Question

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

Answer 1

Since the two zeroes of the given polynomial $p\left(x\right)=3x^4+6x^3-2x^2-10x-5$ are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}}$, therefore,

$(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})=x^2-\frac{5}{3}$ is a factor of the given polynomial.

now, we divide the polynomial $p\left(x\right)=3x^4+6x^3-2x^2-10x-5$ by $x^2-\frac{5}{3}$ as shown below:

Therefore, $3x^4+6x^3-2x^2-10x-5=(x^2-\frac{5}{3})(3x^2+6x+3)$

Now, we factorize $3x^2+6x+3$ as follows:

$3x^2+6x+3=0\Rightarrow 3(x^2+2x+1)=0\Rightarrow x^2+2x+1=0\Rightarrow (x+1{)}^2=0(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow (x+1)(x+1)=0\Rightarrow x+1=0,x+1=0\Rightarrow x=-1,x=-1$

Hence, the zeroes of the polynomial $p\left(x\right)=3x^4+6x^3-2x^2-10x-5$ are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1,-1$.