Question

question #4
determine the value of k such that (x + 6) is a factor of the following polynomial.
f(x)=x^{3}-2x^{2}+kx + 90

k = 17
k=-3
k=-33
k = 10

Explanation:

Step1: Apply factor - theorem

If \(x + 6\) is a factor of \(f(x)=x^{3}-2x^{2}+kx + 90\), then \(f(-6)=0\) according to the factor - theorem.

Step2: Substitute \(x=-6\) into \(f(x)\)

\[

$$\begin{align*} f(-6)&=(-6)^{3}-2(-6)^{2}+k(-6)+90\\ &=-216-2\times36 - 6k+90\\ &=-216 - 72-6k + 90\\ &=-198-6k \end{align*}$$

\]

Step3: Solve for \(k\)

Set \(f(-6) = 0\), so \(-198-6k=0\). Add \(198\) to both sides: \(-6k=198\). Then divide both sides by \(-6\), we get \(k=-33\).

Answer:

\(k = - 33\)