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Obtain all other zeroes of $3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.
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$=x^2-(\frac{5}{3})$
$3x^2-5$)$3x^4+6x^3-2x^2-10x-5$($x^2+2x+1$
$3x^4-5x^2$
---------------------
$6x^3+3x^2-10x-5$
$6x^3-10x$
---------------------
$3x^2-5$
$3x^2-5$
------------
$0$
------------
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 $\sqrt{\frac{5}{3}}$.
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