Question

Find the value of $k$, if $x-1$ is a factor of $p\left(x\right)$ in the following cases:

$p\left(x\right)=k{x}^2-3x+k$

• A. $\frac{3}{2}$
• B. $\frac{1}{2}$
• C. $\frac{3}{3}$
• D. $\frac{4}{3}$

Answer 1

The correct option is A $\frac{3}{2}$

Since, $x-1$ is factor of $p\left(x\right)=k{x}^2-3x+k$

$\Rightarrow p\left(1\right)=0$

Now,$p\left(1\right)=k(1{)}^2-3(1)+k=k-3+k=2k-3$

So, $2k-3=0$

$\Rightarrow k=\frac{3}{2}$