Question

In the expression below, k is an integer.The expression can be simplified as,shown. Assume that the denominatorsare nonzero? x²+kx³ 428x + 12xx-2What is the value of k?A 8B 6C-6D -8

Answer 1

Given the equation:

$\frac{x^4+kx^3}{x^3-8x^2+12x}=\frac{x^2}{x-2}$

Let's assume the denominators are non zero and k is n integer.

Let's find the value of k.

First multiply both sides by the denominator by the left.

$\begin{gathered} \frac{x^4+kx^3}{x^3-8x^2+12x}*x^3-8x^2+12x=\frac{x^2}{x-2}*x^3-8x^2+12x \\ \\ \\ x^4+kx^3=\frac{x^2(x^3-8x^2+12x)}{x-2} \end{gathered}$

Simplify the right side by factoring out x in the parentheses:

$\begin{gathered} x^4+kx^3=\frac{x^2(x(x^2-8x+12)}{x-2} \\ \\ \text{ Now factor the numerator:} \\ x^4+kx^3=\frac{x^2x(x-6)(x-2)}{x-2} \end{gathered}$

Cancel common factors:

$x^4+kx^3=x^3(x-6)$

Solving further:

Apply distributive property on the left

$\begin{gathered} x^4+kx^3=x^4-6x^3 \\ \\ \text{ Subtract x}^4\text{ from both sides:} \\ x^4-x^4+kx^3=x^4-x^4-6x^3 \\ \\ kx^3=-6x^3 \end{gathered}$

Divide both sides by x³:

$\begin{gathered} \frac{kx^3}{x^3}=\frac{-6x^3}{x^3} \\ \\ k=-6 \end{gathered}$

ANSWER:

C. -6