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Math Help - Factorising

  1. #1
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    Smile Factorising

    Hi! I am having trouble factorising these:

    6xsquared - 11x - 35

    6xsquared - 37x + 45

    Can someone please show me worked examples to these. And sorry, I don't know how to use latex but I will learn it soon to make mathematical interpretations easier.

    Any help will be appreciated heaps!!!
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  2. #2
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    Here's a thread that demonstrates how to factor general quadratic equations: Tips on factoring, a not equal to 1

    Go through that and see what you can come up with. If you have any troubles, feel free to post again.
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  3. #3
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    just have to play with them a bit ...

    6x^2 - 11x - 35 = (3x + 5)(2x - 7)

    6x^2 - 37x + 45 = (3x - 5)(2x - 9)
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  4. #4
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    Quote Originally Posted by Joker37 View Post
    Hi! I am having trouble factorising these:

    6xsquared - 11x - 35

    6xsquared - 37x + 45

    Can someone please show me worked examples to these. And sorry, I don't know how to use latex but I will learn it soon to make mathematical interpretations easier.

    Any help will be appreciated heaps!!!
    6x^2-11x-35

    Here's what I do:

    6\times35=210

    Find 2 factors of 210 that add up to -11.

    210=2\cdot3\cdot5\cdot7

    How about -21 and 10?

    Replace -11x with -21x+10x

    6x^2-21x+10x-35

    Group the first two terms and the last two terms

    (6x^2-21x)+(10x-35)

    Factor out common factors in each group

    3x(2x-7)+5(2x-7)

    \boxed{(3x+5)(2x-7)}
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  5. #5
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    Quote Originally Posted by masters View Post



    \boxed{(3x+5)(2x-7)}
    Will it still be considered correct if you put (2x-7)(3x+5) instead of (3x+5)(2x-7)?
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  6. #6
    A riddle wrapped in an enigma
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    Quote Originally Posted by Joker37 View Post
    Will it still be considered correct if you put (2x-7)(3x+5) instead of (3x+5)(2x-7)?
    Absolutely. Multiplication is commutative.

    a \cdot b=b \cdot a
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