# Homework Help: Roots and factor theorem

1. Oct 1, 2012

### safat

1. The problem statement, all variables and given/known data
knowing a,b and c are roots 3x^3-x^2-10x+8=0
show that:
1) 1/a+1/b+1/c=5/4
2)a^2+b^2+c^2=61/9

2. Relevant equations
factor theorem --> (x-a)(x-b)(x-c)

3. The attempt at a solution
can only use factor theorem:
therefore (x-a)(x-b)(x-c)--> up to: x^3-x^2(a+b+c)+x(ac+bc+ab)-abc
no idea where to go now!

Last edited: Oct 1, 2012
2. Oct 1, 2012

### tiny-tim

welcome to pf!

hi safat! welcome to pf!

(try using the X2 button just above the Reply box )

(1) = what / what ?

(2) = (a+b+c)2 minus … ?

3. Oct 2, 2012

### safat

thanks tiny-tim.

that's what I got by confrontating the original equation (for the record:3(x3-1/3x2-10/3x+8/3)=0 )
and x3-x2(a+b+c)+x(ac+bc+ab)-abc=0

1/a+1/b+1/c=(ac+bc+ab)/abc=(-10/3)/(-8/3)=5/4

should I use the quadratic formula or can I go by guessing?
I got (a+b+c)2-2(ac+bc+ab)=(-1/3)2-2(-10/3)=61/9

4. Oct 2, 2012

### tiny-tim

hi safat!

(btw, no such word as "confrontate" … i think you mean "compare"! )
yup!
(i'm not sure what you mean by the quadratic formula, but anyway …)

yes, guessing is fine, so long as you check that your guess works!

5. Oct 2, 2012

### safat

thanks again tiny-tim!
you also provide a really efficient proof reading service!

tbh, I have done it by guessing. But it looks like the great majority of my peers used the quadratic formula (x=-b+-(√b2-4ac)/2a)
I was wondering how it could be done that way.

6. Oct 2, 2012

### tiny-tim

beats me!

Of course if it happens that you can explicitly find all the roots then that is a viable option. In this particular example the roots turned out to be easy to find, but what if the question was slightly different, say $3 x^3 - x^2 -10x +7 = 0$. Your method would still work just as easily, but your class mates would presumably find the going much tougher.