Factorizing polynomials

1. Oct 10, 2007

Trail_Builder

1. The problem statement, all variables and given/known data

factorise the following as far as possible

1) $$x^3 + y^3$$
2) $$x^4 + x^3 - 3x^2 - 4x - 4$$

2. Relevant equations

3. The attempt at a solution

1) Not quite sure really what to do, lol, only just been taught how to divide polynominals, and the factor and remainder theorem. Havn't been taught how to factorise this case yet. I'm not sure how I could work out the answer myself. Ive tried a few ways but they dont lead anywhere :S.

Any help?

2) I was thinking of rearranging it into $$x^4 - 3x^2 - 4 + x^3 - 4x$$. This way I could maybe solve the first half but substituting $$x^2 = y$$ in so get rid of the 4th degree, and then solve the bit on the right after? or would that just plain not work. Other than that I have no ideas :S.

Any help?

thanks
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 10, 2007

HallsofIvy

Staff Emeritus
You first learned to factor polynomials in order to solve equations, didn't you? If x-a is a factor of P(x) then p(a)= 0 It works the other way too. That's the "factor and remainder theorem you mention. For example, it is obvious (I hope!) that x= -y satisfies $x^3+ y^3= 0$ because $(-y)^3+ y^3= -y^3+ y^3= 0$. That tells you that x-(-y)= x+y is a factor of $x^3+ y^3$. Dividing $x^3+ y^3$ by x+ y should give you a remainder of 0 and a quotient which is a quadratic that you can factor (if necessary by using the quadratic formula to solve the equation).

As far as $x^4+ x^3- 3x^2- 4x- 4$ is concerned, unfortunately solving half an equation doesn't help- unless both parts happen to have the same roots. One thing you can try is the "rational root theorem": if a polynomial (set equal to 0) has a rational root: m/n, then n, the denominator must be a divisor of the leading coefficient and m, the numerator, must be a divisor of the constant term. Here, the leading coefficient is 1 and the constant term is -4. That means any rational root must be a factor of 4: the possibilitiees are 1, -1, 2, -2, 4, and -4. I would reccomend that you try each of those. If any of them make the polynomial equal to 0, then x minus that number is a factor. If none of them do, there are no rational roots and you aren't going to be able to factor in any reasonable way! (I've checked a couple of those- they work!)

3. Oct 10, 2007

Trail_Builder

thanks for the help!

for 1) I ended up with (using quadratic formula) $$x^3 + y^3 = (x+y)(x-\frac{y+\sqrt-3y^2}{2})(x-\frac{y-\sqrt-3y^2}{2})$$ seems a lil messy, is that right?

and i got 2) right :D thanks man

4. Oct 11, 2007

HallsofIvy

Staff Emeritus
The original problem "factorize as far as possible" is a little ambiguous. Typically, factorization problems expect integer coefficients. What I had in mind was $(x+y)(x^2+ xy+ y^2)$ but, of course, you can solve the quadratic and get down to completely linear terms as you did.

5. Oct 11, 2007

Feldoh

I think you could also solve number two by $$(x-a)(x-b)(x-c)(x-d)=x^4 - 3x^2 - 4 + x^3 - 4x$$ and Des Cartes Rule of signs. Look at the coefficients and you should be able to find the roots...

Last edited: Oct 11, 2007