Question

Find all the zeroes of polynomial $2x^4-11x^3+7x^2+13x-7$. Given that two of its zeroes are $(3+\sqrt{2})$ and $(3-\sqrt{2})$.

Answer 1

Find the zeroes of the given polynomial

Let $f\left(x\right)=2x^4-11x^3+7x^2+13x-7$

Given that, $(3+\sqrt{2})$ and $(3+\sqrt{2})$ are two zeros.

So $\left\{x-\left(3+\sqrt{2}\right)\right\}$ and $\left\{x-\left(3-\sqrt{2}\right)\right\}$ are factors of $f\left(x\right)$.

Then, $\left\{x-\left(3+\sqrt{2}\right)\right\}\left\{x-\left(3-\sqrt{2}\right)\right\}=x^2-6x+7$is also a factor of $f\left(x\right)$

By applying the long division method, we get

Now, Put $f\left(x\right)=0$

$\Rightarrow 2x^4-11x^3+7x^2+13x-7=0$

$\Rightarrow (x^2-6x+7)(2x^2+x-1)=0$

$$\begin{gathered} \Rightarrow \left\{x-\left(3+\sqrt{2}\right)\right\}\left\{x-\left(3-\sqrt{2}\right)\right\}(2x^2+2x-x-1)=0 \\ \Rightarrow \left\{x-\left(3+\sqrt{2}\right)\right\}\left\{x-\left(3-\sqrt{2}\right)\right\}\left(2x\right(x+1)-1(x+1\left)\right)=0 \end{gathered}$$

$\Rightarrow \left\{x-\left(3+\sqrt{2}\right)\right\}\left\{x-\left(3-\sqrt{2}\right)\right\}\left(2x-1\right)\left(x+1\right)=0$

$\Rightarrow x=\left(3+\sqrt{2}\right)$ or $x=\left(3-\sqrt{2}\right)$ or $x=\frac{1}{2}$ or $x=-1$

Hence, the zeros are $\mathbf{(}\mathbf{3}\mathbf{+}\sqrt{\mathbf{2}}\mathbf{)}\mathbf{,}\mathbf{}\mathbf{(}\mathbf{3}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{)}\mathbf{,}\mathbf{}\frac{\mathbf{1}}{\mathbf{2}}$ and $\mathbf{-}\mathbf{1}$.