A Not-So-Special Cubic Function

Since $f(x)$ is monic, we know that it is of the form $f(x)=x^{3}+ax^{2}+bx+c$ .

Let's start by looking at that first wacky condition, where the $x$ -coordinates of intersections with a line all add up to two, and that the line must pass through $(2,2)$ . Let's define $L(x)=mx-2m+2$ as a line through $(2,2)$ with slope $m$ .

We know that the $x$ values of the intersections between $f(x)$ and $L(x)$ are really just the roots of the equation $f(x)-L(x)=0$ , so we know that $x^{3}+ax^{2}+bx-mx+c+2m-2=0$ has roots $x_{1},x_{2},x_{3}$ .

By Vieta's Formula , we know that $-a=x_{1}+x_{2}+x_{3}$ , which was given to be $2$ by the problem. Therefore, $a=-2$ .

We can deal with the second condition in the same way. We know that $x=0$ is a root of $P(x)$ , therefore we know that $P(x)=x(qx+r)$ for some values $q$ and $r$ . This means that $P(x)=qx^{2}+rx$ . We once again note that the $x$ values of the intersections between $f(x)$ and $P(x)$ are the roots of the equation $f(x)-P(x)=0$ .

We then get that $x^{3}+ax^{2}-qx^{2}+bx-rx+c=0$ has roots $x_{4},x_{5},x_{6}$ . Once again, by Vieta's Formula, we know that $-c=x_{4}x_{5}x_{6}$ , which was given by the problem to be $-3$ . Therefore, we know that $c=3$ .

We now know that $f(x)=x^{3}-2x^{2}+bx+3$ , and so we can apply the last condition of $f(2)=17$ to find that $b=7$ .

And so we get that $f(x)=x^{3}-2x^{2}+7x+3$ , and so $f(5)=125-50+35+3=\boxed{113}$ .

assume f(x) to be general cubic expression. Then for f(x)=x , sum of roots is 2, show that b=-2a. Then for f(x)=lx^2+mx , product of roots is -3,shew that d=3a. Now for f(x)=2,shew that 2c+d=17. Since polynomial is monic put a=1 to get b=-2,c=7and d=3. Then find f(5) as 113