Question

Find the value of$k$, if$x–1$ is a factor of $P\left(x\right)$ in the following case: $P\left(x\right)=k{x}^2–3x+k$.

Answer 1

Step 1: State the given data and the factor theorem

It is given that,

$P\left(x\right)=k{x}^2–3x+k$ is the dividend.

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

According to the factor theorem, a polynomial $P\left(x\right)$ divided by a monomial $g\left(x\right)$ results in the quotient $Q\left(x\right)$ with zero remainders if and only if $g\left(x\right)$ is a factor of$P\left(x\right)$.

Now, let us put the divisor equal to zero.

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

Step 2: Applying factor theorem

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

$\begin{array}{rr}& P\left(1\right)=0\\ \Rightarrow & k{\left(1\right)}^2–3\left(1\right)+k=0\\ \Rightarrow & k–3+k=0\\ \Rightarrow & 2k–3=0\\ \Rightarrow & k=\frac{3}{2}\end{array}$

Hence, the required value of $k=\frac{3}{2}$.