Question

Let $f\left(x\right)=x^3+k{x}^2+hx+6$. Find the value of $h$ and $k$ so that $(x+1)$ and $(x-2)$ are factors of $f\left(x\right)$.

Answer 1

Step 1: Framing the relation between $h$ and $k$.

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

Here, $(x+1)=\left(x-\left(-1\right)\right)$ is a factor of $f\left(x\right)$, then $f\left(-1\right)=0$

$$\begin{gathered} f\begin{array}{l}\left(-1\right)\end{array}\begin{array}{rcl}& =& \end{array}\begin{array}{l}0\end{array} \\ \begin{array}{rcl}& \Rightarrow & {\left(-1\right)}^3+k{\left(-1\right)}^2+h\left(-1\right)+6=0\end{array} \\ \begin{array}{rcl}& \Rightarrow & -1+k-h+6=0\end{array} \\ \begin{array}{rcl}& \Rightarrow & h-k=5\dots \left(1\right)\end{array} \end{gathered}$$

Now, $(x-2)$ is a factor of $f\left(x\right)$, then $f\left(2\right)=0$.

$$\begin{gathered} f\left(2\right)=0 \\ \Rightarrow {\left(2\right)}^3+k{\left(2\right)}^2+h\left(2\right)+6=0 \\ \Rightarrow 8+4k+2h+6=0 \\ \Rightarrow 2k+h=-7\dots \left(2\right) \end{gathered}$$

Step 2: Determine the value of $h$ and $k$.

Subtract equation $\left(2\right)$ from equation $\left(1\right)$.

$\begin{array}{rcl}h-k-\left(2k+h\right)& =& 5-(-7)\\ -3k& =& 12\\ k& =& -4\end{array}$

Substitute $-4$ for $k$ in equation $\left(1\right)$.

$\begin{array}{rcl}h-\left(-4\right)& =& 5\\ h& =& 5-4\\ h& =& 1\end{array}$

Hence, the value of $h$ and $k$ are $1$ and $-4$.