Question

find all the zeroes of polynomial x4-3x3-5x2+21x-41 it two of it's zereos are √7and+√7​

Answer 1

Answer:

$x^4-3x^3-5x^2+21x-14$

We will take $(x + \sqrt{7}),(x-\sqrt{7})$ as the remaining zeroes

So,

   $(x+\sqrt{7}) (x+\sqrt{7}) = x^2+7$

Then,

    $\frac{x^4-3x^3-5x^2+21x-14}{x^2-7} \\= x^2-3x+2$    

    $= x^2- 2x-x-2$$=(x-1)(x-2)$

x=1,2

Therefore,

    The zeroes are $\sqrt{7} , -\sqrt{7}, 2,1$