Question

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

• A. Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},1$ and 1.
• B. Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},1$ and 0.
• C. Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1$ and 0.
• D. Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1$ and -1.

Answer 1

The correct option is D Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1$ and -1.

$f\left(x\right)=3x^4+6x^2-2x^2-10x-5$

Given, $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are the zeros of f(x).

$\therefore (x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})$ are the factors of f(x).

$(x^2-\sqrt{\frac{5}{3}})$ are the factor of f(x) or $(3x^2-5)$ are the factor of f(x).

So, $3x^4+6x^3-2x^2-10x-5$

$=(3x^2-5)(x^2+2x+1)$

$=3(x^2-\frac{5}{3})(x+1{)}^2$

$=3(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})(x+1{)}^2$

$\therefore$ Zeros are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1$ and $-1$.