Question

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

• A. $h=1,k=-4$
• B. $h=-4,k=1$
• C. $h=4,k=-1$
• D. $h=-1,k=-4$

Answer 1

The correct option is A

$h=1,k=-4$

Explanation for correct answer:

Using factor theorem

(Factor theorem states that $f\left(a\right)=0$ if $(x-a)$ is a factor of $f\left(x\right)$.)

Here, $(x+1)$ is a factor of $f\left(x\right)=x^3+k{x}^2+hx+6$

$$\begin{gathered} \therefore f(-1)=0 \\ \Rightarrow {(-1)}^3+k{(-1)}^2+h(-1)+6=0 \\ \Rightarrow -1+k-h+6=0 \\ \Rightarrow k-h+5=0\dots \dots \dots \left(\mathrm{i}\right) \end{gathered}$$

Also, $(x-2)$ is a factor of $f\left(x\right)=x^3+k{x}^2+hx+6$

$$\begin{gathered} \therefore f\left(2\right)=0 \\ \Rightarrow 2^3+k\times 2^2+h\times 2+6=0 \\ \Rightarrow 8+4k+2h+6=0 \\ \Rightarrow 4k+2h+14=0 \\ \Rightarrow 2k+h+7=0\dots \dots \dots \left(\mathrm{ii}\right) \end{gathered}$$

Solving (i) and (ii) and finding $k$ and $h$

Adding $\left(\mathrm{i}\right)$ and $\left(\mathrm{ii}\right)$ we get,

$$\begin{gathered} k-h+5+2k+h+7=0 \\ \Rightarrow 3k+12=0 \\ \Rightarrow 3k=-12 \\ \therefore k=-4 \end{gathered}$$

Substituting the value of $k$ in $\left(\mathrm{i}\right)$ we obtain

$$\begin{gathered} -4-h+5=0 \\ \Rightarrow 1-h=0 \\ \therefore h=1 \end{gathered}$$

Thus, the required values of $h$and $k$ are $h=1,k=-4$

Hence, the correct option is (A).