How to factorise expression without sollution?

**lampak** · October 11th 2009, 11:59 AM

I've got an expression: $x^4+5x^2+6$ . How can I factorise it? All methods I know are based on finding one solution and simplifying the expression to (x-a)(...). But this one is always positive, no matter x, so it's not possible here. It's a polynomial of degree 4, so it must be possible to at least make it to the form $a(x^2+bx+c)(x^2+dx+e)$ How can I do that? (only real numbers, please)

**james_bond** · October 11th 2009, 12:02 PM

Are you sure that there are two variables in your expression ( $x$ and $s$ )? I think they want $\left(2+x^2\right) \left(3+x^2\right)$ ...

**lampak** · October 11th 2009, 12:05 PM

Right, that's a typos.
$x^4+5x^2+6$

**BabyMilo** · October 11th 2009, 12:22 PM

Try look under Application to higher-degree equations.
Quadratic equation - Wikipedia, the free encyclopedia

**Raoh** · October 11th 2009, 12:29 PM

hello

put $x^{2}=u$ ,and factorize.

**lampak** · October 11th 2009, 12:32 PM

The problem is that in this way I get $x^2<0$ and I'd rather not mess up with complex numbers.

edit: To be specific: $x^2=-2 \lor x^2=-3$ if I've got it right.

**jeremyparker10** · October 11th 2009, 12:33 PM

take x^2 = a

your equation will become

a^2 + 5a + 6 = 0

solve this simple quadratic eqn and then put the value of a = x^2 and solve further

**Raoh** · October 11th 2009, 12:38 PM

well, $x^4+5x^2+6$ has only complex roots.

**lampak** · October 11th 2009, 12:44 PM

Originally Posted by Raoh

well, $x^4+5x^2+6$ has only complex roots.

Yes, but as far as I know any polynomial can be written in form of multiplied $(x-a)$ and $(ax^2+bx+c)$ (in this case it would be only the latter). Or at least that's what I was taught.