# factorising x^2 - x + 4

• Dec 17th 2011, 12:02 AM
bermudj
factorising x^2 - x + 4
Hi,

I have been trying to factorise the above, but simply cannot find a solution.

Thanks for any help.
• Dec 17th 2011, 12:07 AM
FernandoRevilla
Re: factorising x^2 - x + 4
Quote:

Originally Posted by bermudj
I have been trying to factorise the above, but simply cannot find a solution.

The roots $z_1,z_2$ of $p(x)=x^2-x+4$ are not real so, $p(x)$ is irreducible on $\mathbb{R}$ . On $\mathbb{C}$ its factorization is $p(x)=(x-z_1)(x-z_2)$ .
• Dec 17th 2011, 04:23 AM
Prove It
Re: factorising x^2 - x + 4
Quote:

Originally Posted by bermudj
Hi,

I have been trying to factorise the above, but simply cannot find a solution.

Thanks for any help.

\displaystyle \begin{align*} x^2 - x + 4 &= x^2 - x + \left(-\frac{1}{2}\right)^2 - \left(-\frac{1}{2}\right)^2 + 4 \\ &= \left(x - \frac{1}{2}\right)^2 - \frac{1}{4} + \frac{16}{4} \\ &= \left(x - \frac{1}{2}\right)^2 + \frac{15}{4} \\ &= \left(x - \frac{1}{2}\right)^2 - \left(\frac{i\sqrt{15}}{2}\right)^2 \\ &= \left(x - \frac{1}{2} - \frac{i\sqrt{15}}{2}\right)\left(x - \frac{1}{2} + \frac{i\sqrt{15}}{2}\right) \end{align*}