Question

Find all the zeros of 2x⁴ – 13x³ + 19x² + 7x – 3 if two of its zeros are $\left(2+\sqrt{3}\right)\mathrm{and}\left(2-\sqrt{3}\right)$.

Answer 1

Let f(x) = 2x⁴ – 13x³ + 19x² + 7x – 3

It is given that $\left(2+\sqrt{3}\right)\mathrm{and}\left(2-\sqrt{3}\right)$ are two zeroes of f(x)

Thus, f(x) is completely divisible by (x $-2-\sqrt{3}$) and (x – $2+\sqrt{3}$).

Therefore, one factor of f(x) is $\left({\left(x-2\right)}^2-3\right)$
$\Rightarrow$one factor of f(x) is (x² – 4x + 1)

We get another factor of f(x) by dividing it with (x² – 4x + 1).

On division, we get the quotient 2x² – 5x – 3.

$$\begin{gathered} \Rightarrow f\left(x\right)=\left(x^2-4x+1\right)\left(2x^2-5x-3\right) \\ =\left(x^2-4x+1\right)\left(2x^2-6x+x-3\right) \\ =\left(x^2-4x+1\right)\left(2x\left(x-3\right)+1\left(x-3\right)\right) \\ =\left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\left(2x+1\right)\left(x-3\right) \\ \mathrm{To}\mathrm{find}\mathrm{the}\mathrm{zeroes},\mathrm{we}\mathrm{put}f\left(x\right)=0 \\ \Rightarrow \left(x-2-\sqrt{3}\right)\left(x-2+\sqrt{3}\right)\left(2x+1\right)\left(x-3\right)=0 \\ \Rightarrow \left(x-2-\sqrt{3}\right)=0\mathrm{or}\left(x-2+\sqrt{3}\right)=0\mathrm{or}\left(2x+1\right)=0\mathrm{or}\left(x-3\right)=0 \\ \Rightarrow x=2+\sqrt{3},2-\sqrt{3,}-\frac{1}{2},3 \end{gathered}$$

Hence, all the zeroes of the polynomial f(x) are $2+\sqrt{3},2-\sqrt{3},-\frac{1}{2}\mathrm{and}3.$