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Math Help - factor radicals

  1. #1
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    factor radicals

    (a2+1)2-7(a2+1)+10

    ans = Its wrong I tried a4-1+7a2+16
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  2. #2
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    Re: factor radicals

    Quote Originally Posted by zbest1966 View Post
    (a2+1)2-7(a2+1)+10
    There are no radicals here. Did you make a mistake when typing the problem?

    Anyway, this is a quadratic in form. Let u=a^2+1. Then

    \left(a^2+1\right)^2-7\left(a^2+1\right)+10 = u^2-7u+10.

    Now factor that trinomial on the right, and then substitute for u to get everything back in terms of a.
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  3. #3
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    Re: factor radicals

    a^4 + 1-7a^2-7+10

    a^4-6-7a^2+10

    a^4-7a^2+4

    Im stuck
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  4. #4
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    Re: factor radicals

    Quote Originally Posted by zbest1966 View Post
    a^4 + 1-7a^2-7+10
    ...
    Im stuck
    Did you not read what I wrote? We have a quadratic in a^2+1. You don't need to multiply everything out. And if you were to multiply everything out, you're doing it wrong: \left(a^2+1\right)^2\neq a^4+1.

    I've shown you how to reduce the problem to factoring u^2-7u+10. Can you factor this trinomial?
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