# Solve the roots of a cubic polynomial?

I have had trouble with this question - mainly due to the fact that I do not fully understand what a 'geometric progression' is:

"Solve the equation $x^3 - 14x^2 + 56x - 64 = 0$" if the roots are in geometric progression.

Any help would be appreciated.

It means the roots are of the form $a, ar, ar^2$.

Here it is not too difficult to see that $2, 4, 8$ are ok, just look at the constant term, which is $- (a \cdot ar \cdot ar^2) = - (a r)^3$, and check that $2, 4, 8$ fit with the other coefficients $$14 = 2 + 4 + 8, \qquad 56 = 2 \cdot 4 + 2 \cdot 8 + 4 \cdot 8.$$

• Thanks, but are there different types of geometric progessions? – missiledragon May 15 '13 at 10:21
• @missiledragon, not that I am aware of, it means a sequence of the form $a_{n} = a r^{n}$ for some $a, r$, see the Wikipedia link. – Andreas Caranti May 15 '13 at 10:23
• Thanks, if you don't mind, could you briefly explain what an arithmetic progression is? I had a similar question with it. – missiledragon May 15 '13 at 10:25
• @missiledragon, the Wikipedia is your friend en.wikipedia.org/wiki/Arithmetic_progression it means a sequence of the form $a_{n} = a + n r$, for fixed $a, r$, and $n \ge 0$. – Andreas Caranti May 15 '13 at 10:27
• Alright, thanks for your help – missiledragon May 15 '13 at 10:32

Let the roots be $a,a\cdot r,a\cdot r^2$

Using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies a(1+r+r^2)=14$

and $a(a\cdot r)+a\cdot r(a\cdot r^2)+a(a\cdot r^2)=56\implies a^2\cdot r(1+r+r^2)=56$

On division, $ar=4$ as $a\cdot r\ne0$

$$\implies a=\frac 4r\implies \frac{4(1+r+r^2)}r=14\implies 2r^2-5r+2=0\implies r=2\text{ or }\frac12$$

Alternatively, using Vieta's formula $a\cdot(a\cdot r) \cdot (a\cdot r^2)=64\implies (ar)^3=64$

So, $a\cdot r$ can be one of $4,4w,4w^2$ where $w$ is the cube root of $1$

Using Polynomial Remainder Theorem, observe that $4$ is a root of the given equation

$\implies a\cdot r=4\iff a=\frac4r$

Again, using Vieta's formula $a+a\cdot r+a\cdot r^2=14\implies 2r^2-5r+2=0$