we have the polynomial
P(x)=x^3+4x^2-20x-48
Remember that
If a binomial is a factor of the given polynomial
then
the value of the polynomial when evaluating the factor must be equal to zero
Verify each binomial
N 1
(x-3)
For x=3
substitute in the given polynomial
P(3) ----> is not equal to zero
N 2
(x+4)
For x=-4
P(-4) ---> is not equal to zero
N 3
(x-2)
For x=2
P(2) ---> is not equal to zero
N 4
(x+6)
For x=-6
substitute
P(-6)=(-6)^3+4(-6)^2-20(-6)-48
P(-6)=-216+144+120-48
P(-6)=0
that means
binomial (x+6) is a factor