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Precalculus

Course: Precalculus > Unit 4

Lesson 6: Multiplying and dividing rational expressions

Multiply & divide rational expressions: Error analysis

Rutendo tried to perform the following multiplication and reduce to lowest terms:

start color #0c7f99, start fraction, minus, 3, x, plus, 15, divided by, 2, x, plus, 16, end fraction, end color #0c7f99, dot, start color #9e034e, start fraction, x, minus, 7, divided by, x, squared, minus, 25, end fraction, end color #9e034e

This is his work:

Rewriting the first factor: start color #0c7f99, start fraction, minus, 3, x, plus, 15, divided by, 2, x, plus, 16, end fraction, end color #0c7f99, equals, start fraction, minus, 3, left parenthesis, x, plus, 5, right parenthesis, divided by, 2, left parenthesis, x, plus, 8, right parenthesis, end fraction

Rewriting the second factor: start color #9e034e, start fraction, x, minus, 7, divided by, x, squared, minus, 25, end fraction, end color #9e034e, equals, start fraction, x, minus, 7, divided by, left parenthesis, x, plus, 5, right parenthesis, left parenthesis, x, minus, 5, right parenthesis, end fraction

Finding excluded values: x, does not equal, plus minus, 5 and x, does not equal, minus, 8

Multiplying and reducing to lowest terms:

$\begin{aligned} &\phantom{=}\blueE{\dfrac{-3x+15}{2x+16}} \cdot \maroonE{\dfrac{x-7}{x^2-25}} \\\\ &=\dfrac{-3(x+5)}{2(x+8)}\cdot\dfrac{x-7}{(x+5)(x-5)} \\\\ &=\dfrac{-3\cancel{(x+5)}(x-7)}{2(x+8)\cancel{(x+5)}(x-5)} \\\\ &=\dfrac{-3(x-7)}{2(x+8)(x-5)} \end{aligned}$

What mistake did Rutendo make?

(Choice A)
He excluded x, equals, minus, 5, which doesn't make the final expression undefined.
(Choice B)
He didn't exclude x, equals, 7, which makes the numerator of the second factor equal to zero.
(Choice C)
He didn't rewrite the first factor correctly.
(Choice D)
He didn't rewrite the second factor correctly.

Stuck?

Precalculus

Course: Precalculus > Unit 4

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