Question

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

Answer 1

The given polynomial $3x^4+6x^3-2x^2-10x-5$ is of degree $4$.

We know that number of roots is equal to the degree of the polynomial.

Thus, there are $4$ roots for the polynomial $3x^4+6x^3-2x^2-10x-5$.

Given zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$

$$\begin{gathered} \left(x-\sqrt{\frac{5}{3}}\right)\left(x-\left(-\sqrt{\frac{5}{3}}\right)\right)=0 \\ \left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=0 \\ \left(x^2-\frac{5}{3}\right)=0 \\ \left(3x^2-5\right)=0 \end{gathered}$$.

Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeroes of polynomial $3x^4+6x^3-2x^2-10x-5$, then $\left(3x^2-5\right)$ must divide the polynomial $3x^4+6x^3-2x^2-10x-5$

$$\begin{gathered} 3x^4+6x^3-2x^2-10x-5=\left(3x^2-5\right)\left(x^2+2x+1\right) \\ 3x^4+6x^3-2x^2-10x-5=\left(3x^2-5\right)\left(x^2+x+x+1\right) \\ 3x^4+6x^3-2x^2-10x-5=\left(3x^2-5\right)\left(x+1\right)\left(x+1\right) \end{gathered}$$

$$\begin{gathered} 3x^4+6x^3-2x^2-10x-5=0 \\ \left(3x^2-5\right)\left(x+1\right)\left(x+1\right)=0 \\ \left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)\left(x+1\right)\left(x+1\right)=0 \\ x=-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}},-1,-1 \end{gathered}$$

Hence, the other two zeroes of polynomial are $-1,-1$.