Results 1 to 3 of 3

Thread: "Factoring Pattern for ax^2 + bx + c"

  1. #1

    "Factoring Pattern for ax^2 + bx + c"

    What we're doing is Factoring by Grouping.

    Example, third question on my sheet.

    2x^2+x-6

    The steps to solving that are:

    1. Multiply the first and last terms, and place the products above the original quadratic.
    2. Find factors of this product that add up to the original quadratic.
    3. Group the first two and last terms together, and continue factoring.

    Er, alright so basically doing that question:

    2x^2+x-6

    We, or I, multiply the beginning and the end.

    (2x^2)(-6) = -12x^2

    Now we find the factor(s) of -12

    -1 and 12
    1 and -12

    -2 and 6
    2 and -6

    -3 and 4
    3 and -4

    Which one of these factors equals x

    ax^2+bx+c is the "Factoring Pattern)

    Well, either I'm stuck here, or the answer is "Prime"

    -Heh, what are your thoughts on it?

  2. #2
    Senior Member
    Join Date
    Dec 2005
    Location
    CO, USA
    Posts
    2,092

    Re: "Factoring Pattern for ax^2 + bx + c"

    Quote Originally Posted by ultrasonicsite
    What we're doing is Factoring by Grouping.

    Example, third question on my sheet.

    2x^2+x-6

    The steps to solving that are:

    1. Multiply the first and last terms, and place the products above the original quadratic.
    2. Find factors of this product that add up to the original quadratic.
    3. Group the first two and last terms together, and continue factoring.

    Er, alright so basically doing that question:

    2x^2+x-6

    We, or I, multiply the beginning and the end.

    (2x^2)(-6) = -12x^2

    Now we find the factor(s) of -12

    -1 and 12
    1 and -12

    -2 and 6
    2 and -6

    -3 and 4
    3 and -4

    Which one of these factors equals x

    ax^2+bx+c is the "Factoring Pattern)

    Well, either I'm stuck here, or the answer is "Prime"

    -Heh, what are your thoughts on it?
    The middle term is 1x. So, you need factors of -12x^2 which add up to 1x. How about (-3x) and (4x)?

    Rewrite the middle term using -3x and 4x:

    2x^2 - 3x + 4x - 6

    Remove the greatest common factor from the first two terms, and the greatest common factor from the last two terms:

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

    Now, (2x - 3) is a common factor; remove it, and you have

    (2x - 3)(x + 2)

    Here's a site you might find really "cool" for working with this type of factoring:

    http://www.purplemath.com/modules/factquad3.htm

  3. #3
    Thanks a lot for your help. Though I don't use that box method(link) =)

    I forgot that x = 1x, that's where I went wrong.

Bookmarks

Posting Permissions

  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •