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Math Help - find factors for (2x+1)^2 - (x-2)^2

  1. #1
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    find factors for (2x+1)^2 - (x-2)^2

    Hi Everyone

    im having trouble with the above equation and would appreciate any help.

    Thanks guys
    Pete
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    a^2-b^2=(a-b)(a+b)
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  3. #3
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by petemg View Post
    Hi Everyone

    im having trouble with the above equation and would appreciate any help.

    Thanks guys
    Pete
    As you have an equation in the form a^2-b^2 it makes sense to use the difference of two squares
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  4. #4
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    i get that i need to use difference of squares and have completed questions with like (x+4)^2 - 3^2 no problem, its the 2 sets of brackets confusing me
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  5. #5
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    a = 2x+1 \text{   and   } b = x-2

    Hence (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] \times [(2x+1)+(x-2)]

    You can simplify further in the brackets
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  6. #6
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    Thanks for that,

    would i be right with my next step,

    (-2x^2-x-2) (2x^2+x-2)
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    No. Read again what -1 wrote you... post #5
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  8. #8
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    where am i going wrong? do i not multiply 2x by -x and get -2x^2
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  9. #9
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by petemg View Post
    where am i going wrong? do i not multiply 2x by -x and get -2x^2
    I think that I know your problem... read hear to understand what factor means... Divisor - Wikipedia, the free encyclopedia
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  10. #10
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    Im still confused

    Do i multiply the left side by the right? and not 2 the two brackets to the left multiplied by each other?
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  11. #11
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by e^(i*pi) View Post
    a = 2x+1 \text{   and   } b = x-2

    Hence (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] \times [(2x+1)+(x-2)]

    You can simplify further in the brackets
    (2x+1)^2 - (x-2)^2 = [(2x+1)-(x-2)] *[(2x+1)+(x-2)]=(2x+1-x+2)*(2x+1+x-2)=(x+3)*(3x-1)

    So... (x+3) and (3x-1) are factors of (2x+1)^2 - (x-2)^2.

    It's like 3 and 2 are factors of 6(=2*3)
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  12. #12
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    would i be left with 6x^2+3x-1?
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  13. #13
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    sorry just seen you wrote the method there, how else could you get the same answer using different method?
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  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
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    Quote Originally Posted by petemg View Post
    sorry just seen you wrote the method there, how else could you get the same answer using different method?

    You can expand (2x+1)^2 - (x-2)^2 and then to find the roots using the famous formula x_1,2={-b+/-sqrt(b^2-4ac)}/2a
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  15. #15
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    and how would that look? its really the 2 brackets confusing me here, i can do the (b^2-4ac)}/2a with 1 bracket
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