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Algebra 1 : How to divide trinomials

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Example Question #1 : How To Divide Trinomials

Divide:

$\frac{3x^2+3x-6}{3x+6}$ $\displaystyle \frac{3x^2+3x-6}{3x+6}$

Possible Answers:

$\displaystyle -6$

$\displaystyle 3$

x+2 $\displaystyle x+2$

x-1 $\displaystyle x-1$

Correct answer:

x-1 $\displaystyle x-1$

Explanation:

Factor the numerator and denominator:

$\frac{3x^2+3x-6}{3x+6}=\frac{3(x+2)(x-1)}{3(x+2)}$ $\displaystyle \frac{3x^2+3x-6}{3x+6}=\frac{3(x+2)(x-1)}{3(x+2)}$

Cancel the factors that appear in both the numerator and the denominator:

$\frac{3(x+2)(x-1)}{3(x+2)}=\frac{x-1}{1}=x-1$ $\displaystyle \frac{3(x+2)(x-1)}{3(x+2)}=\frac{x-1}{1}=x-1$

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Example Question #2 : How To Divide Trinomials

Divide the following trinomials: $\frac{x^2-6x+9}{x^2+5x-24}$ $\displaystyle \frac{x^2-6x+9}{x^2+5x-24}$

Possible Answers:

$-\frac{3}{8}$ $\displaystyle -\frac{3}{8}$

$-\frac{6}{5}$ $\displaystyle -\frac{6}{5}$

$\frac{x+3}{x-8}$ $\displaystyle \frac{x+3}{x-8}$

$\frac{x-3}{x+8}$ $\displaystyle \frac{x-3}{x+8}$

$\frac{1}{x+8}$ $\displaystyle \frac{1}{x+8}$

Correct answer:

$\frac{x-3}{x+8}$ $\displaystyle \frac{x-3}{x+8}$

Explanation:

In order to divide, we must first factor both trinomials on the numerator and denominator.

$\frac{x^2-6x+9}{x^2+5x-24} = \frac{(x-3)(x-3)}{(x-3)(x+8)}$ $\displaystyle \frac{x^2-6x+9}{x^2+5x-24} = \frac{(x-3)(x-3)}{(x-3)(x+8)}$

Notice that we now have common terms in the numerator and denominator that can be divided and cancelled.

Cancel the x-3 $\displaystyle x-3$ terms in the numerator and denominator.

The answer is: $\frac{x-3}{x+8}$ $\displaystyle \frac{x-3}{x+8}$

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Example Question #1 : How To Divide Trinomials

Divide the trinomials: $\frac{x^2-3x+2}{x^2+9x-10}$ $\displaystyle \frac{x^2-3x+2}{x^2+9x-10}$

Possible Answers:

$\frac{1}{5}$ $\displaystyle \frac{1}{5}$

$-\frac{1}{5}$ $\displaystyle -\frac{1}{5}$

$\textup{Cannot be factored.}$

$\frac{x+2}{x-10}$ $\displaystyle \frac{x+2}{x-10}$

$\frac{x-2}{x+10}$ $\displaystyle \frac{x-2}{x+10}$

Correct answer:

$\frac{x-2}{x+10}$ $\displaystyle \frac{x-2}{x+10}$

Explanation:

In order to simplify this, we will need to factorize the numerator and denominator.

$\displaystyle x^2-3x+2 = (x-1)(x-2)$

$\displaystyle x^2+9x-10 =(x-1)(x+10)$

Then, $\frac{x^2-3x+2}{x^2+9x-10}= \frac{(x-1)(x-2)}{(x-1)(x+10)}$ $\displaystyle \frac{x^2-3x+2}{x^2+9x-10}= \frac{(x-1)(x-2)}{(x-1)(x+10)}$ .

Simplify the common terms in the numerator and denominator.

The answer is: $\frac{x-2}{x+10}$ $\displaystyle \frac{x-2}{x+10}$

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Example Question #2 : How To Divide Trinomials

Divide the trinomials: $\frac{x^2+4x-12}{x^2-x-42}$ $\displaystyle \frac{x^2+4x-12}{x^2-x-42}$

Possible Answers:

3x+30 $\displaystyle 3x+30$

$\frac{x+2}{x-7}$ $\displaystyle \frac{x+2}{x-7}$

$\frac{4x-12}{42-x}$ $\displaystyle \frac{4x-12}{42-x}$

$\frac{x-2}{x+7}$ $\displaystyle \frac{x-2}{x+7}$

$\frac{x-2}{x-7}$ $\displaystyle \frac{x-2}{x-7}$

Correct answer:

$\frac{x-2}{x-7}$ $\displaystyle \frac{x-2}{x-7}$

Explanation:

Factor both trinomials on the top of the numerator and denominator.

$\frac{x^2+4x-12}{x^2-x-42} = \frac{(x+6)(x-2)}{(x+6)(x-7)}$ $\displaystyle \frac{x^2+4x-12}{x^2-x-42} = \frac{(x+6)(x-2)}{(x+6)(x-7)}$

Notice that both the top and bottom share the x+6 $\displaystyle x+6$ term, which can be eliminated.

The answer is: $\frac{x-2}{x-7}$ $\displaystyle \frac{x-2}{x-7}$

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Algebra 1 : How to divide trinomials

Study concepts, example questions & explanations for Algebra 1

All Algebra 1 Resources

Example Questions

Example Question #1 : How To Divide Trinomials

Example Question #2 : How To Divide Trinomials

Example Question #1 : How To Divide Trinomials

Example Question #2 : How To Divide Trinomials

All Algebra 1 Resources

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