Question

Find the value of $k$, if $x–1$ is a factor of $P\left(x\right)$ in the following case:

$P\left(x\right)=x^2+x+k$

Answer 1

Step 1: State the given data and the factor theorem

According to the factor theorem, let $f\left(x\right)$ is a polynomial and $a$ is a real number, then $(x-a)$ is a factor of $f\left(x\right)$, if $f\left(a\right)=0$.

It is given that $P\left(x\right)=x^2+x+k$ is the dividend.

And, $g\left(x\right)=x–1$ is the divisor.

Let us put the divisor equal to zero.

$\begin{array}{cr}& g\left(x\right)=0\\ \Rightarrow & x–1=0\\ \Rightarrow & x=1\end{array}$

Step 2: Apply factor theorem

If, $x–1$is a factor of $P\left(x\right)$, then by the factor theorem,

$\begin{array}{lr}& P\left(1\right)=0\\ \Rightarrow & {\left(1\right)}^2+\left(1\right)+k=0\\ \Rightarrow & 1+1+k=0\end{array}$

$\Rightarrow$ $k=-2$

Hence, the required value of $k$ is $-2$.