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Thread: Factorising Quadratic equations

  1. #1
    Junior Member froodles01's Avatar
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    Post Factorising Quadratic equations

    Sooo annoying that I still have trouble with this when I should have moved on.
    Anyway.
    Having trouble factorising;
    3r^2 - 8r -3=0

    2r^2 + r - 6=0

    3r^2 - 5r - 2 = 0

    I continually find it difficult to find the right solutions for r. (or use x if you'd prefer)

    Please help
    Thank you


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    Re: Factorising Quadratic equations

    Quote Originally Posted by froodles01 View Post
    Sooo annoying that I still have trouble with this when I should have moved on.
    Anyway.
    Having trouble factorising;
    3r^2 - 8r -3=0

    2r^2 + r - 6=0

    3r^2 - 5r - 2 = 0

    I continually find it difficult to find the right solutions for r. (or use x if you'd prefer)

    Please help
    Thank you


    After you have looked for common factors (there aren't any in any of them), multiply the a and c values. You will then look for two numbers that multiply to give this product, and add to give the value of b. Then split the b value up into these values, and factorise by grouping. I'll do the first one as an example:

    $\displaystyle \displaystyle \begin{align*} 3r^2 - 8r - 3 &= 0\textrm{ so we are looking for two numbers that multiply to give }3 \cdot (-3) = -9\textrm{ and add to give }-8 \\ 3r^2 + r - 9r - 3 &= 0 \textrm{ since the numbers we are looking for are }1 \textrm{ and } -9 \\ r(3r + 1) - 3(3r + 1) &= 0 \\ (3r + 1)(r - 3) &= 0 \end{align*}$

    And now solve for $\displaystyle \displaystyle \begin{align*}r \end{align*}$ using the Null Factor Law.
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    MHF Contributor Reckoner's Avatar
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    Re: Factorising Quadratic equations

    Quote Originally Posted by froodles01 View Post
    3r^2 - 8r -3=0
    2r^2 + r - 6=0
    3r^2 - 5r - 2 = 0
    Trinomials are slightly more difficult to factor when the leading coefficient is not 1. There are a couple of ways to go about doing this. Either way, you should start by factoring out the greatest common factor if you can (in these problems, the GCF is 1, so we don't have to worry about that).

    One method is to use trial and error. Take $\displaystyle 3r^2-8r-3$. The only way we can factor 3 (over the integers) is $\displaystyle 3\cdot1$. So we look for a factorization of the form $\displaystyle (3r+m)(r+n)$. Now our goal is find $\displaystyle m, n$ such that $\displaystyle mn = -3$ and $\displaystyle m+3n=-8$. From trying a few values for $\displaystyle m$ and $\displaystyle n$, we can see that $\displaystyle m=1$ and $\displaystyle n=-3$ meets this goal. So

    $\displaystyle 3r^2-8r-3=(3r+1)(r-3)$


    There's another method that uses grouping. First multiply the leading coefficient (which we may call $\displaystyle a$) by the constant term (which we may call $\displaystyle c$): $\displaystyle ac=3\cdot(-3)=-9$. Now look for two numbers whose product is -9 and whose sum is -8. Two such numbers are 1 and -9. Now we split up the coefficient of the linear term into these two numbers, and then we factor by grouping:

    $\displaystyle 3r^2-8r-3$

    $\displaystyle =3r^2+r-9r-3$

    $\displaystyle =(3r^2+r)+(-9r-3)$

    $\displaystyle =r(3r+1)-3(3r+1)$

    $\displaystyle =(3r+1)(r-3)$.


    Another example: Factor $\displaystyle 15x^2 + 11x + 2$.

    $\displaystyle ac = 15\cdot2=30$

    Two numbers whose product is 30 and whose sum is 11: 5 and 6

    $\displaystyle 15x^2+11x+2$

    $\displaystyle =15x^2+5x+6x+2$

    $\displaystyle =5x(3x+1)+2(3x+1)$

    $\displaystyle =(3x+1)(5x+2)$.
    Thanks from froodles01
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