Question

Find the value of $k$, if$x-1$ is a factor of $p\left(x\right)$ in the following case: $p\left(x\right)=2x^2+kx+\sqrt{2}$

Answer 1

Apply factor theorem

Given, $p\left(x\right)=2x^2+kx+\sqrt{2}$ and $x-1$ is a factor of the polynomial.

Factor theorem states that, A polynomial $P\left(x\right)$ divided by$Q\left(x\right)$ results in $R\left(x\right)$ with zero remainders if and only if $Q\left(x\right)$ is a factor of $P\left(x\right)$ .

So, by the theorem, $x-1$ is a factor of $p\left(x\right)$.

If $x-1$ is a factor of $p\left(x\right)$, then

$\begin{array}{cc}\Rightarrow & p\left(1\right)=0\\ \Rightarrow & 2{\left(1\right)}^2+k\left(1\right)+\sqrt{2}=0\\ \Rightarrow & 2+k+\sqrt{2}=0\\ \Rightarrow & k=-2-\sqrt{2}\end{array}$

Hence the value of $k$ is $-2-\sqrt{2}$.