Factorable polynomial zeros (ACT). Game on.

Factor: $x^2+x-12=(x+4)(x-3)$. After factoring, set each linear factor equal to zero: $(x+4)=0 \Rightarrow x=-4$; $(x-3)=0 \Rightarrow x=3$. The zeros are $-4$ and $3$.

Factor: $x^2-6x+9=(x-3)^2$. Set the factor equal to zero: $(x-3)^2=0$, so $(x-3)=0 \Rightarrow x=3$. The zero $3$ has multiplicity $2$, so the zeros are $3$ and $3$.

Find two numbers with product $6$ and sum $-5$: $-2$ and $-3$. So $p(x)=(x-2)(x-3)$. After factoring, set each linear factor equal to zero: $(x-2)=0 \Rightarrow x=2$; $(x-3)=0 \Rightarrow x=3$. The zeros are $2$ and $3$.

Since $(x-2)$ is a factor, dividing by $(x-2)$ gives $p(x)=(x-2)(x^2-2x-3)=(x-2)(x-3)(x+1)$. The zeros come from the linear factors, so set each equal to zero: $(x-2)=0 \Rightarrow x=2$; $(x-3)=0 \Rightarrow x=3$; $(x+1)=0 \Rightarrow x=-1$. The zeros are $2$, $3$, and $-1$.

Group: $p(x)=(x^3+3x^2)+(-x-3)=x^2(x+3)-(x+3)=(x+3)(x^2-1)=(x+3)(x+1)(x-1)$. After factoring, set each linear factor equal to zero: $(x+3)=0 \Rightarrow x=-3$; $(x+1)=0 \Rightarrow x=-1$; $(x-1)=0 \Rightarrow x=1$. The zeros are $-3$, $-1$, and $1$.

Since $(x-1)$ is a factor, dividing by $(x-1)$ gives $p(x)=(x-1)(x^2+x-6)=(x-1)(x+3)(x-2)$. The zeros are the roots of the linear factors, so set each equal to zero: $(x-1)=0 \Rightarrow x=1$; $(x+3)=0 \Rightarrow x=-3$; $(x-2)=0 \Rightarrow x=2$. The zeros are $1$, $2$, and $-3$.

Factor out $x$: $p(x)=x(x^2+4x+4)=x(x+2)^2$. Set each factor equal to zero: $x=0 \Rightarrow x=0$; $(x+2)^2=0$, so $(x+2)=0 \Rightarrow x=-2$. The zero $-2$ has multiplicity $2$, so the zeros are $0$, $-2$, and $-2$.

Let $u=x^2$. Then $u^2-10u+9=(u-1)(u-9)$, so $p(x)=(x^2-1)(x^2-9)=(x-1)(x+1)(x-3)(x+3)$. Set each linear factor equal to zero: $(x-1)=0 \Rightarrow x=1$; $(x+1)=0 \Rightarrow x=-1$; $(x-3)=0 \Rightarrow x=3$; $(x+3)=0 \Rightarrow x=-3$. The zeros are $-3$, $-1$, $1$, and $3$.