Question

What are the zeros of the polynomial function f(x) = x3 + x2 − 20x?
A. 0, −5, −4
B. 0, −5, 4
C. 0, 5, −4
D. 0, 5, 4

Answer 1

$x^3 + x^2 - 20x=0\\ x(x^2+x-20)=0\\ x(x^2+5x-4x-20)=0\\ x(x(x+5)-4(x+5))=0\\ x(x-4)(x+5)=0\\ x=0 \vee x=4 \vee x=-5$

Answer 2

Answer:

B. 0, −5, 4

Step-by-step explanation:

Given :

$x^{3} +x^{2} -20x$

Solution:

$x^{3} +x^{2} -20x$

taking x common from the given polynomial

⇒$x(x^{2} +x-20)=0$

⇒$x(x(x+5)-4(x+5))=0$

⇒$x(x+5)(x-4)=0$

⇒ x=0 , x+5=0 , x-4=0

⇒x = 0 , x = -5 , x = 4