## Calculus with Applications (10th Edition)

$$\begin{array}{l}{f(x)=(x-1)(x-2)(x+3)} \\ {g(x)=x^{3}+2 x^{2}-x-2} \\ {h(x)=3 x^{3}+6 x^{2}-3 x-6}\end{array}$$ (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 $( x-a)$.
$$\begin{array}{l}{f(x)=(x-1)(x-2)(x+3)} \\ {g(x)=x^{3}+2 x^{2}-x-2} \\ {h(x)=3 x^{3}+6 x^{2}-3 x-6}\end{array}$$ (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 $( x-a)$.