## Calculus: Early Transcendentals 8th Edition

The solution is $$f(x)=-3x^3+3x^2+6x=-3x(x+1)(x-2)$$
The general expression for the cubic function is $$f(x)=ax^3+bx^2+cx+d,$$ where $a$, $b$, $c$, and $d$ are constants. We can find them from the given conditions: $f(1)=6\Rightarrow a1^3+b1^2+c\cdot1+d=6\Rightarrow a+b+c+d=6;$ $f(-1)=0\Rightarrow a(-1)^3+b(-1)^2+c(-1)+d=-a+b-c+d=0;$ $f(0)=0\Rightarrow a0^3+b0^2+c\cdot0+d=0\Rightarrow d=0.$ $f(2)=a2^3+b2^2+c\cdot2+d=0\Rightarrow8a+4b+2c+d=0$ First, we can put $d=0$ everywhere. $$a+b+c=6\\ -a+b-c=0\\ 8a+4b+2c=0$$ Now add the first and the second equation: $$a+b+c+(-a+b-c)=6+0$$ and this becomes $$2b=6\Rightarrow b=3.$$ Divide the last equation by $2$ and add the second one to it with $b=3$: $$\frac{1}{2}(8a+4\cdot3+2c)+(-a+3-c)=0+0$$ and this becomes $$4a+6+c-a+3-c=0$$ and finally $$3a+9=0\Rightarrow 3a=-9\Rightarrow a=-3.$$ Now put $a=-3$ and $b=3$ into the first equation to find $c$: $$-3+3+c=6\Rightarrow c=6.$$ Now we have $$a=-3,\quad b=3,\quad c=6,\quad d=0$$ so the required cubic function is $$f(x)=-3x^3+3x^2+6x.$$