1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Help with factoring

  1. Aug 28, 2009 #1
    I'm currently taking a Math Analysis class. We started reviewing basic factoring from Algebra 2 before starting new material. I can factor a basic trinomial with the coefficient of x^2 being 1, but if the coefficient is 2 or more, I am not sure how to factor it. I am able to factor these very easily with a graphing calculator, but I'd like to know how to do it by hand.

    1. The problem statement, all variables and given/known data
    Factor: 2x^2 + 5x - 12

    2. The attempt at a solution
    I know, by using a graphing calculator, that the trinomial factors into (2x-3)(x+4). I know that if the trinomial were, for instance, x^2 - 2x - 8, one could say the product of 8 and the coefficient of x is 8, and the factors of 8 that when added together result in negative 2 are -4 and +2, therefore the factors of x^2 - 2x - 8 are (x-4)(x+2).

    Can trinomials with coefficients greater than one be factored in a similar way?

    Using the method described above, I come up with a product of 24. Factors of 24 that result in a sum of 5 when added together are +8 and -3, which by the method I am acquainted with would mean the factors are (x+8)(x-3), but that obviously doesn't equal 2x^2 + 5x - 12 when multiplied. I know I'm missing a step somewhere.

    Thanks for the help in advance. I know this is a simple question, but it plagued me all day.
    Last edited: Aug 28, 2009
  2. jcsd
  3. Aug 28, 2009 #2


    User Avatar
    Homework Helper

    Re: Factoring

    You do it in the same manner

    2x2 + 5x - 12 must be factored into something like (2x+a)(x+b)

    the last times the last should give -12 or ab=-12, so what products can give -12?
    There is -12 and 1, 2 and -6, 3 and -4 (and the others switch around the signs)

    expanding (2x+a)(x+b) in your head, you'd see that the coefficient of x is a+2b. So looking at your choices, you want a+2b=5. Right away 12 and -1 (or 1 and -12) is eliminated.

    2 and -6 is gone since 2+2(-6)≠5 or -2+2(6)≠5

    so you are left with 3 and -4 or -3 and 4

    3+2(-4)= -5 .So we are seeing 5 but negative, so we need to change the signs. So the choice is -3 and 4

    so it is factored as (2x-3)(x+4)

    as you practice, you can quickly do this and eliminate the obvious ones it can't be.

    For ax2+b+c

    Another way is to compute b2-4ac and find the square root of that. If it is an integer, then using the quadratic equation formula

    [tex]x_1,x_2=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/tex]

    In your example b2-4ac=121, so √(b2-4ac)=11

    x1,x2= (-5±11)/2(2)


    So one root is x=3/2 or 2x+3=0

    x2=(-5-11)/4=-4. So the other root is x=-4 or x+4=0

    the product of these two factors will give your original quadratic as being (2x+4)(x-4)
  4. Aug 28, 2009 #3


    User Avatar
    Staff Emeritus
    Science Advisor

    Re: Factoring

    First not all trinomials can be factored with integer coefficients. But trinomials with coefficients of [itex]x^2[/itex] not equal to one are only a little more tedious. To factor [itex]ax^2+ bx+ c[/itex], you need to think about what (mx+ n)(px+ q) would look like: [itex]mx(px+ q)+n(px+q)= mpx^2+ mqx+ npx+ nq[/itex][itex]= mpx^2+ (mq+np)x+ nq[/itex]. So we are looking for a numbers m, n, p, q such that mp= a, nq= c, and mq+np= b. Start by thinking of factors for a and b and look at all the different ways you could form mq+ np.

    With you example, [itex]2x^2 + 5x - 12[/itex], there are two ways to factor 2: (1)(2) and (2)(1). But there are many ways to factor 12: (1)(12), (2)(6), (3)(4), (4)(3), (6)(2), and (12)(1). Since there are 2 ways to factor 2 and 6 ways to factor 12, there are (2)(6)= 12 possible combinations. Also, since the last term is negative we know the last term in the two factors must have opposite sign:
    (1)(2) and (1)(12): [itex](x+ 1)(2x- 12)= 2x^2- 10x- 12[/itex]
    (2)(1) and (1)(12): [itex](2x+ 1)(x- 12)= 2x^2- 23x- 12[/itex]
    (1)(2) and (2)(6): [itex](x+ 2)(2x- 6)= 2x^2- 2x- 12[/itex]
    (2)(1) and (2)(6): [itex](2x+ 2)(x- 6)= 2x^2- 10x- 12[/itex]
    (1)(2) and (3)(4): [itex](x+ 3)(2x- 4)= 2x^2+ 2x- 12[/itex]
    (2)(1) and (3)(4): [itex](2x+ 3)(x- 4)= 2x^2- 5x- 12[/itex]
    (1)(2) and (4)(3): [itex](x+ 4)(2x- 3)= 2x^2+ 5x- 12[/itex] !!!
    (2)(1) and (4)(3): [itex](2x+ 4)(x- 3)= 2x^2- 2x- 12[/itex]
    (1)(2) and (6)(2): [itex](x+ 6)(2x- 2)= 2x^2+ 10x- 12[/itex]
    (2)(1) and (6)(2): [itex](2x+ 6)(x- 2)= 2x^2+ 2x- 12[/itex]
    (1)(2) and (12)(1): [itex](x+ 12)(2x- 1)= 2x^2+ 23x- 12[/itex]
    (2)(1) and (12)(1): [itex](2x+12)(x- 1)= 2x^2+ 10x+ 12[/itex]
    Whew! Of course, we could have stopped when we got "(1)(2) and (4)(3): [itex](x+ 4)(2x- 3)= 2x^2_ 5x- 12[/itex]" but I wanted to show what they all looked like.
  5. Aug 28, 2009 #4
    Re: Factoring

    Thank you both for your in depth replies. I'm able to do this now. Again, thanks a lot :)
  6. Aug 29, 2009 #5
    Re: Factoring

    There is a clever factorization method (I believe developed by Viete) for non-monic quadratic trinomials with integer coefficients. (i.e. of the form [itex]ax^2 + bx + c \text{ for }a,b,c \text{ integers where } a \neq 1 \text{ or } 0[/itex]).

    Step 0.
    Factor out any common divisors of a, b, and c and work with the cofactor.

    Step 1.
    Multiply a and c and call the product d. This is the key number to unlocking the factorization.

    Step 2.
    Determine if there is a pair of factors of d whose sum is b. If such a factor pair exists, then the polynomial is factorable, otherwise it is not factorable (and you'd stop here).

    If a factor pair exists, call them m and n.

    Step 3.
    Rewrite [itex]ax^2 + bx + c \text{ as }(ax^2 + mx) + (nx + c)[/itex] and factor the latter expression by grouping. Be careful when n is negative.

    Step 4.
    You should have your factorization.

    The advantage to this method is that it cuts down the "try-this-then-try-this-other-way"-ness of other methods.

    Example: Factor [itex]12x^2 -31x-30[/itex].

    Step 0. No common divisors exist so I'm working with it as is.

    Step 1. The product of 12 and -30 gives a key number of -360.

    Step 2. The factos of -360 the add to -31 are -40 and 9.

    Step 3.





Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Help with factoring
  1. Polynomial factor help (Replies: 3)

  2. Quick Factoring Help (Replies: 10)