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Thread: Factoring, please help, step by step

  1. #1

    Factoring, please help, step by step

    My parents had me move, and I am in a new school and they're currently learning factoring. I need to learn this in order to catch up with my new class. I have a very hard time with Algebra, so please explain in an easy step by step process for me. I think if I get help with this one problem I can figure out the others.

    The equation I need to factor is:

    2x^2+x-6

  2. #2
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    Quote Originally Posted by aviguerias View Post
    My parents had me move, and I am in a new school and they're currently learning factoring. I need to learn this in order to catch up with my new class. I have a very hard time with Algebra, so please explain in an easy step by step process for me. I think if I get help with this one problem I can figure out the others.

    The equation I need to factor is:

    2x^2+x-6
    For a detailed step-by-step review - go to:

    http://www.purplemath.com/modules/factquad.htm
    “... mathematics is only the art of saying the same thing in different words” - B. Russell

  3. #3
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    Quote Originally Posted by aviguerias View Post
    My parents had me move, and I am in a new school and they're currently learning factoring. I need to learn this in order to catch up with my new class. I have a very hard time with Algebra, so please explain in an easy step by step process for me. I think if I get help with this one problem I can figure out the others.

    The equation I need to factor is:

    2x^2+x-6
    Before you start "factoring" polynomials you should have learned to multiply them!

    [tex](ax+ b)(cx+ d)= acx^2+ (ad+ bc)x+ bd[/tex]
    So you are looking for numbers, a, b, c, and d, such that ac= 2, ad+ bc= 1, and bc= -6.
    Now, that would be nearly impossible except that when we talk about "factoring" in this sense, we mean "factoring with integer coefficients".

    You proceed by "educated" trial and error.

    "2" can only be factored, with integers, as (1)(2) so either a=1 and c= 2 or a= 2 and c= 1.
    Similarly 6 can be factored as (1)(6), (6)(1), (2)(3), or (3)(2). Also the factors of -6 must have opposite signs.
    So we have quite a few, but still only "a few" possible values for b and d:
    b= 1, d= -6 or b= -6, d= 1, or b= 2, d= -3, or b= -3, d=2, or b= -1, d= 6, or b= 6, d= -1, or b= -2, d= 3, or b= 3, d= -2.

    That is 2 possible cases for a and c and 8 cases for b and d so (2)(8)= 16 cases altogether. We determine which is correct (if any- most polynomials don't factor with integer coefficients) by checking ad+ bc= 1.

    The "educated" part is recognizing that we don't have to try all of them. Obviously, if have b or d equal to 6 or -6, we wont be able to get "ad+ bc= 1" so we should try just the "3""2" factors for -6. One thing that should jump out at you is that ad+ bc= (2)(2)+ (1)(-3)= 1 so a= 2, d= 2, b= -3, c= 1 should work.

    Check by multiplying (ax+ b)(cx+ d)= (2x- 3)(x+ 2) to see what you get

  4. #4
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    Factor out the greatest term possible from the entire polynomial. In this case, it's 1, leaving the same. This method can be used here:
    You need 2 #s that when multiplied, equal the first coefficient times the last (2*-6 or -12), & also when added, equal the 2nd (1). If there are no such #s; the polynomial is probably prime and cannot be factored. If there are; you convert the polynomial to 4 terms using those 2 #s in place of the 2nd term
    (2x^2 - dx + ex -6). Factor the 1st 2 terms & the last 2 separately. ( (2x^2 - dx) (ex - 6) ). Factor out a term from each to leave you with a common binomial (example: x(2x -d) ). Put the 2 terms you factored out together in a binomial. That, and the common binomial are your factors. ( x might be 1 of the 2 terms, &
    2x - d might be the common binomial) You may not be able to use the 1st & last pairs. You may have to rearrange the terms to make it work. Check by multiplying your factors.

  5. #5
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    Quote Originally Posted by Denis View Post

    problem: factor 2x^2+x-6

    SO:

    (x + 2)(x - 3/2)

    in most cases [simpler] to use quadratic formula right off the bat
    If we expand the factorization above, we don't recover the given polynomial. Forgetting about the leading coefficient (after dividing it out), may result, when jumping to finding roots.

    But, if an exercise is to find the roots, then Quadratic Formula is fine (unless the instructions specify "by factoring").

    For a strictly-factoring exercise, I would suggest one of the methods mentioned earlier (eg: groupings, charts). Teacher likely wants to see workings of a factoring method taught in class.

    Ciao

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