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Thread: Factoring help, again...

  1. #1
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    Factoring help, again...

    $\displaystyle -(x-2)(x^2+2x+5)$
    $\displaystyle -(x^3+2x^2+5x-2x^2+4x-10)$
    $\displaystyle -(x^3+x-10)$
    $\displaystyle -x^3-x+10$

    How do you factor this? I tried adding in a $\displaystyle 0x^2$ and grouping, but that didn't work out. The other group can't factor out to $\displaystyle x-0$...
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  2. #2
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    Re: Factoring help, again...

    Hey Nervous.

    Try factoring the quadratic by getting the roots through the quadratic formula.
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  3. #3
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    Re: Factoring help, again...

    I thought quadratics were only second degree polynomials...?


    EDIT: Ohhhhh I get it, you meant the first trinomial... OK...
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  4. #4
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    Re: Factoring help, again...

    Why in the world would you first multiply and then factor? You start with $\displaystyle -(x- 2)(x^2+ 2x+ 5)$ so obviously, one factor is x- 2. In order to factor $\displaystyle x^2+ 2x+ 5$ into factors with integer coefficients, I would note that 5 factors only as 1(5) but neither 5-1 nor 5+ 1 is equal to 2 so we can't factor with integer coefficients. Not every quadratic can be factored with integer (or even real) coefficients.

    So instead, complete the square! $\displaystyle x^2+ 2x+ 1= (x+ 1)^2$ so that $\displaystyle x^2+ 2x+ 5= x^2+ 2x+ 1+ 4= (x+1)^2+ 4= (x+1)^2- (-4)= (x+ 1)^2- (2i)^2$. Now use the identity $\displaystyle a^2- b^2= (a- b)(a+ b)$.
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    Re: Factoring help, again...

    That's a beautiful idea... God, I still have so much to learn...
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    Re: Factoring help, again...

    Another one:
    $\displaystyle 2x^3+11x^2-12x-36$

    I tried grouping, I tried factoring out an x and using quadratic formula on $\displaystyle x^+11x-12$ I don't know how to do this one.
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    Re: Factoring help, again...

    Edit
    Quote Originally Posted by nervous View Post
    another one:
    $\displaystyle 2x^3+11x^2-12x-36$

    i tried grouping, i tried factoring out an x and using quadratic formula on $\displaystyle x^2+11x-12$ i don't know how to do this one.
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  8. #8
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    Re: Factoring help, again...

    Have you tried the "Rational Root Test?"

    -Dan
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