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Math Help - Factoring polynomials with exponents

  1. #1
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    Factoring polynomials with exponents

    I'm currently stuck on 2 problems.
    4(3-x)^2-(3-x)^3+3(3-x) I got 3-x(6(3-x)(3-x)^2)
    That doesn't seem right but what went wrong.

    also

    15(2z+1)^3+10(2z+1)^2-25(2z+1) => 5(2z+1)(3(2z+1)^2+2(2z+1)-5) => 5(2z+1)((2z+1)^2(2z+1))
    which also doesn't seem right.
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  2. #2
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    Quote Originally Posted by m0nkeysensei View Post
    I'm currently stuck on 2 problems.
    4(3-x)^2-(3-x)^3+3(3-x) I got 3-x(6(3-x)(3-x)^2)
    That doesn't seem right but what went wrong.

    also

    15(2z+1)^3+10(2z+1)^2-25(2z+1) => 5(2z+1)(3(2z+1)^2+2(2z+1)-5) => 5(2z+1)((2z+1)^2(2z+1))
    which also doesn't seem right.
    check your steps ,you are doing mistake somewhere

    check spoiler if stuck
    Spoiler:

    4 { \color{blue}(3-x)^2 }- { \color{blue} (3-x)^3}+3 { \color{blue} (3-x)}={ \color{blue} (3-x)} \{ 4{ \color{blue} (3-x) }- { \color{blue} (3-x)^2 }+3 \}
    solving
    = (3-x) \{ 4(3-x)- (9+x^2-6x)+3 \}
    = (3-x) ( 12-4x- 9-x^2+6x+3)
    =(3-x) (-x^2+2x+6)
     = - (3-x) (x-1+ \sqrt {7})(x-1-\sqrt {7}) =(x-3)(x-1+ \sqrt {7})(x-1-\sqrt {7})
    Last edited by ramiee2010; October 25th 2009 at 10:41 AM. Reason: typo correction
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  3. #3
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    Quote Originally Posted by ramiee2010 View Post

    Spoiler:

    4 { \color{blue}(3-x)^2 }- { \color{blue} (3-x)^3}+3 { \color{blue} (3-x)}={ \color{blue} (3-x)} \{ 4{ \color{blue} (3-x) }- { \color{blue} (3-x)^2 }+{ \color{blue} (3-x)} \}
    solving
    = (3-x) \{ 4(3-x)- (9+x^2-6x)+(3-x) \}
    = (3-x) ( 12-4x- 9-x^2+6x)+3-x )
    =(3-x) (-x^2+2x+6)
     = - (3-x) (x-1+ \sqrt {7})(x+1+\sqrt {7}) =(x-3)(x-1+ \sqrt {7})(x+1+\sqrt {7})
    but how does
    if you are taking out a 3-x from all terms, wouldn't that cancel out the last 3-x?
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  4. #4
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    Quote Originally Posted by m0nkeysensei View Post
    but how does
    if you are taking out a 3-x from all terms, wouldn't that cancel out the last 3-x?
    that was mistyped by me. i have been corrected that
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