# Find all zeroes of the polynomial $f(x)\ =\ 2x^4\ ‚Äì\ 2x^3\ ‚Äì\ 7x^2\ +\ 3x\ +\ 6$, if two of its zeroes are¬Ý $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$.

Given:

$f(x)\ =\ 2x^4\ –\ 2x^3\ –\ 7x^2\ +\ 3x\ +\ 6$ and the two of its zeroes are $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$.

To do:

We have to find all the zeros of $f(x)$.

Solution:

If $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$ are zeros of $f(x)$ then $(x+\sqrt{\frac{3}{2}})(x-\sqrt{\frac{3}{2}})$ is a factor of $f(x)$.

This implies,

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

Therefore,

Dividend$f(x)\ =\ 2x^4\ –\ 2x^3\ –\ 7x^2\ +\ 3x\ +\ 6$

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

$x^2-(\frac{3}{2})$)$2x^4-2x^3-7x^2+3x+6$($2x^2-2x-4$

$2x^4 -3x^2$
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$-2x^3-4x^2+3x+6$
$-2x^3 +3x$
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$-4x^2+6$
$-4x^2+6$
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$0$

‚Ää

Quotient$=2x^2-2x-4$

$f(x)=(x^2-\frac{3}{2})(2x^2-2x-4)$

To find the other zeros put $2x^2-2x-4=0$.

$2x^2-2x-4=0$

$2x^2-4x+2x-4=0$

$2x(x-2)+2(x-2)=0$

$(x-2)(2x+2)=0$

$x-2=0$ and $2x+2=0$

$x=2$ and $2x=-2$

$x=2$ and $x=-1$

All the zeros of $f(x)$ are $-1$, $2$, $-\sqrt{\frac{3}{2}}$ and $\sqrt{\frac{3}{2}}$.

Updated on: 10-Oct-2022

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