Find all zeroes of the polynomial $f(x)\ =\ 2x^4\ β\ 2x^3\ β\ 7x^2\ +\ 3x\ +\ 6$, if two of its zeroes areΒ $-\sqrt{\frac{3}{2}}$ 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$=x^2-(\frac{3}{2})$
$x^2-(\frac{3}{2})$)$2x^4-2x^3-7x^2+3x+6$($2x^2-2x-4$
$2x^4 -3x^2$
-------------------------------
$-2x^3-4x^2+3x+6$
$-2x^3 +3x$
---------------------------
$-4x^2+6$
$-4x^2+6$
-------------
$0$
β
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 $\sqrt{\frac{3}{2}}$.
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