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Factoring without trial and erorr?

  1. Jul 19, 2008 #1
    Is there a formula for factoring simple second order polynomials with zero trial and error? For example:


    I know the answer is (3X-2)(X-3) and that can be checked by foiling obviously. I know 3X and X have to be the first terms, and that the last two have to be multiples of 6. I'm just curious if there's a way to quickly put them together so that the middle term is correct, without checking the answer by foiling. Thanks.
  2. jcsd
  3. Jul 19, 2008 #2
    There are other methods for factoring quadratics....but sometimes trial and error is the fastest way.
    In your particular problem, what we can try to do is multiply the leading coefficient (3) by the constant term on the end (6) giving 18.

    Using this method, you say to yourself "I have no way of knowing which factors will work, so I'll just throw them all in together and then later, get rid of the extras....."
    I need factors of 18 that ADD to give me 11 (you have to understand what the plus and minus signs and their location means in the original equation)
    so we list them off:
    1 18
    2 9
    3 6 (obvious)

    I like 2 and 9...2+9=11

    since we included ALL factors, the method says that you write this (fully aware that you are NOT done:)

    Clearly, if you FOIL that, It doesn't work, but since we started out not knowing what factors would work, we have extra factors in there, so we need to get rid of the unwanted ones. The way this is done is to look at each binomial factor separately and see if there is something that can be divided out and thrown away:

    we have:
    (3x-2) and (3x-9)
    so we look at (3x-2). There is no common factor in the terms there, so we leave it.

    Then we have (3x-9). We can divide both terms by 3....throw the extra 3 away.

    You are left with (x-3).

    So your final answer is the first binomial (3x-2) and the second one with the extra 3 thrown out, (x-3).

    All of that being said, MOST of the time, trial and error works. There is also another method that involves rewriting the original equation as:
    [tex]3x^2-2x-9x+6[/tex] and then using factoring by grouping to arrive at the solution.
    you get:
    and then:
    I prefer the trial and error method 95% of the time.
    All of these methods require lots of practice and still take up paper and time...and if your leading coefficient is 30 and your constant is 11, you have to multiply 30 x 11 and then list off all of those factors....can be way worse than trial and error.
    best of luck
    Last edited: Jul 19, 2008
  4. Jul 20, 2008 #3
    Thanks, I like this way.
  5. Jul 20, 2008 #4


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    Of course, happyg1's method requires some cleverness in seeing that grouping. For quadratic polynomials, a method that requires no cleverness or trial and error is to use the quadratic formula to find the zeros of the polynomial, then use those as factors.

    In your example, 3x2- 11x+ 6, the quadratic formula gives
    [tex]x= \frac{-(-11)\pm\sqrt{(-11)^2- 4(3)(6)}}{2(3)}= \frac{11\pm\sqrt{121-72}}{6}= \frac{11\pm\sqrt{49}}{6}[/tex]
    Taking the positive sign we have
    [tex]x= \frac{11+ 7}{6}= \frac{18}{6}= 3[/tex]
    Taking the negative sign we have
    [tex]x= \frac{11-7}{6}= \frac{4}{6}= \frac{2}{3}[/tex]
    That leads us to (x-3)(x- 2/3)= x2- (11/3)x+ 2 and, multiplying by 3, 3x2- 11x+ 6= (x-3)(3x- 2).

    But what do you have against "trial and error"? It is an old and respected mathematical method and is typically much easier than this method.

    The method above, by the way, has the advantage of factoring a polynomial even when it "cannot be factored" (in the sense of factors with integer coefficients).

    No amount of "trial and error" will lead you to a factorization of x2- 3x- 5 simply because it, like almost all polynomials cannot be factored (with integer coefficients). But it is easy to determine that the zeros are
    [tex]x= \frac{-(-3)\pm\sqrt{(-3)^2- 4(1)(5)}}{2(1)}= \frac{3\pm\sqrt{9+ 20}}{2}= \frac{3\pm\sqrt{29}}{2}[/itex]
    Of course, [itex]\sqrt{29}[/itex] is not an integer, nor even rational, but we can, none the less, write
    [itex]x^2-3x- 5= (x- (3/2)- \sqrt{29}/2)(x- (3/2)+ \sqrt{29}/3)[/itex]

    And that works even when the roots are complex.
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