Question

question #3
which of the following would be the third step in writing the rational expression in simplest form?
\frac{7x^{2}-21x + 14}{x - 6}cdot\frac{1}{7x - 7}

○ \frac{7(x^{2}-3x + 2)}{x - 6}cdot\frac{1}{7(x - 1)}
○ \frac{7(x - 2)(x - 1)}{x - 6}cdot\frac{1}{7(x - 1)}
○ \frac{x - 2}{x - 6}
○ \frac{7(x - 2)(x - 1)}{7(x - 1)(x - 6)}

Explanation:

Step1: Factor out the greatest - common factor from the numerator of the first fraction

Factor $7x^{2}-21x + 14$:
$7x^{2}-21x + 14=7(x^{2}-3x + 2)$
So the expression becomes $\frac{7(x^{2}-3x + 2)}{x - 6}\cdot\frac{1}{7x - 7}$

Step2: Factor the quadratic and the linear - factor

Factor $x^{2}-3x + 2=(x - 2)(x - 1)$ and $7x-7 = 7(x - 1)$
The expression is now $\frac{7(x - 2)(x - 1)}{x - 6}\cdot\frac{1}{7(x - 1)}$

Step3: Multiply the two fractions

When multiplying two fractions $\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}$, we have $\frac{7(x - 2)(x - 1)\cdot1}{(x - 6)\cdot7(x - 1)}=\frac{7(x - 2)(x - 1)}{7(x - 1)(x - 6)}$

Answer:

$\frac{7(x - 2)(x - 1)}{7(x - 1)(x - 6)}$