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Thread: 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.

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

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

    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.

    $\displaystyle 5t^2 +7t =9$
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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.

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

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

    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.

    $\displaystyle 5t^2 +7t =9$
    $\displaystyle cd + 6ce$

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

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

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


    $\displaystyle 5x^4 - 3x^3 + 4x^2$

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

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

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


    $\displaystyle 5a + ax - 2b + by$

    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...

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

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


    $\displaystyle 4s^2 -11s -3 =0$

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

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

    $\displaystyle ac = -12$ and $\displaystyle b = -11$.

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

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

    $\displaystyle 4s^2 - 11s - 3 = 0$

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

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

    $\displaystyle 4s(s - 3) + 1(s - 3) = 0$

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

    $\displaystyle (s - 3)(4s + 1) = 0$.

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

    So $\displaystyle s - 3 = 0$ or $\displaystyle 4s + 1 = 0$.

    Can you tell me what $\displaystyle s$ 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; Dec 31st 2009 at 05:48 AM. Reason: Showing Thanks.
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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 $\displaystyle s$ 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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