ACT Math : How to divide polynomials

Study concepts, example questions & explanations for ACT Math

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Example Questions

Example Question #2 : Polynomials

What is   \(\displaystyle \frac{x^{2}-x-30}{x^{2}-4x-12}\)  equal to?

Possible Answers:

\(\displaystyle \frac{x+5}{x+2}\)

\(\displaystyle \frac{x-5}{x-2}\)

\(\displaystyle \frac{x-6}{x+2}\)

\(\displaystyle \frac{x+5}{x-6}\)

Correct answer:

\(\displaystyle \frac{x+5}{x+2}\)

Explanation:

1. Factor the numerator:

\(\displaystyle x^{2}-x-30= (x+5)(x-6)\)

 

2. Factor the denominator:

\(\displaystyle x^{2}-4x-12=(x-6)(x+2)\)

 

3. Divide the factored numerator by the factored denominator:

\(\displaystyle \frac{(x+5)(x-6)}{(x-6)(x+2)}\)

 

You can cancel out the \(\displaystyle x-6\) from both the numerator and the denominator, leaving you with:

\(\displaystyle \frac{x+5}{x+2}\)

Example Question #2 : Polynomials

Simplify:

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

Possible Answers:

\(\displaystyle \frac{x-9}{x-1}\)

\(\displaystyle \frac{x+9}{x-5}\)

\(\displaystyle \frac{x-1}{x+5}\)

\(\displaystyle \frac{8x-9}{-6x+5}\)

\(\displaystyle \frac{x+9}{x-1}\)

Correct answer:

\(\displaystyle \frac{x+9}{x-5}\)

Explanation:

In order to divide these polynomials, you need to first factor them.

\(\displaystyle x^2+8x-9=(x+9)(x-1)\)

and

\(\displaystyle x^2-6x+5=(x-1)(x-5)\)

Now, the expression becomes \(\displaystyle \frac{(x+9)(x-1)}{(x-5)(x-1)}\)

\(\displaystyle \frac{x-1}{x-1}=1\)

so,

\(\displaystyle \frac{(x+9)(x-1)}{(x-5)(x-1)}=\frac{x+9}{x-5}\)

Example Question #1 : Polynomial Operations

Simplify the following division of polynomials:

\(\displaystyle \frac{x^2+3x+3}{x+1}\)

Possible Answers:

\(\displaystyle x+2+\frac{1}{x+1}\)

\(\displaystyle x+\frac{3}{x+1}\)

It cannot be simplified any further.

\(\displaystyle x+3\)

Correct answer:

\(\displaystyle x+2+\frac{1}{x+1}\)

Explanation:

The leading term of the numerator is one exponent higher than the leading term of the denominator. Thus, we know the result of the division is going to be somewhere close to \(\displaystyle \frac{x^2}{x}=x\). We can separate the fraction out like this: 

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

The first term is easily seen to be \(\displaystyle \frac{x(x+1)}{(x+1)}\), which is equal to \(\displaystyle x\). The second term can also be written out as:

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

and combining these, we get our final answer,

\(\displaystyle x+2+\frac{1}{x+1}\)

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