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Factoring quickly

  1. Jul 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Factor this quadratic expression completely.
    126x² - 15x - 66

    2. Relevant equations

    3. The attempt at a solution
    126x² - 15x - 66

    I'm a good ways past this subject, but I find that it always takes me longer than some to complete these problems. If the coefficients are low numbers, I can "guess and check" and get the right combination. When the coefficients get into the higher numbers I run into trouble.

    I always resort to the "box" method when the factors are not really apparent, but my professor says that I should be using "guess and check" as it's much more efficient.

    Here is how I would solve the problem..
    126x² - 15x - 66
    3(42x² -5x - 22) //factor out the GCF
    Now I take 42*-22, and think of factors of that which would make the -5.
    Ok, so 33-28=-5

    Now I use the "box method"

    42x² | 33x
    -28x | -66

    Factoring out the "box"

    14x 11
    42x² | 33x 3x
    -28x | -66 -2

    So now I know the sets of numbers that I can use, I just adjust the signs to work..
    3(14x-11)(3x+2) or (9x+6)(14x-11)

    I know that I don't have to factor out the GCF, I just did it that way to show my steps.

    I am told by everyone that the "guess and check" method is much faster, easier, and generally better to use. Am I guessing-and-checking wrong or something? At least this box method is systematic, I know I will arrive at the right answer, and not based on chance.

    Any other tips on how to do this would be greatly appreciated.
  2. jcsd
  3. Jul 24, 2010 #2
    First of all, all coefficients are divisible by 3:

    3 \left(42 x^{2} - 5 x - 22\right)

    Then, find the discriminant:
    D = (-5)^{2} - 4 \cdot 42 \cdot (-22) = 25 + 3696 = 3721 = 61^{2}
    so the solutions are:

    \frac{5 \pm 61}{84} = \left\{\begin{array}{l}\frac{11}{14} \\ -\frac{2}{3}\end{array}\right.

    Then, the factoring of the trinomial in the parenthesis is according to the formula:

    a (x - x_{1})(x - x_{2})

    where a is the coefficient before the quardratic term (here 42). Don't forget the factor of 3.
    Last edited: Jul 24, 2010
  4. Jul 24, 2010 #3
    Edit, I just saw you are still editing that post....
  5. Jul 24, 2010 #4
    Could you clarify this part? I'm not familiar with the "discriminant", other than the definition (which I just looked up).
  6. Jul 24, 2010 #5
    The quadratic formula for the quadratic equation [itex]a x^{2} + b x + c = 0[/itex] is:

    x_{1/2} = \frac{-b \pm \sqrt{D}}{2a}, \; D = b^{2} - 4 \, a \, c
  7. Jul 24, 2010 #6
    I see, there seems to be many ways to do this. I'll read up on that. Thank you.
  8. Jul 24, 2010 #7


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    If a method works for you, use it. I wouldn't worry about what other people say you should be using as long as you get the right answer. I would add, however, that you should at least understand how to use alternate methods. Don't just stick with one method because you don't want to figure out and understand other, possibly more efficient techniques.
  9. Jul 24, 2010 #8
    I've seen many high school textbooks "teaching" the so called "box method". The box method is nothing more than a guessing game of trying to solve the system of equations given by the Vieta's Formulas:

    x_{1} + x_{2} = -\frac{b}{a}

    x_{1} \, x_{2} = \frac{c}{a}

    In fact, you should be using these formulas the other way around. Having found the solutions to the quadratic equation [itex]a x^{2} + b x + c = 0[/tex] by the quadratic formula (which is an intellectual achievement of previous generations to solve an arbitrary quadratic equation), you express the coefficients b and c as:

    b = -a(x_{1} + x_{2})

    c = a x_{1} x_{2}

    and write the trinomial as:

    a \left[x^{2} - (x_{1} + x_{2}) x + x_{1} x_{2}\right]

    a \left( x^{2} - x_{1} x - x_{2} x + x_{1} x_{2}\right)

    a \left[ x(x - x_{1}) - x_{2}(x - x_{1}) \right]

    a (x - x_{1})(x - x_{2})

    Once you have the general formula, keep substituting numbers and it will always work, provided that the discriminant is nonnegative.
  10. Jul 24, 2010 #9
    Well, the so-called "box" method is the easiest way for me to come up with the correct numbers to use in the factors. Then it take 2 seconds of thinking to figure out the signs. Dickfore's method creates a great deal more work than either of the two methods I knew how to use :P (Still, thank you for the information).

    Suppose you have something like..

    If A is a relatively large number, and I cannot factor anything else out of the expression, I have a hard time finding the product of two numbers that would equal C, and add/multiply up to equal B, given that a factor of A is going to work on one (or both) of the numbers. Is there an easier way to look at it and determine how to factor A?

    It seems to me that if A has 6 ways of being factored, and then C has 6 ways of being factored, that is a possible 36 combinations (not including how the signs are arranged). My professor preaches "guess and check" for this situation, which I think is a bit unreasonable unless the coeficcient of x² is a really small number.
  11. Jul 24, 2010 #10
    One rule I heard of is to guess and check if the leading coefficient is prime (I think this should also work fine if c is prime), otherwise use the "box method" or the quadratic formula. You can still try guessing and checking if a has only a couple of different pairs of factors, but as the number of different pairs of factors increases, you should really consider using another method.
  12. Jul 25, 2010 #11

    By Approx(smart guess)

    *Ignore my last step:redface:

    It is better to estimate the distance between the roots: 5/3

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    Last edited: Jul 25, 2010
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