Question

What are the zeros of the polynomial function
F(x) = x^3-5x^2-6x

A. x= -2, x=0 and x=3
B. x=-1, x=0 and x=6
C. x=-3, x=0 and x=2
D. x=-6, x=0 and x=1

Answer 1

Option B

The zeros of the polynomial function are x = 0 , x = 6, x = -1

Solution:

Given function is:

$f(x) = x^3 -5x^2 - 6x$

To find: zeros of the polynomial function

To find the zeros of polynomial function, set the function equal to zero and then solve for x.

$x^3 -5x^2 - 6x = 0$

Taking "x" as common term, we get

$x(x^2 - 5x - 6) = 0$

Equating to zero,

$x = 0 \text{ and } x^2 - 5x - 6 = 0$

So one of the zeros of polynomial is x = 0

Let us solve $x^2 - 5x - 6 = 0$ to find other zeros

Let us solve using quadratic formula,

$\text {For a quadratic equation } a x^{2}+b x+c=0, \text { where } a \neq 0\\\\x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

Using the above formula,

$\text{ for } x^2 - 5x - 6 = 0 \text{ we have } a = 1 ; b = -5 ; c = -6$

$\text{ The discriminant } b^2 - 4ac > 0 , \text{ so, there are two real roots }$

Substituting the values of a, b, c in above formula,

$x=\frac{-(-5) \pm \sqrt{(-5)^{2}-4(1)(-6)}}{2(1)}\\\\x=\frac{5 \pm \sqrt{25+24}}{2}=\frac{5 \pm \sqrt{49}}{2}\\\\x=\frac{5 \pm 7}{2}\\\\x=\frac{5+7}{2} \text{ or } x=\frac{5-7}{2}\\\\x=\frac{12}{2} \text{ or } x=\frac{-2}{2}\\\\x=6 \text{ or } x=-1$

Thus the zeros of the polynomial function are x = 0 , x = 6, x = -1