Question

Find all the zeros of 2x⁴ – 3x³ – 3x² + 6x – 2 if it is given that two of its zeros are 1 and $\frac{1}{2}$.

Answer 1

Let f(x) = 2x⁴ – 3x³ – 3x² + 6x – 2

It is given that 1 and $\frac{1}{2}$ are two zeroes of f(x).

Thus, f(x) is completely divisible by (x – 1) and (x – $\frac{1}{2}$).

Therefore, one factor of f(x) is $\left(x-1\right)\left(x-\frac{1}{2}\right)$
$\Rightarrow$one factor of f(x) is $\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)$

We get another factor of f(x) by dividing it with $\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)$.

On division, we get the quotient 2x² – 4.

$$\begin{gathered} \Rightarrow f\left(x\right)=\left(x^2-\frac{3}{2}x+\frac{1}{2}\right)\left(2x^2-4\right) \\ =\left(x-1\right)\left(x-\frac{1}{2}\right)\left(2x^2-4\right) \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zeroes},\mathrm{we}\mathrm{put}f\left(x\right)=0 \\ \Rightarrow \left(x-1\right)\left(x-\frac{1}{2}\right)\left(2x^2-4\right)=0 \\ \Rightarrow \left(x-1\right)=0\mathrm{or}\left(x-\frac{1}{2}\right)=0\mathrm{or}\left(2x^2-4\right)=0 \\ \Rightarrow x=1,\frac{1}{2},\pm \sqrt{2} \end{gathered}$$

Hence, all the zeroes of the polynomial f(x) are $1,\frac{1}{2},\sqrt{2}\mathrm{and}-\sqrt{2}.$