Answer:
f(−2)=0
(x+2) is a factor of f(x) .
Step-by-step explanation:
f(x)=x^3+5x^2+6x
LEts factor f(x)
f(x)= x(x^2 +5x+6)
product is 6 and sum is 5
3*2= 6 and 3+2=5
f(x)= x(x+2)(x+3)
Now we check with each option
f(-2)= 0
LEts plug in -2 for x in f(x)
f(x)=x^3+5x^2+6x
f(-2)= (-2)^3 + 5(-2)^2 + 6(-2)= -8 +20-12= 0
f(x) divided by (x−2) has a remainder of 0
f(x)= x(x+2)(x+3)
f(x) does not have factor x-2 so f(x) is not divisible by (x-2)
(x+2) is a factor of f(x)
f(x)= x(x+2)(x+3)
f(x) have factor x+2 so (x+2) is a factor of f(x)
when x=2, f(x)=0
LEts plug in 2 for x in f(x)
f(x)=x^3+5x^2+6x
f(-2)= (2)^3 + 5(2)^2 + 6(2)= 8 +20+12= 40
