Question

Find the value of $\mathrm{k}$ if $(\mathrm{x}+3)$ is a factor of $3{\mathrm{x}}^3+\mathrm{kx}+6$.

Answer 1

Factor theorem:

Let $\mathrm{p}\left(\mathrm{x}\right)$ be any polynomial greater than or equal to 1 and $\mathrm{a}$ be any real number.

If $\mathrm{p}\left(\mathrm{a}\right)=0$, then $(\mathrm{x}-\mathrm{a})$ is a factor of $\mathrm{p}\left(\mathrm{x}\right)$.

Solution:

Let $\mathrm{p}\left(\mathrm{x}\right)=3{\mathrm{x}}^3+\mathrm{kx}+6$. We are given that $(\mathrm{x}+3)$ is a factor of $\mathrm{p}\left(\mathrm{x}\right)$. Therefore, $\mathrm{p}(-3)=0$.

then,

$$\begin{gathered} \mathrm{p}(-3)=3{(-3)}^3+\mathrm{k}(-3)+6 \\ \mathrm{p}(-3)=-81-3\mathrm{k}+6 \\ 0=-75-3\mathrm{k} \\ \mathrm{k}=-25 \end{gathered}$$

Hence, the value of $\mathrm{k}=-25$.