# If two zeroes of the polynomial $x^4 - 6x^3 - 26x^2 + 138x - 35$ are $2 \pm \sqrt3$, find other zeroes.

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Given:

Two zeroes of the polynomial $x^4 - 6x^3 - 26x^2 + 138x - 35$ are $2 \pm \sqrt3$.

To do:

We have to find other zeroes.

Solution:

Two zeroes are $2 + \sqrt3$ and $2 - \sqrt3$

This implies,

$[x-(2 + \sqrt3)] [x- (2 - \sqrt3)] = (x-2- \sqrt3)(x-2 + \sqrt3)$

$= (x-2)^2– (\sqrt3)^2$

$x^2 – 4x + 1$ is a factor of the given polynomial.

Dividing the given polynomial by $x^2 – 4x + 1$, we get,

$x^2-4x+1$)$x^4-6x^3-26x^2+138x-35$($x^2-2x-35$

$x^4-4x^3+x^2$

----------------------------------

$-2x^3-27x^2+138x-35$

$-2x^3+8x^2-2x$

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$-35x^2+140x-35$

$-35x^2+140x-35$

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$0$

To get the other zeroes,

$x^2-2x-35=0$

$x^2-7x+5x-35=0$

$x(x-7)+5(x-7)=0$

$(x+5)(x-7)=0$

$x+5=0$ or $x-7=0$

$x=-5$ or $x=7$

Hence, the other zeroes of the given polynomial are $7$ and $-5$.

Updated on 10-Oct-2022 13:19:38

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