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Thread: How to factorise expression without sollution?

  1. #1
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    Question How to factorise expression without sollution?

    I've got an expression: $\displaystyle 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 $\displaystyle a(x^2+bx+c)(x^2+dx+e)$ How can I do that? (only real numbers, please)
    Last edited by lampak; Oct 11th 2009 at 12:05 PM. Reason: typos
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  2. #2
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    Are you sure that there are two variables in your expression ($\displaystyle x$ and $\displaystyle s$)? I think they want $\displaystyle \left(2+x^2\right) \left(3+x^2\right)$...
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  3. #3
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    Right, that's a typos.
    $\displaystyle x^4+5x^2+6$
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  4. #4
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    Try look under Application to higher-degree equations.
    Quadratic equation - Wikipedia, the free encyclopedia
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  5. #5
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    Smile

    hello
    put $\displaystyle x^{2}=u$,and factorize.
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  6. #6
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    The problem is that in this way I get $\displaystyle x^2<0$ and I'd rather not mess up with complex numbers.

    edit: To be specific: $\displaystyle x^2=-2 \lor x^2=-3$ if I've got it right.
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  7. #7
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    Re: How to factorise expression without sollution?

    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
    Last edited by CaptainBlack; Oct 11th 2009 at 01:22 PM.
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  8. #8
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    well, $\displaystyle x^4+5x^2+6$ has only complex roots.
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  9. #9
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    Quote Originally Posted by Raoh View Post
    well, $\displaystyle x^4+5x^2+6$ has only complex roots.
    Yes, but as far as I know any polynomial can be written in form of multiplied $\displaystyle (x-a)$ and $\displaystyle (ax^2+bx+c)$ (in this case it would be only the latter). Or at least that's what I was taught.
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  10. #10
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    well in $\displaystyle \mathbb{R} $,the only factorization you can get is $\displaystyle (x^2+2) (x^2+3)$.
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  11. #11
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    in your $\displaystyle 4^{th}$ post,$\displaystyle (a,b,c) \in \mathbb{R}^{3}$.
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  12. #12
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    Quote Originally Posted by Raoh View Post
    well in $\displaystyle \mathbb{R} $,the only factorization you can get is $\displaystyle (x^2+2) (x^2+3)$.
    Ahh, what a fool I am! That's what I was looking for
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