# Polynomial factor help

1. Apr 11, 2010

### Mentallic

1. The problem statement, all variables and given/known data
In the polynomial $$x^5+22x^3-34x^2+117x-306$$ given that the roots on the real and imaginary plane are all integers, factorize the polynomial into real linear and quadratic factors.

3. The attempt at a solution
I was able to find the real factor, which is x=2 and then I reduced the polynomial to this:

$$(x-2)(x^4+2x^3+26x^2+18x+153)$$

But this quartic has all complex roots, and I'm unsure how to find them given the information above. Please help!

2. Apr 11, 2010

### Mentallic

Re: Polynomial

Oh and I forgot to mention that since the polynomial has real coefficients, the complex roots will have complex conjugate roots also, so this can be expressed as follows:

$$x^4+2x^3+26x^2+18x+153 \equiv (x-(a+ib))(x-(a-ib))(x-(c+id))(x-(c-id))$$

where, a,b,c,d all integers (from the information in the question).

Expanding these complex linear factors into real quadratic factors gives:

$$(x^2-2a+a^2+b^2)(x^2-2c+c^2+d^2)$$

I could try expand this and equate the coefficients with the original quartic, but this is quite a lot of effort considering the question should have been answered relatively quickly.

3. Apr 11, 2010

### Mentallic

Re: Polynomial

bump.

4. Apr 12, 2010

### Bohrok

Re: Polynomial

The last term in each of the quadratic factors have to be factors of 153, so either 3 and 51 or 9 and 17, but the last pair is the obvious choice. Then you could write out the quadratic factors as (x2 + ax + 9)(x2 + bx + 17)
Now you don't have to multiply out the whole thing but only part of it. Multiplying it out, you would get two terms with x3, three with x2, and two with x. Multiply the terms that give you the x3 and x terms and equate them with the coefficients in your quartic. You'll easily get a and b in the two quadratic factors above and you're done.

5. Apr 13, 2010

### Mentallic

Re: Polynomial

Oh I was just told today that it is acceptable to use Maple to solve this problem, considering no one else could answer it within the 4 minute allocated time limit.

Anyway, not just with guessing but from the quadratics I expressed, we also have:

$$(a^2+b^2)(c^2+d^2)=153$$ where $$a,b,c,d E Z$$

then looking at the possible factors -

1 and 153
3 and 51
9 and 17

It can be concluded that the only possible pair that can be expressed as above is 9 and 17, with $$a=0, b=\pm 3$$ (since it has to be an imaginary root) and $$c,d=\pm 4,\pm 1$$.

From there, using the other relationships, the results can be found.