numbers that multiply to 60 and add to 17.

Factors of $60$	Sum of factors
$1, 60$	$1 + 60 = 61$
$2, 30$	$2 + 30 = 32$
$3, 20$	$3 + 20 = 23$
$4, 15$	$4 + 15 = 19$
$5, 12$	$5 + 12 = 17 *$
$6, 10$	$6 + 10 = 16$

$\begin{matrix} Use 5 and 12 as the last terms. & (y + 5) (y + 12) \\ Check. \\ (y + 5) (y + 12) \\ (y^{2} + 12 y + 5 y + 60) \\ (y^{2} + 17 y + 60) ✓ \end{matrix}$

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Source: OpenStax, Elementary algebra. OpenStax CNX. Jan 18, 2017 Download for free at http://cnx.org/content/col12116/1.2

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7.2 Factor quadratic trinomials with leading coefficient 1

Factor trinomials of the form x 2 + bx + c

How to factor trinomials of the form x 2 + b x + c

Solution

Factor trinomials of the form x 2 + b x + c .

Solution

Solution

Questions & Answers

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Factor trinomials of the form x ² + bx + c

How to factor trinomials of the form $x^{2} + b x + c$

Factor trinomials of the form $x^{2} + b x + c$ .