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I was doing some integration by partial fractions exercises and I found this equation:$$\int\nolimits_{0}^{1}\frac{x^{3}+1}{x^{4}+4x+3},$$ and I don't know how to factor that in order to compute the partial fractions method, I've done some research and I got this as the answer:$$(x+1)^{2}(x^2-2x+3)$$ but I don't know what is the procedure to get there. Thanks for the answers, and sorry my English is not my first language.

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First, $x^3 + 1$ is a sum of cubes, which has nice factorization. Second, $(-1)^4 + 4(-1) + 3 = 0$, and so $-1$ is a root of the polynomial in the denominator. Then, just divide $x^4 + 4x + 3$ by $x - (-1) = x+1$. – JavaMan Oct 24 '11 at 2:03
This particular example can be easily done by the change of variables $u=x^4+4\,x+3$, since the numerator is, up to a factor, the derivative of the denominator. Always be alert for possible alternative (simpler) methods of solution. – Julián Aguirre Oct 24 '11 at 11:10
up vote 3 down vote accepted

First, try to find roots for the denominator. The Rational Root Test can be useful. Here, it will quickly give you that $-1$ is a root.

Once you find a root $a$, you can always factor $(x-a)$ from the polynomial. Then you can continue working with what's left.

Here, once you discover that $a=-1$ is a root, you take $x^4+4x+3$ and divide by $x-(-1) = x+1$. We get $$x^4+4x+3 = (x+1)(x^3-x^2+x+3).$$ Then we work with $x^3-x^2+x+3$. Again, $a=-1$ should be easy to find, and dividing we get $$x^3 - x^2 + x + 3 = (x+1)(x^2 -2x +3).$$

At this point, you want to work with $x^2-2x+3$. This is a quadratic, so you can apply the quadratic formula to recognize it is irreducible. Putting it all together, you get the factorization $$x^4+4x+3 = (x+1)(x^3-x^2+x+3) = (x+1)(x+1)(x^2-2x+3) = (x+1)^2(x^2-2x+3).$$

It's possible that you can find no real roots of the denominator (if it is a product of irreducible quadratics). You may want to look for complex roots then.

In general, finding exact roots of a polynomial may be difficult. In practice, you usually use something like Newton's Method to approximate them.

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