How to divide trinomials

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Algebra › How to divide trinomials

Questions 1 - 4

1

Divide:

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

Explanation

Factor the numerator and denominator:

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

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

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

2

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

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

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

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

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

Explanation

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

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

Notice that both the top and bottom share the term, which can be eliminated.

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

3

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

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

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

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

$-\frac{3}{8}$

$-\frac{6}{5}$

Explanation

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

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

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

Cancel the terms in the numerator and denominator.

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

4

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

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

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

$-\frac{1}{5}$

$\frac{1}{5}$

$\textup{Cannot be factored.}$

Explanation

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

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

Simplify the common terms in the numerator and denominator.

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

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