Question

question 8 (1 point)

if \\(x \
eq 4\\), then
\\\frac{4x^2 - 13x - 12}{-x^2 + 9x - 20} =\\

\\(\bigcirc \frac{4x + 3}{x - 5}\\)

\\(\bigcirc \frac{4x + 3}{5 - x}\\)

\\(\bigcirc \frac{2x - 3}{5 - x}\\)

\\(\bigcirc \frac{2(2x - 3)}{x - 5}\\)

question 9 (1 point)

if \\(x \
eq -3, -2, -\frac{3}{2}, 0, 3\\) then
\\\frac{3x^2 + 2x - 8}{2x^2 - 18} \div \frac{x^2 - x - 6}{2x^2 + 3x} =\\

\\(\bigcirc x(3x - 4)\\)

\\(\bigcirc \frac{x(3x - 4)}{x - 3}\\)

\\(\bigcirc \frac{x(3x - 4)(2x + 3)}{2(x + 3)(x - 3)^2}\\)

\\(\bigcirc \frac{x(3x - 4)}{(x - 3)^2}\\)

Explanation:

Factor the numerator and denominator of Question 8

We factor the quadratic expressions in the numerator and denominator.
For the numerator:

$$ 4x^2 - 13x - 12 = (4x + 3)(x - 4) $$

For the denominator:

$$ -x^2 + 9x - 20 = -(x^2 - 9x + 20) = -(x - 4)(x - 5) = (x - 4)(5 - x) $$

Simplify the expression for Question 8

We substitute the factored forms back into the rational expression.
Given \(x
eq 4\), we can cancel the common factor \((x - 4)\):

$$ \frac{4x^2 - 13x - 12}{-x^2 + 9x - 20} = \frac{(4x + 3)(x - 4)}{(x - 4)(5 - x)} = \frac{4x + 3}{5 - x} $$

This matches the second option.

Factor all components of Question 9

We factor each polynomial in the division problem.
First term numerator:

$$ 3x^2 + 2x - 8 = (3x - 4)(x + 2) $$

First term denominator:

$$ 2x^2 - 18 = 2(x^2 - 9) = 2(x - 3)(x + 3) $$

Second term numerator:

$$ x^2 - x - 6 = (x - 3)(x + 2) $$

Second term denominator:

$$ 2x^2 + 3x = x(2x + 3) $$

Simplify the division in Question 9

We rewrite the division as multiplication by the reciprocal.

$$ \frac{3x^2 + 2x - 8}{2x^2 - 18} \div \frac{x^2 - x - 6}{2x^2 + 3x} = \frac{(3x - 4)(x + 2)}{2(x - 3)(x + 3)} \cdot \frac{x(2x + 3)}{(x - 3)(x + 2)} $$

Given the domain restrictions, we cancel the common factor \((x + 2)\):

$$ = \frac{x(3x - 4)(2x + 3)}{2(x - 3)^2(x + 3)} $$

This matches the third option.

Answer:

Question 8

• (A) \(\frac{4x+3}{x-5}\)
• (B) \(\frac{4x+3}{5-x}\) (Correct answer)
• (C) \(\frac{2x-3}{5-x}\)
• (D) \(\frac{2(2x-3)}{x-5}\)

Question 9

• (A) \(x(3x-4)\)
• (B) \(\frac{x(3x-4)}{x-3}\)
• (C) \(\frac{x(3x-4)(2x+3)}{2(x+3)(x-3)^2}\) (Correct answer)
• (D) \(\frac{x(3x-4)}{(x-3)^2}\)