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Math Help - middle term / factor question..

  1. #1
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    middle term / factor question..

    I'm kinda banging my head for this one. Is this one has middle term?

    3x^2+6x-3 ?

    Help me plz.

    tx
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  2. #2
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    Re: middle term / factor question..

    Well, divide by 3 to start: x^2 + 2x - 1
    Now explain your question!
    If it was x^2 + 2x + 1, then that would factor to (x + 1)^2
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  3. #3
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    Re: middle term / factor question..

    As Wilmer said, you factor out the "3" in each term: 3(x^2+ 2x- 1). Other than that, it cannot be factored with integer coefficients. The fact is that [b]most[b] polynomials cannot be factored with integer coefficients.

    (I keep saying "with integer coefficients" because we could say that x^2+ 2x- 1= x^2+ 2x+ 1- 2= (x+1)^2- 2. And now we can use the simple rule a^2- b^2= (a- b)(a+ b) so that x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2}). Of course, to do that, I had to, effectively, solve the equation x^2+ 2x- x= 0 by 'completing the square' and, since a main application of factoring is to solve equations, we usually mean "with integer coeffcients". But, once again, most polynomial equations can't be factored that way.)
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    Re: middle term / factor question..

    thanks,

    Hei HallsofIvy

    How do you write maths characters like that?
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    Re: middle term / factor question..

    Quote Originally Posted by ameerulislam View Post
    How do you write maths characters like that?
    You have to use LaTeX coding.
    [tex]x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})[/tex] gives x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2}).
    Thanks from ameerulislam
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    Re: middle term / factor question..

    Quote Originally Posted by Plato View Post
    You have to use LaTeX coding.
    [tex]x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2})[/tex] gives x^2+ 2x- 1= (x+1-\sqrt{2})(x+1+\sqrt{2}).
    where can I get codes for all the characters I'm looking for. I want the code for under line, I mean 2/3 .. Two over three here.... and code of the line between them.
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  7. #7
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    Re: middle term / factor question..

    3x^2+6x-3

    first you have to convert it into quadratic form of equation
    i.e a x^2 + b x + c =0

    then find D = b^2 - 4ac

    if D>0 then factor of equation is possible
    otherwise use

    X1 = [ -b + sqrt(D) ] / 2a
    X2 = [ -b - sqrt(D) ] / 2a

    ..... try it also helps you Quadratic Equationmiddle term / factor question..-quadratic-equation.jpg
    Last edited by Neeraj; July 26th 2012 at 04:17 AM.
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    Re: middle term / factor question..

    Quote Originally Posted by ameerulislam View Post
    where can I get codes for all the characters I'm looking for.
    See the LaTeX Help subforum. Remember to use [tex] ... [/tex] tags. See also the LaTeX Wikibook or search the web.
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    Re: middle term / factor question..

    Thanks emakarov

    I came across this formula, I heard this one is used when there is no easy way to factor. What is the story with this formula and how does it works?

    \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}

    I'm a programmer and I am liking La Text

    Edit: My bad, I fixed the sign before 4ac.
    Last edited by ameerulislam; July 26th 2012 at 07:28 AM.
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    Re: middle term / factor question..

    Quote Originally Posted by ameerulislam View Post
    I came across this formula, I heard this one is used when there is no easy way to factor. What is the story with this formula and how does it work?

    \frac{-b \pm\sqrt{b^2 + 4ac}}{2a}
    The expression under the square root (called discriminant) should be b^2-4ac. I personally use this formula instead of factoring. It gives the two roots of the quadratic equation ax^2+bx+c=0.
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    Re: middle term / factor question..

    I have this equation to factor
    x^2 - 4x +5


    \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}

    within square root it comes -16.. So does that means I can't factor that equation?
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    Re: middle term / factor question..

    Quote Originally Posted by ameerulislam View Post
    I have this equation to factor
    x^2 - 4x +5 \frac{-b \pm\sqrt{b^2 - 4ac}}{2a}
    within square root it comes -16.. So does that means I can't factor that equation?
    Actually it is \sqrt{-4}. Using complex numbers:
    (x-2+i)(x-2-i)
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  13. #13
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    Re: middle term / factor question..

    yeah only complex number will fit there. I think I can ignore it as I'm just doing simple calculus (only real numbers)..
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    Re: middle term / factor question..

    Quote Originally Posted by ameerulislam View Post
    yeah only complex number will fit there. I think I can ignore it as I'm just doing simple calculus (only real numbers)..
    Well then, why isn't the expression x^2 - 4x - 5 ?
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