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Math Help - Factorise Equation

  1. #1
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    Factorise Equation

     <br />
f(x) = 2(x-3)(x+2)^2 + 2(x+2)(x-3)^3<br />
    Factorise f(x) completely.

    Mind to tell me how I going to start to solve this ?
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  2. #2
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
     <br />
f(x) = 2(x-3)(x+2)^2 + 2(x+2)(x-3)^3<br />
    Factorise f(x) completely.

    Mind to tell me how I going to start to solve this ?
    Start by taking out the common factors 2, (x - 3) and (x + 2)...
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  3. #3
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    Re: Factorise Equation

    Quote Originally Posted by Prove It View Post
    Start by taking out the common factors 2, (x - 3) and (x + 2)...

    is it look like this ?


    f(x)=2(x-3)(x^2+4x+4) + 2(x+2)(x-3)(x^2-6x+9)

    =2((x^3+4x^2+4x)+(-3x^2-12x-12))+2((x^2-3x+2x-6)(x^2-6x+9))
    Last edited by gilagila; October 3rd 2011 at 01:25 AM.
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  4. #4
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    is it look like this ?


    f(x)=2(x-3)(x^2+4x+4) + 2(x+2)(x-3)(x^2-6x+9)

    =2((x^3+4x^2+4x)+(-3x^2-12x-12))+2((x^2-3x+2x-6)(x^2-6x+9))
    No.

    Can you factorise 2ab^2 + 2ba^3 ? If so, please do so, showing all steps.
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  5. #5
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    Re: Factorise Equation

    Quote Originally Posted by mr fantastic View Post
    No.

    Can you factorise 2ab^2 + 2ba^3 ? If so, please do so, showing all steps.
    Sorry sir, not understand what you mean, mind to explain ? TQ

    2ab(b+a^2) or just stop like that only ?
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    Sorry sir, not understand what you mean, mind to explain ? TQ

    2ab(b+a^2) or just stop like that only ?
    2ab(b+a^2) is correct.

    Now note that a = x - 3 and b = x + 2 in the question that you posted ....
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  7. #7
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    Re: Factorise Equation

    Quote Originally Posted by mr fantastic View Post
    2ab(b+a^2) is correct.

    Now note that a = x - 3 and b = x + 2 in the question that you posted ....
    ooh...ic

    meaning that

    f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

    =2ab^2 + 2ba^3 = 2ab(b+a^2)

    =2(x-3)(x+2)[(x+2)+(x-3)^3]

    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^3]

    or any other format ?
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  8. #8
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    ooh...ic

    meaning that

    f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

    =2ab^2 + 2ba^3 = 2ab(b+a^2)

    =2(x-3)(x+2)[(x+2)+(x-3)^3]

    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^3]

    or any other format ?
    No, you still have (x-3)^3 inside the right most bracket, it should be squared not cubed. When you have corrected that expand the right most bracket and verify that it can(not) be further factorised.

    CB
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  9. #9
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    Re: Factorise Equation

    Quote Originally Posted by CaptainBlack View Post
    No, you still have (x-3)^3 inside the right most bracket, it should be squared not cubed. When you have corrected that expand the right most bracket and verify that it can(not) be further factorised.

    CB
    mind to show me how to do it, not realy sure how to do
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  10. #10
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    mind to show me how to do it, not realy sure how to do
    Oh come on! You were shown exactly how to do the question. You made a careless mistake and that was also pointed out to you. Fixing the error is simple - you were told exactly what to do. Sorry, but you have been given more than enough help here to get the final answer.

    Fix the mistake, post your answer.
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  11. #11
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    Re: Factorise Equation

    f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

    =2ab^2 + 2ba^3 = 2ab(b+a^2)

    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    oops, sorry for the reply. Not noticed that I had changed the squared to cube, now I rectified.

    Is it correct now ?
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  12. #12
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    ...Is it correct now ?
    Yes, oui, si, ...
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  13. #13
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    f(x)=2(x-3)(x+2)^2 + 2(x+2)(x-3)^3

    =2ab^2 + 2ba^3 = 2ab(b+a^2)

    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    oops, sorry for the reply. Not noticed that I had changed the squared to cube, now I rectified.

    Is it correct now ?
    It's correct, but it's not complete. You now need to expand everything contained in the square brackets, collect like terms, and see if that factorises as well...
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  14. #14
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    Re: Factorise Equation

    Quote Originally Posted by Prove It View Post
    It's correct, but it's not complete. You now need to expand everything contained in the square brackets, collect like terms, and see if that factorises as well...
    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    = 2(x-3)(x+2)[(x+2)+(x^2-6x+9)]

    = 2(x-3)(x+2)(x^2-5x+11)

    Is that the final answer, or still need to factorise the (x^2-5x+11) ?

    Wondering how to factorise that, (x - p)(x-q)
    p+q=5, pq=11 ??
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  15. #15
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    Re: Factorise Equation

    Quote Originally Posted by gilagila View Post
    Factorise f(x) completely
    =2(x-3)(x+2)[(x+2)+(x-3)^2]

    = 2(x-3)(x+2)[(x+2)+(x^2-6x+9)]

    = 2(x-3)(x+2)(x^2-5x+11)

    Is that the final answer, or still need to factorise the (x^2-5x+11) ?

    Wondering how to factorise that, (x - p)(x-q)
    p+q=5, pq=11 ??
    If you check the discriminant (-5)^2 - 4(1)(11) = 25 - 44 = -19, since this is negative that means it does not factorise over the reals. So yes, that's the final answer.
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