Question

Determine all the zeros of $4x^3+5x^2-180x-225$ if one of its zeros is $-\frac{5}{4}$.

• A. $3\sqrt{5},-3\sqrt{5}$
• B. $-\frac{5}{4},3\sqrt{7},-3\sqrt{5}$
• C. $-\frac{5}{4},3\sqrt{5},-3\sqrt{5}$
• D. $5,-3$

Answer 1

The correct option is C $-\frac{5}{4},3\sqrt{5},-3\sqrt{5}$

Let, $p\left(x\right)=4x^3+5x^2-180x-225$
Since, $-\frac{5}{4}$ is zero of $p\left(x\right)$
$\Rightarrow (4x+5)$divides $p\left(x\right)$ ($\because Factorthm.$)
$4x+5)\overline{4x^3+5x^2-180x-225}$( $x^2-45$
$\underset{–}{\underset{-}{4}{x}^3\underset{-}{+}5x^2}$
$-180x-225$
$\underset{–}{\underset{+}{-}180x\underset{+}{-}225}$
$0$
Now, the other roots can be obtained from $x^2-45=0$
$\Rightarrow x^2=45$
$\Rightarrow x=\pm \sqrt{45}=\pm 3\sqrt{5}$
Hence, the zeros are $\frac{-5}{4},-3\sqrt{5},3\sqrt{5}$