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Math Help - How to factorize 4z^2 -4z+17 Over C?

  1. #1
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    How to factorize 4z^2 -4z+17 Over C?

    By over C I mean it has a solution that is a complex number

    My Attempt:

    4z^2-4z+17=(4z^2-4z+4)-4+17
    ................=(2z-2)^2 +13
    ................=(2z-2)^2 - (Squareroot of (13)*i)^2
    ................=((2z+2)+(Squareroot of 13 *i))((2z+2)-(Squareroot of 13 *i))

    However, the real answer is (2z-1+4i)(2z-1-4i)

    I can factorise questions such as z^2-4z+17 but when there is a number infront of z^2 I do not know how to do it.

    Thank you for any help .
    Last edited by anees7112; January 29th 2011 at 01:50 AM. Reason: Wrong Signs
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If z_1,z_2 are the solutions of :

    4z^2-4z+17=0

    then,

    4z^2-4z+17=4(z-z_1)(z-z_2)


    Fernando Revilla
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  3. #3
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    Quote Originally Posted by anees7112 View Post
    By over C I mean it has a solution that is a complex number

    My Attempt:

    4z^2-4z+17=(4z^2-4z+4)-4+17
    ................=(2z+2)^2 +13
    ................=(2z+2)^2 - (Squareroot of (13)*i)^2
    ................=((2z+2)+(Squareroot of 13 *i))((2z+2)-(Squareroot of 13 *i))

    However, the real answer is (2z-1+4i)(2z-1-4i)

    I can factorise questions such as z^2-4z+17 but when there is a number infront of z^2 I do not know how to do it.

    Thank you for any help .
    Alternatively

    4z^2-4z+17=4z^2-4z+1+16=(2z-1)(2z-1)+16=(2z-1)^2-(4i)^2

    which is the difference of two squares
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  4. #4
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    Quadratic Formula works nicely here

    \displaystyle z = \frac{4 \pm \sqrt{(-4)^2 - 4\cdot 4\cdot 17}}{2\cdot 4}.

    Simplify
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  5. #5
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    Quote Originally Posted by Archie Meade View Post
    Alternatively

    4z^2-4z+17=4z^2-4z+1+16=(2z-1)(2z-1)+16=(2z-1)^2-(4i)^2

    which is the difference of two squares
    Hello.
    I am just wondering how you got 4z^2-4z+1+16. My teacher told me to take the -4 from -4z , half it, and then square it. (-4)/2=-2 2^2=4.
    Thank you.
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  6. #6
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    Quote Originally Posted by anees7112 View Post
    Hello.
    I am just wondering how you got 4z^2-4z+1+16. My teacher told me to take the -4 from -4z , half it, and then square it. (-4)/2=-2 2^2=4.
    Thank you.
    That method applies well when the "squared term" has a coefficient of 1.

    For example

    z^2-4z+17=z^2-4z+4+13=(z-2)^2+13

    However, because of the 4z^2 term

    (2z-1)(2z-1)=2z(2z-1)-1(2z-1)=4z^2-2z-2z+1=4z^2-4z+1

    so I took 1 from 17 instead.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    That method applies well when the "squared term"

    so I took 1 from 17 instead.
    Alright thanks for that I think I understand now.

    But one more thing. How did you chose the numbers 1 and 17 exactly. I know they work but how did you pick these 2.

    I tried your method with 9z-24z+32
    and came up with 9z-24z+16 +16
    But it took me a while to come up with 16 and 16.

    Is there a formula like the (x*0.5)^2 for the case where the coefficient of z^2 is 1, to work out 1 and 17 or 16 and 16.

    Thanks
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  8. #8
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    Quote Originally Posted by anees7112 View Post
    Alright thanks for that I think I understand now.

    But one more thing. How did you chose the numbers 1 and 17 exactly. I know they work but how did you pick these 2.

    I tried your method with 9z^2-24z+32
    and came up with 9z^2-24z+16 +16
    But it took me a while to come up with 16 and 16.

    Is there a formula like the (x*0.5)^2 for the case where the coefficient of z^2 is 1, to work out 1 and 17 or 16 and 16.

    Thanks
    Ok,
    I will lead you through both of those.

    (1)

    4z^2-4z+17

    will factor with (4z+?)(z+?)\;\;\;or\;\;\;(2z+?)(2z+?)

    I must end up with -4z as my centre term after multiplying out the factors,
    so that rules out the first pair if I'm looking for integers.
    Hence

    (2z+?)(2z+?) must give me 4z^2-4z

    which is 4z^2-2z-2z so now I see what my "?" must be. I need -1 there.

    Then I just break up 17 into 16+1, since (-1)^2=1.

    Another way to look at would have been to figure that 16 is the nearest square to 17
    and so is a good place to start.


    (2)

    9z^2-24z+32

    The most obvious place to start is

    (3z)^2=9z^2

    and

    -24z=-12z-12z

    3(4)=12\Rightarrow\ (3z-4)(3z-4)=3z(3z-4)-4(3z-4)=9z^2-12z-12z+16

    Hence we take 16 from 32 to get

    9z^2-24z+32=9z^2-24z+16+16=(3z-4)^2-(4i)^2=(3z-4+4i)(3z-4-4i)
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