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Math Help - Factorising Quadratics

  1. #1
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    Factorising Quadratics

    Hi all, I started this subject today and I have many questions of a similar nature. It would be much appreciated if someone could give me a step-by-step tutorial on how to answer a question such as this:

    Factorize these quadratic expressions:
    3x^2 - 7x + 2

    or

    9x^2 + 30x + 25

    I've never used this forum before so I'm sorry if I did anything wrong.
    Thanks heaps!
    Sam
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  2. #2
    Senior Member topher0805's Avatar
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    The easiest way is to set the expressions equal to 0 then find the roots using the quadratic equation.

    Do you know the quadratic equation?
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  3. #3
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    I don't sorry
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  4. #4
    Senior Member topher0805's Avatar
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    Ok, your equations are:

    3x^2 - 7x + 2

    and:

    9x^2 + 30x + 25

    They are of the form:

    ax^2 + bx + c where a, b, and c are all constants.

    The quadratic equation is:

    x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}

    So, you plug in the numbers, and voila! You have the roots of the expression.
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  5. #5
    Super Member angel.white's Avatar
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    Quote Originally Posted by Ziandia View Post
    Hi all, I started this subject today and I have many questions of a similar nature. It would be much appreciated if someone could give me a step-by-step tutorial on how to answer a question such as this:

    Factorize these quadratic expressions:
    3x^2 - 7x + 2

    or

    9x^2 + 30x + 25

    I've never used this forum before so I'm sorry if I did anything wrong.
    Thanks heaps!
    Sam
    I learned this 2 semesters ago, so forgive me if I err (someone feel free to correct any misconceptions I have, I tried to look it up but didn't find anything)

    Look at the polynomial 3x^2 - 7x + 2

    The first coefficient is 3, and the last is 2. There is a property that if you can remove a factor, it will take the form of (x+p/q) or (x+q/p) where p is a factor of the first coefficient, and q is a factor of the second coefficient. So potential factors are (x+\frac 12), (x+\frac 11), (x+\frac {-1}2), (x+\frac {-1}2), (x+\frac 32), (x+\frac 31), (x+\frac {-3}2), (x+\frac {-3}1), (x+\frac 13), (x+\frac {-1}3), (x+\frac 23), (x+\frac {-2}3), (x+\frac 21), (x+\frac {-2}1)

    In this case, I found that (x+\frac {-2}1) = (x-2) works.

    To pull it out, I used division:

    Perform the division:

    \begin{array}{rr}<br />
&3x-1\\<br />
(x-2) & \overline{\begin{array}{|r}3x^2-7x+2\end{array}}<br />
\end{array}

    So 3x^2 - 7x + 2 can be factored into (x-2)(3x-1)

    ----------

    The second one is pretty easy, because it's obvious. The first and last coefficients are perfect squares, so you could look at it like this:
    9x^2 + 30x + 25 = (3x)^2+30x+5^2

    From here it should be rather clear you are looking at a perfect square, if it is not, try playing around with perfect squares a bit so that you can learn to recognize them.
    (3x)^2+30x+5^2 = (3x+5)^2
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