Help Getting the roots of a polynomial

[degs2k4]

Homework Statement

I am trying to get the roots of:

$$(x-a)^3 - (x-a)b^2 - 2b^3 -2b^2(x-a)$$

which I know they are $$x = a-b$$ and $$x = a+2 b$$

the problem is, how can I reach that solution ?

The Attempt at a Solution

At first I thought of separating the independent term (the one without x's) and manipulating the expression in order to guess x, but I haven't been successful...

$$x^3 - xb^2 - 3xb^2 - a^3 + 3ab^2 - 2b^3 = 0$$

$$x^3 - xb^2 - 3xb^2 = a^3 - 3ab^2 + 2b^3$$

How could I solve this?

Thanks...

 

[owlpride]

The first thing to notice is that x and a always appear together. I would solve for the quantity (x-a) first instead of x directly. With that in mind, let's substitute z = (x-a).

When you are trying to guess the answer for a polynomial, factoring it is usually much better than expanding. In this particular case, there are two nice quantities you could factor out of the equation, either z=(x-a) or b^2. This yields:

$$z (z^2 - 3 b^2) = 2 b^3$$

$$b^2 (3 z + 2b) = z^3$$

You should be able to guess at least one solution for z from these equations. Then you can factor out a (z-solution) term and use the quadratic formula to find the remaining solutions. Once you have all solutions for z, undo the substitution to get x.

 

[Mentallic]

To add onto that, what you can try is the rational root theorem instead. Because for this question to be do-able without a calculator of some sort, the roots must be simple in some sense!

Say we have a cubic x³+ax²+bx+12=0

We would use the rational root theorem test by trying all values that are factors of 12, i.e. 1,2,3,4,6,12. In this sense, we can try and assume b is an integer, which means that the factors of -2b³
Oh, btw, the cubic is z³-3b²z-2b³=0, where z=x-a

Anyway, the possible simple factors of -2b³ are going to be b,-b,2b,-2b. There could be more using this logic, such as b³ but since by testing these, we are going to have to substitute this into z and the z³ will just make this term b⁹ and that's obviously not going to cancel with the others.

So trying the above possible roots:

z=b, b³-3b³-2b³$\neq$0
z=-b, -b³+3b³-2b³=0
z=2b, 8b³-6b³-2b³=0
z=-2b, -8b³+6b³-2b³$\neq$0

So this means we have roots z=-b,2b.

This means to find the last root we factor these roots and make it equivalent to the initial cubic:

(z+b)(z-2b)(z-k)$\equiv$z³-3b²z-2b³

Now equate.

 

[epenguin]

Seems you have missed that you have two terms in (x-a)b² that you should combine.

Then you may also see that you could further simplify, or make it look easier and more obvious, by dividing by b³ and expressing in terms of the new unknown, call it X, X = (x - a)/b.

After which I hope the whole factorisation should look fairly obvious, if not come back with your result after those operations.

 

[degs2k4]

$$z (z^2 - 3 b^2) = 2 b^3$$

Thanks for your reply.

However, factoring z = x-a I get the following:

$$z (z^2 - z - 2 b^2) = 2 b^3$$

Which is quite different to what you wrote...

 

[owlpride]

$$z (z^2 - z - 2 b^2) = 2 b^3$$

The z inside the parentheses should be a b^2. There's no z^2 term in the original polynomial.

 

[epenguin]

Sorry - I am hasty and error prone. See my edited version. I think this is the simplest approach. Unless I still have another error. :uhh:

 

[owlpride]

I like epenguin's approach. I totally missed that the powers of z and b always add up to three. Thanks for pointing that out!

 

[gomunkul51]

This is the form from which you WILL solve it :)

z = (x-a)

z^3 - z(3b^2) - 2b^3 = 0

like Mentallic said, IF integer roots exist, they must be the integer divisors of - 2b^3.

Try to find in this way at least ONE root, then you can lower the degree of the polynomial (e.g. by poly long division) and then solve a second degree polynomial :P

 

[owlpride]

I think epenguin's suggestion results in the simplest equation to solve:

$$z^3 - 3 z - 2 = 0$$

with z = (x-a)/b

 

[degs2k4]

Thank you very much everyone for your help.

I finally reached the solution with z = (x-a), but I also tried epenguin's substitution (z=(x-a)/b) and everything gets nicely simplified.

Thanks!

 

[Mentallic]

Ahh yes epenguin that's a sharp eye you have there!

 

[ehild]

$$(x-a)^3 - (x-a)b^2 - 2b^3 -2b^2(x-a)$$

One can notice at once the factor 2 in the last two terms. The expression offers the grouping


(x-a)((x-a)^2 - b^2 )- 2b^2(b +x-a).

From here, it is straightforward to proceed.

ehild

 

[Mentallic]

That is quite clever too! I can honestly admit I didn't spot that before grouping, so I wouldn't even start with grouping it in the first place, but if I did... that is definitely the way to go.