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    In C1, you get inequalties with very big numbers, I know you can times the first and last number to then to factorise, but it takes to long to think of numbers that times together to sub back in, so is there another way?

    example
    9k^2 +4k -28
    or 4k^2 -9k-9 (not has hard but stil
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    I have a pretty new method despite being in a2. nonetheless I firstly times 9 by 28 to get -252. No we know two factors of -252 are 9 and 28 but we can find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    18 and -14 are the forget we are looking for. As we can add to 4


    Pm. If you need further help.
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    (Original post by Guls)
    I have a pretty new method despite being in a2. nonetheless I firstly times 9 by 28 to get -252. No we know two factors of -252 are 9 and 28 but we can find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    18 and -14 are the forget we are looking for. As we can add to 4


    Pm. If you need further help.
    find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    I understand that when spliting, you just halfed the original numbers, however I don't get in the end where you got 18 from?
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    (Original post by am99)
    find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    I understand that when spliting, you just halfed the original numbers, however I don't get in the end where you got 18 from?
    Take the two numbers and write them in their prime factorisation:

    \displaystyle 

\begin{equation*}9 \times 28 =2^2 \times 3^2  \times 7 = (2 \times 7) \times (2\times 3^2) = 14 \times (2 \times 9) = 14 \times 18 \end{equation*}

    (with the negative signs, of course, I've ommitted them here.)
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    (Original post by am99)
    find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    I understand that when spliting, you just halfed the original numbers, however I don't get in the end where you got 18 from?
    You know 9 and 28 are factors, so you double the 9, and half the 28 to get 18 and 14
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    (Original post by am99)
    find more factors by splitting 9 and 28. Eg 3(3)x(14)2=252 or 3(3)x(-7)((2)(2) now seeing as that were multiplying everything we can mix and match to get (18)x(-14)=-252.

    I understand that when spliting, you just halfed the original numbers, however I don't get in the end where you got 18 from?
    To reiterate Zacken. Try writing all the primes: 2,2,3,3,7. Now to get 18 it's simply 3(3)(2).
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    (Original post by Zacken)
    Take the two numbers and write them in their prime factorisation:

    \displaystyle 

\begin{equation*}9 \times 28 =2^2 \times 3^2  \times 7 = (2 \times 7) \times (2\times 3^2) = 14 \times (2 \times 9) = 14 \times 18 \end{equation*}

    (with the negative signs, of course, I've ommitted them here.)
    Okay, thank you !, that was very helpful
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    (Original post by KaylaB)
    You know 9 and 28 are factors, so you double the 9, and half the 28 to get 18 and 14
    Thank you, very simple explnation !
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    (Original post by am99)
    Thank you, very simple explnation !
    No problem :hat2:
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    (Original post by Guls)
    To reiterate Zacken. Try writing all the primes: 2,2,3,3,7. Now to get 18 it's simply 3(3)(2).
    I understand now, Thank you for your help, good luck with your exams !
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    You could always 'cheat' and use quadratic formula to find out one of the factors. And then for a quadratic it's very simple to find the other factor.
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    (Original post by am99)
    In C1, you get inequalties with very big numbers, I know you can times the first and last number to then to factorise, but it takes to long to think of numbers that times together to sub back in, so is there another way?

    example
    9k^2 +4k -28
    or 4k^2 -9k-9 (not has hard but stil
    have you learnt completing the square yet in class?
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    You can always use the quadratic forumula. If the question is to factorise, find the roots then the factor will be (x-α)
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    (Original post by am99)
    I understand now, Thank you for your help, good luck with your exams !
    Thanks and good luck
 
 
 
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