We can factor the polynomial by grouping.
$$p(x)=x^3+x^2-4x-4 $$
$$ p(x)=x^2(x+1)-4(x+1) $$
$$ p(x)=(x^2-4)(x+1) $$
$$ p(x)=(x-2)(x+2)(x+1) $$
In this form, we clearly see that it is divisible by \(x+1\) and \(x+2\).
If a binomial term is a factor of a polynomial, it refers to the zero or \(x\)-intercept of the polynomial. For example, if \(x-a\) is a factor of a polynomial, \(a\)
must be a zero or \(x\)-intercept to the polynomial. We can check whether the three choices are \(x\)-intercepts:
$$p(x)=x^3+x^2-4x-4 $$
$$0=x^3+x^2-4x-4 $$
$$x=-1\ $$
$$0=(-1)^3+(-1)^2-4(-1)-4 $$
$$0=-1+1+4-4 $$
$$ 0=0 \checkmark $$
$$x=-2\ $$
$$0=(-2)^3+(-2)^2-4(-2)-4 $$
$$0=-8+4+8-4 $$
$$ 0=0 \checkmark $$
$$x=-3\ $$
$$0=(-3)^3+(-3)^2-4(-3)-4 $$
$$0=-27+9+12-4 $$
$$ 0=-10 ✖ $$