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Factoring a trinomial

  1. Nov 9, 2006 #1

    DaveC426913

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    I'm helping edit a math textbook and have been asked to provide the answer for this:

    Factor to find the value of b and c in the trinomial:
    -0.5x2 + x -1= -0.5 (x2 - bx + c)

    Here is what I've done:

    -0.5x^2 + x -1= -0.5 (x^2 - bx + c)
    -0.5x^2 + x -1= x^2 + .5bx - .5c
    -0.5x^2 + x -1= x^2 + .5bx - .5c
    x -1= 1.5x^2 + .5bx - .5c
    -1 = 1.5x^2 - .5bx - .5c
    1.5x^2 - .5bx - .5c +1 = 0

    or, in latex:
    [tex]-0.5x^2 + x -1= -0.5 (x^2 - bx + c)[/tex]
    [tex]-0.5x^2 + x -1= x^2 + .5bx - .5c[/tex]
    [tex]-0.5x^2 + x -1= x^2 + .5bx - .5c[/tex]
    [tex]x -1= 1.5x^2 + .5bx - .5c[/tex]
    [tex]-1 = 1.5x^2 - .5bx - .5c[/tex]
    [tex]1.5x^2 - .5bx - .5c +1 = 0[/tex]

    Ultimately, I'm going to end up with factors of the form (ax^2+b+?)(x^2+c+?) ? Oh shoot . I don't know what I'm supposed to end up with. The question wants values for b and c. It's been 25 years!

    So, am I then supposed to use trial and error?

    if b=1, c=3
    1.5x^2 - .5bx - .5c +1 = 0
    then 1.5x^2 – .5x - .5 = 0

    So (1.5x + .5)(x - 1) (This isn't quite right, since it actually makes 1.5x2 – x -1=0, but I think I'm close.)

    Even if I get the right answer, I'll need to hsow my work, I can't tellt he students to guess can I?
     
    Last edited: Nov 9, 2006
  2. jcsd
  3. Nov 9, 2006 #2

    Office_Shredder

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    You've overcomplicated stuff. Go back to

    -0.5x^2 + x -1= x^2 + .5bx - .5c

    First, it's wrong. It should read

    -0.5x^2 + x -1 = -.5x^2 + .5bx - .5c

    Then, if it's true for all x, the coefficients next to each different x term must be the same. So -.5c = -1 for example. Or .5b = 1
     
  4. Nov 9, 2006 #3

    DaveC426913

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    Oh shoot, I made a mistake. I'll rewrite.

    -0.5x^2 + x -1= -.5x^2 + .5bx - .5c
    x -1=.5bx - .5c
    -1 = .5bx -.5c –x

    So .5bx-x+.5c+1=0

    But I didn't follow anything you said in the last line.

    (bx-1)(.5c-1)=0. Nope, still lost.
     
    Last edited: Nov 9, 2006
  5. Nov 9, 2006 #4

    Office_Shredder

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    Dude, you need to calm down on the algebraic errors :P

    -1 = .5bx -.5c –x

    So .5bx-x+.5c+1=0

    should be
    .5bx-x -.5c + 1 = 0

    So it's true for all x. Suppose x=0. Then -.5c+1=0, and c=2. Now we have

    .5bx-x-1+1=0 Or .5bx-x=0

    EDIT: It should be noted you never actually factored. It should be

    -0.5x2 + x -1 = -.5(-.5x^2/-.5 + x/-.5 -1/-.5) = -.5(x^2 -2x + 2)

    which gives b and c trivially

    So let x=1. .5b-1=0, or b=2.
     
  6. Nov 9, 2006 #5

    DaveC426913

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    Sorry 'bout that. Need to get this back tomorrow - which was fine - until I got a phone call to go to a job interview TONIGHT. Sort of distracted me AND ate up my time.


    Oh @#$!&*, I feel like I'm asking to be spoonfed. See, I need to give an authoritative answer (it's the answer that I'm providing for the textbook), and I need to show the work. But before begging for you to peel my grapes for me, I will try to take what you've demonstrated and write it out as a correct and succinct answer to the problem....


    -.5x^2 + x - 1 = -0.5 (x^2 - bx + c)
    -.5x^2 + x - 1 = -.5x^2 + .5bx - .5c
    x^2 + 2x - 2 = x^2 + bx - c
    2x - 2 = bx - c
    0 = bx - c - 2x - 2
    bx - 2x - c + 2 = 0

    At x=0:
    b(0)-2(0)-c+2=0
    -c+2=0
    c=2

    At x=-1:
    b(-1)-2(-1)-c+2=0
    -b+2-c+2=0
    -b-c+4=0
    -b-(2)+4=0
    -b+2=0
    b=2

    OK, well that gets us the answer, but
    1] it doesn't do it by factoring, which is what was asked. How are students supposed to figure this out??
    2] Have I written the full form of the answer correctly? This is going in the answers section of the book.
     
    Last edited: Nov 9, 2006
  7. Nov 13, 2006 #6

    Gokul43201

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    This question is meaningless (for starters, a trinomial is an expression, not an equation...but that's hardly all that's wrong with it) and certainly NOT what you want to put into a textbook. If it's not too late, please ask the author to remove it entirely.
     
  8. Nov 13, 2006 #7

    Office_Shredder

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    1.) That's where the last part of my post came in. Factor -.5 from the expression -.5x2 + x - 1. You get -.5*(x2 - 2x + 2).

    But this needs to equal -.5*(x2 - bx + c). Now, it should instinctually leap to you, that this implies x2 - bx + c = x2 - 2x + 2. And from here, let's analyze polynomials in general.

    If I told you ax+b = 3x+2, what are a and b? Trivially, if a=3, and b=2, we're done. Now, it can be shown this is a unique solution, but I don't think you're worried about that. Going back to our example up here, x2 - bx + c = x2 - 2x + 2 is true if b=2 and c=2 (plug it in, it should be obvious). This is because -b is next to the 'x' term, and so is -2 on the other side. And 2 is the constant term on one side, and so is c. So -b=-2, and c=2. This is probably what they're trying to drive at
     
  9. Nov 13, 2006 #8

    JasonRox

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    I thought answers in a textbook only need the final solution and not the work itself, so why concern yourself with that?

    Get Maple, and let it do it for you.
     
  10. Nov 14, 2006 #9

    DaveC426913

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    You thought wrong.:rolleyes:

    Anyway, it's OK. Thnaks everyone. The work has been passed to the appropriate person for the job (I was not it).
     
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