Show that $\dfrac{b}{a} = -(A+B+C)$ and $\dfrac{c}{a} = (AB+AC + BC)$

Question

Let´s assume we have the polynomial: $ax^3+bx^2+cx+d$

This polynomial has three zero points: $A$, $B$ and $C$.

How can I show that $$\dfrac{b}{a} = -(A+B+C)$$

and

$$\dfrac{c}{a} = (AB + AC + BC)$$?

Thanks for your help!

Ok, I tried like this: ax^3+bx^2+cx+d = (x-A)(x-B)(x-C) = x^3 - Cx^2 - Bx^2 + BCx - Ax^2 + ABx - ABC — Alexander West, Mar 30 '18 at 08:50

Dr. Sonnhard Graubner · Answer 1 · 2018-03-30T08:58:59.923

0

HINT: note that if $A,B,C$ are the zero points of $$ax^3+bx^2+cx+d$$ then you can write $$ax^3+bx^2+cx+d=a(x-A)(x-B)(x-C)$$ so $b=-a(A+B+C)$ etc.

edited Mar 30 '18 at 08:58

answered Mar 30 '18 at 08:49

Yes, and a(x - A)(x - B)(x - C) = ax^3 - aCx^2 - aBx^2 + aBCx - a^2x^2 + a^2Cx + a^2Bx - a^2BC – Alexander West Mar 30 '18 at 08:54

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