Question

Find the zeroes of the polynomial p(x) = x^4- x^3 - 5x^2+ 3x + 6 if two of its
zeroes are
$\sqrt{3}$
and
$\sqrt[ - ]{3}$
​

Answer 1

Answer:

$\bold\red{-1\:and\:2}$

Step-by-step explanation:

Given,

$p(x) = {x}^{4}- {x}^{3} - 5{x}^{2}+ 3x + 6$

Also,

√3 and -√3 are zeroes of the Polynomial

Therefore,

(x + √3 ) and (x - √3) are factors .

Therefore,

[(x+√3)(x-√3)] is also a factor of Polynomial.

$= > ({x}^{2} - 3) \: is \: a \: factor \: of \: polynomial$

Therefore,

It will leave remainder equal to zero when divides the Polynomial.

$\bold\red{Note:-}$ Refer to the attachment for division process.

Now,

we get,

$= > p(x) = ( {x}^{2} - 3)( {x}^{2} - x - 2)$

To find the othere zeroes,

$= > {x}^{2} - x - 2 = 0 \\ \\ = > {x}^{2} - 2x + x - 2 = 0 \\ \\ = > x(x - 2) + 1(x - 2) = 0 \\ \\ = > (x + 1)(x - 2) = 0 \\ \\ = > x = - 1 \\ \\ and \\ \\ = > x = 2$

Hence,

the other zeroes of Polynomial are -1 and 2.

Answer 2

This is how u have to do it so thank u for asking this question .