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Question

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

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Solution

## The given polynomial $3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5$ 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 $3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5$.Given zeroes are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ $\left(x-\sqrt{\frac{5}{3}}\right)\left(x-\left(-\sqrt{\frac{5}{3}}\right)\right)=0\phantom{\rule{0ex}{0ex}}\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)=0\phantom{\rule{0ex}{0ex}}\left({x}^{2}-\frac{5}{3}\right)=0\phantom{\rule{0ex}{0ex}}\left(3{x}^{2}-5\right)=0$.Since $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeroes of polynomial $3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5$, then $\left(3{x}^{2}-5\right)$ must divide the polynomial $3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5$$3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5=\left(3{x}^{2}-5\right)\left({x}^{2}+2x+1\right)\phantom{\rule{0ex}{0ex}}3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5=\left(3{x}^{2}-5\right)\left({x}^{2}+x+x+1\right)\phantom{\rule{0ex}{0ex}}3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5=\left(3{x}^{2}-5\right)\left(x+1\right)\left(x+1\right)$$3{x}^{4}+6{x}^{3}-2{x}^{2}-10x-5=0\phantom{\rule{0ex}{0ex}}\left(3{x}^{2}-5\right)\left(x+1\right)\left(x+1\right)=0\phantom{\rule{0ex}{0ex}}\left(x-\sqrt{\frac{5}{3}}\right)\left(x+\sqrt{\frac{5}{3}}\right)\left(x+1\right)\left(x+1\right)=0\phantom{\rule{0ex}{0ex}}x=-\sqrt{\frac{5}{3}},\sqrt{\frac{5}{3}},-1,-1$Hence, the other two zeroes of polynomial are $-1,-1$.

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