Question

Obtain all the zeros of the polynomial x⁴ + x³ – 14x² – 2x + 24 if two of its zeros are $\sqrt{2}\mathrm{and}-\sqrt{2}$.

Answer 1

Let f(x) = x⁴ + x³ – 14x² – 2x + 24

It is given that $\sqrt{2}\mathrm{and}-\sqrt{2}$ are two zeroes of f(x)

Thus, f(x) is completely divisible by (x + $\sqrt{2}$) and (x – $\sqrt{2}$).

Therefore, one factor of f(x) is (x² – 2).

We get another factor of f(x) by dividing it with (x² – 2).

On division, we get the quotient x² + x – 12.

$$\begin{gathered} \Rightarrow f\left(x\right)=\left(x^2-2\right)\left(x^2+x-12\right) \\ =\left(x^2-2\right)\left(x^2+4x-3x-12\right) \\ =\left(x^2-2\right)\left(x\left(x+4\right)-3\left(x+4\right)\right) \\ =\left(x^2-2\right)\left(x+4\right)\left(x-3\right) \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zeroes},\mathrm{we}\mathrm{put}f\left(x\right)=0 \\ \Rightarrow \left(x^2-2\right)\left(x+4\right)\left(x-3\right)=0 \\ \Rightarrow \left(x^2-2\right)=0\mathrm{or}\left(x+4\right)=0\mathrm{or}\left(x-3\right)=0 \\ \Rightarrow x=\pm \sqrt{2},-4,3 \end{gathered}$$

Hence, all the zeroes of the polynomial f(x) are $\sqrt{2},-\sqrt{2},-4\mathrm{and}3.$