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Math Help - Factoring completely.

  1. #1
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    Question Factoring completely.

    Here is my question. It's from a pre-calc class, but feels more appropriate here in the algebra section.

    Factor the expression completely.
    5 * (x^2+4)^4 * (2x) * (x-2)^4 + (x^2+4)^5 * 4 * (x-2)^3

    The answer in the back of the book is:
    2 * (x^2+4)^4 * (x-2)^3 * (7x^2-10x+8)

    I'm aware of special factoring forms. I could do all the math for (x+y)^4 etc, etc, but it gets really long . I was wondering if there was a more elegant solution. After scouring my book for any clues with no luck I feel really lost! Any help would be greatly appreciated!
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  2. #2
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    Re: Factoring completely.

    You have 2 terms here that I can see.

    From each you can divide out (x^2+4) and (x-2)
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  3. #3
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    Re: Factoring completely.

    I'm still lost. How does dividing out (x^2+4) and (x-2) help? A step by step solution would really help me understand what's going on here. Thanks for your time and effort!
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  4. #4
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    Re: Factoring completely.

    You won't get a step-by-step solution here. As you have been advised, start by taking out \displaystyle 2(x^2 + 4)^4(x - 2)^3 asyour highest common factor. See what you can do with it. If you get stuck again, post what you have done and we will help you from there. You are expected to show some effort.
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  5. #5
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    Re: Factoring completely.

    Ok! So I think I got it...or at least I'm on the right track. Here's what I did:

    Factor the expression completely.
    5 * (x^2+4)^4 * (2x) * (x-2)^4 + (x^2+4)^5 * 4 * (x-2)^3

    1. Find the smallest exponent I can factor out from both sides: (x^2+4)^4 and (x-2)^3 and 2 <--(Getting the 2 as the GCF of 10 and 4.)

    2. Divide each term by 2 * (x^2+4)^4 * (x-2)^3 getting: 5x(x-2) + 2(x^2+4) (distributed out becomes)--> 7x^2-10x+8

    3. Add the term I used to simplify the factor back in, getting a final answer of: 2 * (x^2+4)^4 * (x-2)^3 * (7x^2-10x+8)

    I'm a bit confused on step 2/3. I saw that it worked...but am I really factoring out the GCF instead of dividing like I did? (Because I can't change an expression like I would an equation, having only one side.)

    Thanks again for all the help. It's midnight here and I think I'm in need of a good sleep .
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  6. #6
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    Re: Factoring completely.

    Quote Originally Posted by perryc View Post
    Ok! So I think I got it...or at least I'm on the right track. Here's what I did:

    Factor the expression completely.
    5 * (x^2+4)^4 * (2x) * (x-2)^4 + (x^2+4)^5 * 4 * (x-2)^3

    1. Find the smallest exponent I can factor out from both sides: (x^2+4)^4 and (x-2)^3 and 2 <--(Getting the 2 as the GCF of 10 and 4.)

    2. Divide each term by 2 * (x^2+4)^4 * (x-2)^3 getting: 5x(x-2) + 2(x^2+4) (distributed out becomes)--> 7x^2-10x+8

    3. Add the term I used to simplify the factor back in, getting a final answer of: 2 * (x^2+4)^4 * (x-2)^3 * (7x^2-10x+8)

    I'm a bit confused on step 2/3. I saw that it worked...but am I really factoring out the GCF instead of dividing like I did? (Because I can't change an expression like I would an equation, having only one side.)

    Thanks again for all the help. It's midnight here and I think I'm in need of a good sleep .
    You are basically correct, but factoring and dividing aren't the same thing. Factoring means to write an equivalent expression in terms of the product of its factors.
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