Question

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

Answer 1

Interpret the given data and find the factor of the given polynomial.
Since this is a polynomial of degree 4, there will be total 4 roots $\sqrt{\frac{5}{3}}\text{and}-\sqrt{\frac{5}{3}}$ are zeros of the polynomial $f\left(x\right)$.

$\therefore (x-\sqrt{\frac{5}{3}})(x+\frac{5}{3})=0$
$x^2-\left(\frac{5}{3}\right)=0$
$(3x^2-5)=0,$ is a factor of given polynomial.

To find unknown factor, divide the given polynomial by known factor.

Now, when we will divide $f\left(x\right)$ by $(3x^2-5)$ the quotient obtained will also be factor of $f\left(x\right)$ and the remainder will be $0$.

$x^2+2x+1$
$3x^2-5\sqrt{3x^4+6x^3-2x^2-10x-5}$
$3x^4+0x^3-5x^2$
$--+$

$6x^3+3x^2-10x-5$
$6x^3+0x^2-10x$
$--+$
$3x^2+0x-5$

Find the other zero by factorizing the factor. Therefore, $3x^4+6x^3-2x^2-10x-5=(3x^2-5)(x^2+2x=1)$

Now, on further factorizing $(x^3+2x+1)$ we get,
$\Rightarrow x^2+2x+1=x^2=x+x+1=0$
$\Rightarrow x(x+1)+1(x+1)=0$
$\Rightarrow (x+1)(x+1)=0$

So, its zeros are given by $x=-1$ and $x=-1$.
Therefore, all four zeros of given polynomial equation are :
$\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1\text{and}-1.$