Thread: How to factorise expression without sollution?

1. How to factorise expression without sollution?

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)

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

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

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

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

6. 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.

7. Re: How to factorise expression without sollution?

take x^2 = a

a^2 + 5a + 6 = 0

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

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

9. 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.

10. well in $\mathbb{R}$,the only factorization you can get is $(x^2+2) (x^2+3)$.

11. in your $4^{th}$ post, $(a,b,c) \in \mathbb{R}^{3}$.

12. Originally Posted by Raoh
well in $\mathbb{R}$,the only factorization you can get is $(x^2+2) (x^2+3)$.
Ahh, what a fool I am! That's what I was looking for