# A cubic equation and its roots

1. Apr 28, 2015

### mooncrater

1. The problem statement, all variables and given/known data
The question says that :
Find the value of $a$ so that the equation $$x^3-6x^2+11x+a-6=0$$ has exactly three integer solitions.

2. Relevant equations
IF $p$,$q$,$r$ are the roots of this equation then:
$p+q+r=6$
$pq+pr+rq=11$
$pqr=6-a$

3. The attempt at a solution
I just don't know how to use these equations to find $a$.
So how to do this question?Are these equations of any use here?

2. Apr 28, 2015

### brainpushups

I would simply factor it.

3. Apr 28, 2015

### mooncrater

But how can you simply factor it when there is an unknown $a$ in the equation?

4. Apr 28, 2015

### haruspex

So factor the rest of it.

5. Apr 28, 2015

### Dick

So that's pretty much just guessing that $a=0$, right? Or is there a better reason to do it? Suppose the constant were $a-7$?

6. Apr 28, 2015

### haruspex

It just struck me as something worth doing to obtain insight. It doesn't presume a=0, in fact it reveals an infinity of solutions. A minor modification solves a-7.

7. Apr 29, 2015

### ehild

Find possible integer solutions which sum up to 6. Then select those for which pg+pr+qr=11.
You can eliminate one of the roots. Try completing squares.

Last edited: Apr 29, 2015
8. Apr 29, 2015

### mooncrater

Thank you....Did what you said.
Leaving the zero values of $p,q,r$ , three possibilities were present:
$$p=1,q=1,r=4$$, $$p=1,q=2,r=3$$,$$p=q=r=2$$
Among which the second one satisfies the equation $$pq+qr+pr=11$$
So $pqr=6=6-a$
Therefore $a=0$.
Is this how you meant it to be done?

9. Apr 29, 2015

### ehild

Yes, it was my first idea. There might be other solutions, including negative roots, but the problem text did not want you give or exclude more.

10. Apr 29, 2015

### haruspex

It says integer solutions, not natural numbers.
Factorising the terms other than a makes it obvious what all the solutions are.

11. Apr 29, 2015

### Dick

I still don't see it. You aren't just saying that $-a$ is the product of three consecutive integers, are you? That suffices to give one integer root. Doesn't say there are three unless $a=0$.

12. Apr 29, 2015

### haruspex

Ah yes, not thinking straight.

13. Apr 29, 2015

### SammyS

Staff Emeritus
I have puzzled over how to use those equations for p, q, & r.

Use the factor theorem, plus perhaps Descartes's Sign Rule.

At most one negative root, that's if a > 6 . If -1 is a root, then a = 24.

If 0 is a root, a = 6. Look at the resulting quadratic.

If 1 is a root, a = __ , ...

14. Apr 30, 2015

### epenguin

You could use psychology. Why are they giving you the equation in the form

?

The constant term is peculiar. It would be equivalent but normally more natural to ask what values of b give integral solutions of

$$x^3-6x^2+11x+b=0$$ .

It is because you might have to be a tad smarter to see how to do it when espressed that way. They are giving you a hint! I'll give you another. In the first equation the cubic without the a is quite easily factorised. Just knowing that it is makes it easier. Then if you think what a cubic looks like in a graph, you will see there is only a limited range of b within which it can have 3 real solutions and only a limited number of values for which it can have integer solutions.

Last edited: May 1, 2015
15. Apr 30, 2015

### ehild

You meant factorising the part without a ?

16. May 1, 2015

### epenguin

Yes. Thank you, I have corrected that now.

17. May 1, 2015

### mooncrater

Okay then , now I understand that this equation can also be solved without trial and error method. So I did it as:
$x^3-6x^2+11x-6=-a$
$x^3-5x^2-x^2+5x+6x-6=-a$
Which gives the equation:
$(x-1)(x-2)(x-3)+a=0$
Which has integer roots only if $a=0$
Is that what you want me to understand?

18. May 1, 2015

### epenguin

Up to the last two lines. It has integer roots for a = 0 . The equation asks you find THE value of a that gives you that, so I suppose you have answered the question. You might worry a bit about whether this is the only possible value of a to give integer roots and how you would handle it if there were more than one value of a giving integer roots.

Perhaps there are cases like that, say for a cubic. Perhaps you could construct such cases. Or show that there can't be any. Depending on your time and interest. Polya principle of 'How to solve it' - when you have solved it, it's not finished. But at least try see if you can see in this case if 0 is the only value.

19. May 1, 2015

### mooncrater

Okay I think I got what you want to say. We can have infinite values of $a$ to satisfy the equation. But since it's given that the equation has 3 roots therefore $a$ can have only one value that is $0$.
Is this what you want to say to me?

20. May 1, 2015

### epenguin

No , I am saying it is not demonstrated and not all that obvious that there are no other values of a to give integral roots, and for someone wanting to develop a bit of mathematical muscle (of course you may consider you have better things to do with your time) it is of value to try and see further inside this.