Question

Find all the zeros of $2x^4-9x^3+5x^2+3x-1$, if two of its zeros are $2+\sqrt{3}$ and $2-\sqrt{3}$

• A. 2, 3, 1
• B. 4, 7, -1
• C. 6, 5, 8
• D. None of these

Answer 1

The correct option is D None of these

Let, $p\left(x\right)=2x^4-9x^3+5x^2+3x-1$
Since, $2+\sqrt{3}$ and $2-\sqrt{3}$ are zeros of $p\left(x\right)$
$\Rightarrow (x-2-\sqrt{3})and(x-2+\sqrt{3})$ divides $p\left(x\right)$ ($\because Factorthm.$)
$\Rightarrow (x-2-\sqrt{3})(x-2+\sqrt{3})$ divides $p\left(x\right)$
$\Rightarrow (x^2-4x+1)$ divides $p\left(x\right)$
$x^2-4x+1)\overline{2x^4-9x^3+5x^2+3x-1}$( $2x^2-x-1$
$\underset{–}{\underset{-}{2}{x}^4\underset{+}{-}8x^3\underset{-}{+}2x^2}$
$-x^3+3x^2+3x-1$
$\underset{–}{\underset{+}{-}{x}^3\underset{-}{+}4x^2\underset{+}{-}x}$
$-x^2+4x-1$
$\underset{–}{\underset{+}{-}{x}^2\underset{-}{+}4x\underset{+}{-}1}$
$0$
Now, the other zeros can be obtained from on solving $2x^2-x-1=0$
$\Rightarrow 2x^2-2x+x-1=0$
$\Rightarrow 2x(x-1)+1(x-1)=0$
$\Rightarrow (2x+1)(x-1)=0$
$\Rightarrow x=-\frac{1}{2},1$
Hence, all the zeros are $-\frac{1}{2},1,2+\sqrt{3},2-\sqrt{3}$