Question

Find the zeros of ƒ(x) = x3 + 6x2 + 8x.

Answer 1

$x^3+6x^2+8x = 0\\\\\implies x^3+2x^2 +4x^2+8x =0\\\\\implies x^2(x+2) +4x (x+2) =0\\\\\implies (x+2)(x^2 +4x) = 0\\\\\implies x (x+2) (x+4)=0\\\\\text{Hence,} \\\\x = 0 ,-2,-4$

Answer 2

Answer:

Option A: $x=0$, $x=-2$, and $x=-4$

Step-by-step explanation:

$f(x)=x^3+6x^2+8x$

$0=x^3+6x^2+8x$

$0=x(x^2+6x+8)$

$0=x(x+2)(x+4)$

$x=0$, $x=-2$, and $x=-4$