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Math Help - Factoring expressions raised to powers

  1. #1
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    Factoring expressions raised to powers

    The book I am using gave factoring expressions raised to powers a very brief run through and a few examples, but the authors examples dont make any sense.

    Could some one please provide detailed, clear instructions to walk me through this problem and explain whats going on?

    expression
    2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2

    the book gives the answer as  [2(x^2-6)^7+(x^2-6)^2+4x^2-23](x^2-6)^2
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  2. #2
    Senior Member Twig's Avatar
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    Hi

    <br />
2(x^2-6)^9+(x^2-6)^4+4(x^2-6)^3+(x^2-6)^2<br />

    Note that the common factor (x^{2}-6) appears in all the terms. This means you can factor this whole parenthesis out, which is what is done in the answer.

    You know that  (x^{2}-6)^{9} = (x^{2}-6)^{7} \cdot (x^{2}-6)^{2}

    Now you get:  (x^{2}-6)^{2}(1 + 4(x^{2}-6) + ...)

     1 + 4(x^{2}-6) = 4x^{2}-23

    So, the answer just leavs the other terms as they are and we get the same answer as you wrote.

    If you want me to explain better let me know!
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  3. #3
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    I am still confused unfortunately .
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  4. #4
    Senior Member Twig's Avatar
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    hi

    Ok. What part is most confusing?

    You do know that  x^{a} \cdot x^{b} = x^{a+b} ?

    Example)

     a + a^2 + a^3 + a^4 = a(1+a+a^2+a^3)

    Now think of our whole paraenthesis (x^{2}-6) as the 'a' from the example above.

    They do the exact same thing in your question.

    \boxed{(x^{2}-6)^{2}}[1+4\boxed{(x^{2}-6)} + \boxed{(x^{2}-6)^{2}} + 2\boxed{(x^{2}-6)^{7}}]

    Do you see it? The boxes act as our 'a' from the above example.
    YouŽll have to focus to read it here, the parenthesis show kinda bad.
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