Obtain all the zeores 2x power …

we are given that the two zeroes of given polynomial are (3 + {tex}\sqrt{2}{/tex}) and (3 - {tex}\sqrt{2}{/tex}).
The given quadratic polynomial is
p(x) = 2x⁴ - 11x³ + 7x² + 13x - 7
Sum of (3 + {tex}\sqrt{2}{/tex}) and (3 - {tex}\sqrt{2}{/tex}) = (3 + {tex}\sqrt{2}{/tex}) + (3 - {tex}\sqrt{2}{/tex}) = 6
Product of (3 + {tex}\sqrt{2}{/tex}) and (3 - {tex}\sqrt{2}{/tex})  = (3 + {tex}\sqrt{2}{/tex})(3 - {tex}\sqrt{2}{/tex})  = 9 - 2 = 7
Polynomial whose zeros are 3 + {tex}\sqrt{2}{/tex} and 3 - {tex}\sqrt{2}{/tex}  is 
x² - (sum of zeros)x + (Product of zeros) = x² - 6x + 7
Now we Divide p(x) by x² - 6x + 7 as:

Therefore, Quotient = 2x² + x - 1 and remainder = 0.
Other two zeros of polynomial p(x) are also the zeros of q(x)
i.e.,  q(x) = 2x² + x - 1 = 2x² + 2x - x - 1  (by splitting the midddle term)
= 2x(x + 1) - (x + 1) = (x + 1 ) (2x - 1)
In order to find the values of x, put(x) = 0
{tex}\Rightarrow{/tex} (x + 1) (2x - 1) = 0
{tex}\Rightarrow{/tex} Either x + 1 = 0 or 2x - 1 = 0
{tex}\Rightarrow{/tex} Either x = -1 or x = {tex}\frac{1}{2}{/tex}
{tex}\therefore{/tex} The zeros of given polynomial p(x) are
{tex}\frac{1}{2}{/tex}, -1, (3 + {tex}\sqrt{2}{/tex}) and (3 - {tex}\sqrt{2}{/tex})