Answer
$$
\begin{array}{l}{f(x)=(x-1)(x-2)(x+3)} \\ {g(x)=x^{3}+2 x^{2}-x-2} \\ {h(x)=3 x^{3}+6 x^{2}-3 x-6}\end{array}
$$
(a)
$$
f(1)=(0)(-1)(4)=0
$$
(b)
$$
f(x) =0
$$
when $x=2$and $x=-4$.
(c)
$$
\begin{aligned} g(-1)&=(-1)^{3}+2(-1)^{2}-(-1)-2 \\ &=-1+2+1-2\\ &=0 \\ g(1) &=(1)^{3}+2(1)^{2}-(1)-2 \\ \quad &= 1+2-1-2 \\ &= 0 \\ g(-2) &=(-2)^{3}+2(-2)^{2}-(-2)-2 \\ &=-0 \end{aligned}
$$
(d)
$$
\begin{aligned} g(x) &=[x-(-1)](x-1)[x-(-2)] \\ g(x) &=(x+1)(x-1)(x+2) \end{aligned}
$$
(e)
$$
\begin{aligned} h(x) &=3 g(x) \\ &=3(x+1)(x-1)(x+2) \end{aligned}
$$
(f)
If $f$is a polynomial and $f(a)=0$ for some number $a$ then one factor of the polynomial is $( x-a)$.
Work Step by Step
$$
\begin{array}{l}{f(x)=(x-1)(x-2)(x+3)} \\ {g(x)=x^{3}+2 x^{2}-x-2} \\ {h(x)=3 x^{3}+6 x^{2}-3 x-6}\end{array}
$$
(a)
$$
f(1)=(0)(-1)(4)=0
$$
(b)
$$
f(x) =0
$$
when $x=2$and $x=-4$.
(c)
$$
\begin{aligned} g(-1)&=(-1)^{3}+2(-1)^{2}-(-1)-2 \\ &=-1+2+1-2\\ &=0 \\ g(1) &=(1)^{3}+2(1)^{2}-(1)-2 \\ \quad &= 1+2-1-2 \\ &= 0 \\ g(-2) &=(-2)^{3}+2(-2)^{2}-(-2)-2 \\ &=-0 \end{aligned}
$$
(d)
$$
\begin{aligned} g(x) &=[x-(-1)](x-1)[x-(-2)] \\ g(x) &=(x+1)(x-1)(x+2) \end{aligned}
$$
(e)
$$
\begin{aligned} h(x) &=3 g(x) \\ &=3(x+1)(x-1)(x+2) \end{aligned}
$$
(f)
If $f$is a polynomial and $f(a)=0$ for some number $a$ then one factor of the polynomial is $( x-a)$.
