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. $\frac{\sqrt{5}}{\sqrt{3}},-\frac{\sqrt{5}}{\sqrt{3}},-1,-1$
• B. $\frac{\sqrt{2}}{\sqrt{3}},-\frac{\sqrt{2}}{\sqrt{3}},1,1$
• C. $\frac{\sqrt{7}}{\sqrt{2}},-\frac{\sqrt{7}}{\sqrt{2}},2,-1$
• D. $\frac{\sqrt{9}}{\sqrt{2}},-\frac{\sqrt{9}}{\sqrt{2}},-2,-2$

Answer 1

The correct option is A $\frac{\sqrt{5}}{\sqrt{3}},-\frac{\sqrt{5}}{\sqrt{3}},-1,-1$

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

Since, $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeros of $p\left(x\right)$

$\Rightarrow (x-\sqrt{\frac{5}{3}})and(x+\sqrt{\frac{5}{3}})$ divides $p\left(x\right)$ ($\because Factorthm.$)

$\Rightarrow (x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})$ divides $p\left(x\right)$

$\Rightarrow (x^2-\frac{5}{3})$ divides $p\left(x\right)$

$\Rightarrow (3x^2-5)$ divides $p\left(x\right)$

$3x^2-5)\overline{3x^4+6x^3-2x^2-10x-5}$( $x^2+2x+1$

$\underset{–}{\underset{-}{3}{x}^4\underset{+}{-}5x^2}$

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

$\underset{–}{\underset{-}{6}{x}^3\underset{+}{-}10x}$

$3x^2-5$

$\underset{–}{\underset{-}{3}{x}^2\underset{+}{-}5}$

$0$

$\therefore 3x^4+6x^3-2x^2-10x-5=(x-\sqrt{\frac{5}{3}})(x+\sqrt{\frac{5}{3}})(x^2+2x+1)$

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

Hence, zeros of $3x^4+6x^3-2x^2-10x-5$ are $\sqrt{\frac{5}{3}},-\sqrt{\frac{5}{3}},-1,-1$