$x^3-8x^2+30x-20=0$ has roots $a$, $b$, $c$. Find the equation with roots $a+2$, $b+2$, $c+2$

Question

I have the equation $$x^3-8x^2+30x-20=0$$ let's call the roots $a,b,c$

It's easy to find the equation with roots $(a+2)(b+2)(c+2)$ these are the steps in my books

so the equation is $$ aX^3+ bX^2 + cX + D = 0 $$ $$ x^3 -14x^2 +74x -120 = 0$$

my problem is, I can't understand why divide by $-2$ not $2$ how do I do this division with long division? like what am I dividing by if the symbols were here?!

Answer 1

You are finding $P(x-2)$, given $P(x)$. If $a$ is a root of $P(x)$, clearly $a+2$ is a root of $P(x-2)$.

Answer 2

Your book is using synthetic division. As you are dividing by $(x+2)$, the divisor must be $-2$. Generalized, if you are dividing polynomial $f(x)$ by $x+a$, for synthetic division, you use $-a$, which is exactly what your book did.

Answer 3

Alternatively: for the roots $a,b,c$ of $x^3-8x^2+30x-20=0$, by Vieta's we have: $$a+b+c=8\\ ab+bc+ca=30\\ abc=20$$ Now we want the roots $a+2,b+2,c+2$ of $X^3+kX^2+lX+m=0$. Hence: $$a+2+b+2+c+2=a+b+c+6=14=-k\\ (a+2)(b+2)+(b+2)(c+2)+(c+2)(a+2)=\\ (ab+bc+ca)+4(a+b+c)+12=74=l\\ (a+2)(b+2)(c+2)=\\ abc+2(ab+bc+ca)+4(a+b+c)+8=120=-m \Rightarrow \\ x^3-14x^2+74x-120=0.$$