Given the equation:
[tex]\frac{x^4+kx^3}{x^3-8x^2+12x}=\frac{x^2}{x-2}[/tex]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.
[tex]\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}[/tex]Simplify the right side by factoring out x in the parentheses:
[tex]\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}[/tex]Cancel common factors:
[tex]x^4+kx^3=x^3(x-6)[/tex]Solving further:
Apply distributive property on the left
[tex]\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}[/tex]Divide both sides by x³:
[tex]\begin{gathered} \frac{kx^3}{x^3}=\frac{-6x^3}{x^3} \\ \\ k=-6 \end{gathered}[/tex]ANSWER:
C. -6
