Find all the zeroes of the polynomial $x^4\ +\ x^3\ –\ 34x^2\ –\ 4x\ +\ 120$, if the two of its zeros are $2$ and $-2$.

Given:

Given polynomial is $x^4\ +\ x^3\ –\ 34x^2\ –\ 4x\ +\ 120$ and two of its zeroes are $2$ and $-2$.

To do:

We have to find all the zeros of the given polynomial.

Solution:

If $2$ and $-2$ are zeros of the given polynomial then $(x-2)(x+2)$ is a factor of it.

This implies,

$(x-2)(x+2)=x^2-(2)^2=x^2-4$

Therefore,

Dividend$=x^4+x^3-34x^2-4x+120$

Divisor$=x^2-4$

$x^2-4$)$x^4+x^3-34x^2-4x+120$($x^2+x-30$

 

Quotient$=x^2+x-30$

To find the other zeros put $x^2+x-30$.

$x^2+x-30=0$

$x^2+6x-5x-30=0$

$x(x+6)-5(x+6)=0$

$(x+6)(x-5)=0$

$x+6=0$ and $x-5=0$

$x=-6$ and $x=5$

All the zeros of the given polynomial are $-6$, $-2$, $2$ and $5$.

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Updated on: 10-Oct-2022

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