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Math Help - factoring x^4+5x^2+4

  1. #1
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    factoring x^4+5x^2+4

    Hi, I needed to factor x^4+5x^2+4. I understand the answer is (x-2)(x-1)(x+1)(x+2)

    Can anyone tell me how to go about answering this though? How do I figure it out?
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  2. #2
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    let w^2=x^4
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  3. #3
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    Are you sure it is not x^4 - 5x^2 + 4?

    Substitute y = x^2 to get y^2 - 5y + 4.

    First factor this, then resubstitue in x^2 and continue factoring.
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  4. #4
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    In other words, let \displaystyle w = x^2.
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    If all else fails, search for a root and then use synthetic division.
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    Quote Originally Posted by Random Variable View Post
    If all else fails, search for a root and then use synthetic division.
    No need, as a quadratic it factorises nicely as \displaystyle 4 \cdot 1 = 4 and \displaystyle 4 + 1 = 5.

    Also, there aren't any real roots anyway, so it would be pretty hard to find one and then divide...
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    Or complete the square:

    x^4-5x^2+4 = \left(x^2+\frac{5}{2}\right)^2-\frac{9}{2}.

    Of course you know that y^2-t = (y-\sqrt{t})(y+\sqrt{t}).
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    Quote Originally Posted by Prove It View Post
    No need, as a quadratic it factorises nicely as \displaystyle 4 \cdot 1 = 4 and \displaystyle 4 + 1 = 5.

    Also, there aren't any real roots anyway, so it would be pretty hard to find one and then divide...
    Usually the functions aren't this nice.

    In any case  x=i is clearly a root. And since the function has real coefficients, you know that  x=-i is also a root. Then a factor is  x^{2}+1 . Now you can use synthetic division to find the other two.

    But he might have meant  x^{4}-5x^{2}+4 .
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    Quote Originally Posted by frankinaround View Post
    Hi, I needed to factor x^4+5x^2+4. I understand the answer is (x-2)(x-1)(x+1)(x+2)

    Can anyone tell me how to go about answering this though? How do I figure it out?
    (x-2)(x-1)(x+1)(x+2) is zero for x=2, 1, -1, -2.

    x^4+5x^2+4 is not zero for any of these values of x, since the powers of x are all positive,
    hence it is a different expression.

    On the other hand... by factoring over 2 stages, the 2nd stage being "difference of squares"

    x^4-5x^2+4=\left(x^2-4\right)\left(x^2-1\right)=(x+2)(x-2)(x+1)(x-1)
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  10. #10
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    thanks everyone. That substitution is great, but in the future how can I know to use that kind of substitution? meaning what if it was a little bit different right? then how would you solve it? or how did you know to substitute y=x^2 ? The quadratic equation didnt work for x^4 thats why im asking. So does that mean you just need to find any way to make it ax^2+bx+c?
    Last edited by frankinaround; January 10th 2011 at 07:57 PM.
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  11. #11
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    Quote Originally Posted by frankinaround View Post
    thanks everyone. That substitution is great, but in the future how can I know to use that kind of substitution? meaning what if it was a little bit different right? then how would you solve it? or how did you know to substitute y=x^2 ? The quadratic equation didnt work for x^4 thats why im asking. So does that mean you just need to find any way to make it ax^2+bx+c?
    Experience teaches you to recognise the underlying quadratic form.
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