Question

Determine all the zeros of $x^4-x^3-8x^2+2x+12$ if two of its zeros are $\sqrt{2}$ and $-\sqrt{2}$

• A. $,-\sqrt{2},3,-2$
• B. $\sqrt{2},-\sqrt{2},3,-2$
• C. $-\sqrt{2},-3,-2$
• D. $-\sqrt{2},\sqrt{2},-3,2$

Answer 1

The correct option is B $\sqrt{2},-\sqrt{2},3,-2$

Let, $p\left(x\right)=x^4-x^3-8x^2+2x+12$
Since, $\sqrt{2}$ and $-\sqrt{2}$ are zeros of $p\left(x\right)$
$\Rightarrow (x-\sqrt{2})and(x+\sqrt{2})$ divides $p\left(x\right)$ ($\because Factorthm.$)
$\Rightarrow (x-\sqrt{2})(x+\sqrt{2})$ divides $p\left(x\right)$
$\Rightarrow (x^2-2)$ divides $p\left(x\right)$
$x^2-2)\overline{x^4-x^3-8x^2+2x+12}$( $x^2-x-6$
$\underset{–}{{\underset{-}{x}}^4\underset{+}{-}2x^2}$
$-x^3-6x^2+2x+12$
$\underset{–}{\underset{+}{-}{x}^3\underset{-}{+}2x}$
$-6x^2+12$
$\underset{–}{\underset{+}{-}6x^2\underset{-}{+}12}$
$0$
Now, the other zeros can be obtained from on solving $x^2-x-6=0$
$\Rightarrow x^2-3x+2x-6=0$
$\Rightarrow x(x-3)+2(x-3)=0$
$\Rightarrow (x-3)(x+2)=0$
$\Rightarrow x=-2,3$
Hence, all the zeros are $-2,3,\sqrt{2},-\sqrt{2}$