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Homework Help: Simple factoring question here: Thanks!

  1. Aug 30, 2012 #1

    nukeman

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    1. The problem statement, all variables and given/known data

    Ok, just doing a re-fresher here!

    2x^2 - 7x - 2

    Factor it...

    First, let me ask you guys, when you see this, what is the FIRST thing you think of doing?


    3. The attempt at a solution

    For me... And I do not think this is correct.

    I would first automatically put the following:

    (2x ) (x )

    ?

    Then, all I need to do is put in the values that add to -4???
     
    nukeman, Aug 30, 2012
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  3. Aug 30, 2012 #2

    Kolmin

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    I take a look at the variable and I check where it is around.
    So, here I go for:

    [tex]x(2x-7)-2[/tex]
     
    Kolmin, Aug 30, 2012
  4. Aug 30, 2012 #3

    nukeman

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    Ok thanks for the input.

    The answer from my text says:

    (2x + 1) (x - 4)
     
    nukeman, Aug 30, 2012
  5. Aug 30, 2012 #4

    Mark44

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    So far, so good.
    Next, you want to find two factors of -2 so that the x terms add up to -7.

    Here are the possibilities:
    (2x - 2) (x + 1) -- x term is -2x + 2x = 0x
    (2x + 2) (x - 1) -- x term is 2x - 2x = 0x
    (2x + 1) (x - 2) -- x term is x - 4x = -3x
    (2x - 1) (x + 2) -- x term is -x + 4x = 3x

    So, no good, since none of the combinations gave us an x term with a coefficient of -7.

    Your quadratic trinomial doesn't factor into "nice" factors (i.e., factors with integer coefficients).

     
    Mark44, Aug 30, 2012
  6. Aug 30, 2012 #5

    Mark44

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    ???
    This is not what it means to factor a trinomial.
     
    Mark44, Aug 30, 2012
  7. Aug 30, 2012 #6

    Mark44

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    There's a mistake somewhere. If you expand this product, you get 2x2 - 7x - 4. In post 1 you wrote 2x2 - 7x - 2. Did you make a mistake or is it a typo in your book?
     
    Mark44, Aug 30, 2012
  8. Aug 30, 2012 #7

    nukeman

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    Ohh CRAP!!! Sorry, dam!

    Its: 2x^2 - 7x - 4
     
    nukeman, Aug 30, 2012
  9. Aug 30, 2012 #8

    Kolmin

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    :redface:

    Linguistic problems...

    Sorry for the epic fail.
     
    Kolmin, Aug 30, 2012
  10. Aug 30, 2012 #9

    Kolmin

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    Btw, having now an idea about the problem, I go for the solution of the corresponding equation and then I follow this principle:

    [itex]ax^2 + bx + c = a(x-u)(x-v)[/itex]

    where [itex]u[/itex] and [itex]v[/itex] are the solutions of the equation.

    In this case we indeed have

    [tex]2(x-4)(x+ \frac{1}{2})[/tex]

    which is equal to the result posted by OP with some manipulation.
     
    Kolmin, Aug 30, 2012
  11. Aug 30, 2012 #10

    nukeman

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    Don't quite understand the steps here :(

    2x^2 - 7x - 4

    So, like I said, I automatically put:

    (2x ) (x )

    Then, I need to find 2 factors of -4 that add up to -7 ?????
     
    nukeman, Aug 30, 2012
  12. Aug 30, 2012 #11

    Mark44

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    No, you need to find two factors of -4 so that the two cross terms (the terms in x) have a coefficient of -7.

    You're probably thinking of a method where you multiple a times c (the coefficient of the x^2 term times the constant. Then you factor that number into factors that add up to b.

    In your problem, a = 2, b = -7, and c = -4
    a * c = -8

    We look for factors of -8 that add to -7, and these would be 1 and -8.

    Now, write the trinomial using these numbers.
    2x2 + x - 8x - 4

    Now form two groups:
    (2x2 + x) + (-8x - 4)

    Factor each group:

    x(2x + 1) - 4(2x + 1)

    Now, use the distributive property to write as (x - 4)(2x + 1).
     
    Mark44, Aug 30, 2012
  13. Aug 30, 2012 #12

    SammyS

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    nukeman,

    Here's a little trick that I actually picked up from some of my students:

    Multiply by 2 to make the leading term a perfect square. (In the en you must divide the result by 2, but that will work fine, because both terms in one of the factors will be multiples of 2.)

    [itex]\displaystyle 2x^2 - 7x - 4[/itex]
    [itex]
    \displaystyle =\frac{4x^2 - 7(2x)-8}{2}[/itex]

    [itex]\displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}[/itex]​
    Now threat the 2x like it is some other variable such as u.

    u2 - 7u - 8 = (u -  ?  )(u +  ?  )

    Of course those should be filled in with 8 and 1 .

    u2 - 7u - 8 = (u - 8)(u + 1)

    Then back to the original:

    [itex]\displaystyle 2x^2 - 7x - 4[/itex]
    [itex]
    \displaystyle =\frac{(2x)^2 - 7(2x)-8}{2}[/itex]

    [itex]\displaystyle =\frac{(2x-8)(2x+1)}{2}[/itex]

    [itex]\displaystyle =\frac{2(x-4)(2x+1)}{2}[/itex]​
    Giving you the desired result.
     
    SammyS, Aug 30, 2012
  14. Aug 30, 2012 #13

    nukeman

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    Fantastic! Thank you Mark! :)

    Sammy, pretty cool way of doing, never seen that!

     
    nukeman, Aug 30, 2012
  15. Aug 31, 2012 #14

    Mentallic

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    Learn the cross method, it'll save you a heap of heartache. I still remember it even after years of not using it (don't be discouraged by that, it's just because most the quadratics I see can't be factorized with integer coefficients).

    http://www.youtube.com/watch?v=2dqAPKgC-KI
     
    Mentallic, Aug 31, 2012
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