# Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

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Given:

$3x^4 + 6x^3 - 2x^2 - 10x - 5$ and the two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.

To do:

We have to find all the other zeroes

Solution:

If $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeros of the given polynomial then $(x+\sqrt{\frac{5}{3}})(x-\sqrt{\frac{5}{3}})$ is a factor of it.

This implies,

$(x+\sqrt{\frac{5}{3}})(x-\sqrt{\frac{5}{3}})=x^2-(\sqrt{\frac{5}{3}})^2=x^2-(\frac{5}{3})$

Therefore,

Dividend$f(x)\ =\ 3x^4\ + \ 6x^3\ –\ 2x^2\ -\ 10x\ -\ 5$

Divisor$=x^2-(\frac{5}{3})$

$3x^2-5$)$3x^4+6x^3-2x^2-10x-5$($x^2+2x+1$

$3x^4-5x^2$

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$6x^3+3x^2-10x-5$

$6x^3-10x$

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$3x^2-5$

$3x^2-5$

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$0$

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Quotient$=x^2+2x+1$

$f(x)=(x^2-\frac{5}{3})(x^2+2x+1)$

To find the other zeros put $x^2+2x+1=0$.

$x^2+x+x+1=0$

$x(x+1)+1(x+1)=0$

$(x+1)(x+1)=0$

$x+1=0$ and $x+1=0$

$x=-1$ and $x=-1$

All the zeros of $f(x)$ are $-1$, $-1$, $-\sqrt{\frac{5}{3}}$ and $\sqrt{\frac{5}{3}}$.

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