Question

Two zeros of $x^3+x^2-5x-5$ are $\sqrt{5}$ and $-\sqrt{5}$ then the third zero is ______.

• A. $-2$
• B. $-1$
• C. $2$
• D. $1$

Answer 1

The correct option is B $-1$

$a{x}^3+b{x}^2+cx+d$ has zeros $\alpha ,\beta$ and $\gamma$.

Then $\alpha +\beta +\gamma =-\frac{b}{a}$

Comparing $x^3+x^2-5x-5$

with $a{x}^3+b{x}^2+cx+d$

We have, $a=1,b=1,c=-5$ and $d=-5$

Now, $x^3+x^2-5x-5$ has two zeros $\sqrt{5}$ and $-\sqrt{5}$

So, $\alpha =\sqrt{5},\beta =-\sqrt{5},\gamma =?$

$\therefore \alpha +\beta +\gamma =-\frac{b}{a}$

$\therefore \sqrt{5}+(-\sqrt{5})+\gamma =-1$

$\therefore \gamma =-1$

Therefore, the third zero is $-1$.