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Math Help - factorising expressions.

  1. #1
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    Post factorising expressions.

    Hi, I'm new to the forums and pretty much fail at maths.
    General (Basic) maths I can do, but algebra etc. is alien to me.
    I'm looking for some help in factorising the following expressions.

    cd + 6ce
    5x^4 - 3x^3 + 4x^2
    5a + ax -2b + by

    and finding the solution to this equation through factorisation:
    4s^2 -11s -3 =0

    I also have another problem regarding solving an equation using the Formula Method?
    I'll include the question in case it will help you, help me identify the formula.

    5t^2 +7t =9
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  2. #2
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    Quote Originally Posted by tubruk View Post
    Hi, I'm new to the forums and pretty much fail at maths.
    General (Basic) maths I can do, but algebra etc. is alien to me.
    I'm looking for some help in factorising the following expressions.

    cd + 6ce
    5x^4 - 3x^3 + 4x^2
    5a + ax -2b + by

    and finding the solution to this equation through factorisation:
    4s^2 -11s -3 =0

    I also have another problem regarding solving an equation using the Formula Method?
    I'll include the question in case it will help you, help me identify the formula.

    5t^2 +7t =9
    cd + 6ce

    Can you see that each term has been multiplied by c?

    That means that you can take c out as a common factor.

    So cd + 6ce = c(d + 6e).


    5x^4 - 3x^3 + 4x^2

    Can you see that each term has x^2 in it? (note that x^4 = x^2\cdot x^2 and x^3 = x\cdot x^2)

    So we can take x^2 out as a common factor.

    5x^4 - 3x^3 + 4x^2 = x^2(5x^2 - 3x + 4).


    5a + ax - 2b + by

    Can you see anything that goes into every term in this expression? I can't...

    The best you could hope for in this case is to factorise groups of two...

    5a + ax - 2b + by = a(5 + x) + b(-2 + y).

    We still don't get any common factors though...


    4s^2 -11s -3 =0

    This is a quadratic equation of the form ax^2 + bx + c = 0.

    We can factorise the quadratic by finding two numbers that multiply to be ac and add to be b.

    ac = -12 and b = -11.

    What numbers multiply to be -12 and add to be -11? Surely the two numbers must be -12 and 1.

    Watch what happens when I break up the middle term...

    4s^2 - 11s - 3 = 0

    4s^2 - 12s + 1s - 3 = 0.

    If we factorise by groups of two, we can find a common factor...

    4s(s - 3) + 1(s - 3) = 0

    Can you see that s - 3 is a common factor?

    (s - 3)(4s + 1) = 0.

    Now we have two "things" multiplying to become 0. This means at least one of those "things" had to be 0.

    So s - 3 = 0 or 4s + 1 = 0.

    Can you tell me what s has to be?
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  3. #3
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    Thank you for your Quick Reply and thorough Help.

    Yes, 's' has to be 3 in the case of 's -3 =0'
    But can't it also be '-0.25' if its '4s+1=0'?
    Last edited by tubruk; December 31st 2009 at 05:48 AM. Reason: Showing Thanks.
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  4. #4
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    Quote Originally Posted by tubruk View Post
    Thank you for your Quick Reply and thorough Help.

    Yes, 's' has to be 3 in the case of 's -3 =0'
    But can't it also be '-0.25' if its '4s+1=0'?
    Yes that's exactly right. There are two values that s could be to make the equation true.

    Check it by putting those values into the original equation and see if you get 0 each time...
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