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Help Factoring Patterns

  1. Apr 13, 2006 #1
    Help!!! Factoring Patterns

    ok i need help asap! test tomarrow and have no clue what to do!! :bugeye: i need someone to help me understand the steps!!
    ex problem
    10x[squared]+6x-28
    answer
    2(5x-7)(x+2)

    what i need to know is what steps do i take to get the answer?
     
  2. jcsd
  3. Apr 13, 2006 #2
    This is quite a simple factoring problem...You just need to play with the numbers and find something that works!

    (I don't know how to use the LaTex Typesetting yet, so it might be scruffy!)

    0 = 10x^2 + 6x - 28 (first take a common factor of 2 out)
    0 = 2(5x^2 + 3x - 14) (Now, factorise the bracket)
    0 = 2(5x - ?)(x + ?) ( I know there is a 5x and x there because they need to multiply together to form 5x^2 and that is the only solution there. The + and - are obvious from how the constants need to be multiplied together to form -14. A minus can only be formed by multiplying a - and +.

    Now you need 2 constants that multiply together to form minus 14, 7 and 2. Put these into the equation and watch it work.

    0 = 2(5X-7)(x+2)

    You can check it by multiplying it back out and getting the same answer.
     
  4. Apr 13, 2006 #3
    true, but you don't know (unless you already know the answer) which brackets they go in..

    obviously trial/error works here, but it's a reasonable bet that because we want a positive x coefficient it will be that way around due to the 5 and the 1.
     
  5. Apr 13, 2006 #4
    Whenever you are factoring, always first ask yourself, is there a common factor? If there is, factor it out. The remaining expression will be that much simpler. If your leading coefficient (the one in front of x2) is 1, find a pair of numbers that sum to the middle coefficient and have the constant term as their product. I like to write a table of all the possible pairs of integers that have the right product. Once you find the pair of numbers, plug them into (x+a)(x+b) in place of the a and b.

    If your leading coefficient is not 1, multiply it by the constant term. This will be the product which has factors that add up to the middle coefficient. When you find the pair of numbers, split up your middle term using them. Then factor by grouping.

    Ex: 5x2+3x-14.
    5(-14) = -70.

    -1*70
    -2*35
    -3*(doesn't work)
    -4*(doesn't work)
    -5*14
    -6*(doesn't work)
    -7*10
    -8*(doesn't work)
    -9*(doesn't work)
    -10*7 (now you can rewrite the table in reverse order since you already had a 10)
    -14*5
    -35*2
    -70*1

    And the one that sums to a positive 3 is in red. Now, split the middle term up using -7 and 10: 5x2-7x +10x-14 = x(5x-7) +2(5x-7). There is a common factor of (5x-7) now, so you can continue: (5x-7)(x+2).
     
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