How can I factorize these polynomials?

[Trail_Builder]

Homework Statement

factorise the following as far as possible

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

Homework Equations

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. I've tried a few ways but they don't 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

 

[HallsofIvy]

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 recommend 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!)

 

[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

 

[HallsofIvy]

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.

 

[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...