Question

If two zeroes of the polynomial $(x^3-5x^2-16x+80)$ are equal in magnitude but opposite in sign, then zeroes are

• A. $4,-4,5$
• B. $3,-3,5$
• C. $2,-2,5$
• D. $1,-1,5$

Answer 1

The correct option is A $4,-4,5$

Let $f\left(x\right)=x^3-5x^2-16x+80$
$f\left(x\right)=0\Rightarrow x^3-5x^2-16x+80=0$
$\Rightarrow x^2(x-5)-16(x-5)=0$
$\Rightarrow (x-5)(x^2-16)=0$
$\Rightarrow (x-5)(x^2-4^2)=0$
$\Rightarrow (x-5)(x+4)(x-4)=0$
$\Rightarrow x=4,-4,5$
$\therefore$ the zeroes of the polynomial are $4,-4,5$
The correct answer is A.