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Thread: Questions about factoring an equation with 3rd degree and 2nd degree

  1. #1
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    Question Questions about factoring an equation with 3rd degree and 2nd degree

    $\displaystyle 2x^3+3x^2-1$

    This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

    I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

    next I did division



    Questions about factoring an equation with 3rd degree and 2nd degree-math1.gif


    I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. any help appreciated.
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    "I figured out that p(-1) = 0 that would mean that x+1 is a factor of this equation." Good, so you have: $\displaystyle 2x^3+3x^2-1=2x^3+2x^2+x^2+x-x-1=(x+1)(2x^2+x-1)$.
    I don't understand what you did in that picture, but from here, in general $\displaystyle ax^2+bx+c=0$ can be factored: $\displaystyle a(x-x_1)(x-x_2)$.

    $\displaystyle (x^2-1)(x^2-x)$ ~ $\displaystyle a^2-b^2=(a+b)(a-b)$, for the second bracket x is common factor.
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  3. #3
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    Quote Originally Posted by sara213 View Post
    $\displaystyle 2x^3+3x^2-1$

    This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

    I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

    next I did division



    Click image for larger version. 

Name:	math1.gif 
Views:	52 
Size:	1.8 KB 
ID:	21393


    I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. any help appreciated.
    Your idea is sound but you need to remember to include all powers of x when you do division

    To clarify you started with

    $\displaystyle 2x^3+3x^2-1=2x^3+3x^2+0x-1$

    $\displaystyle \begin{array}{crrrrrrr}
    & 2x^2&+& x &-&1 &&\\ \cline{2-8}
    (x+1) \vline & 2x^3 &+& 3x^2&+&0&-&1 \\
    &-(2x^3 & + & 2x^2) & &&&\\
    &&& x^2 & + & 0x &&\\
    &&&-(x^2 & + & x) &&\\
    &&&&&-x & - & 1 \\
    &&&&&-(-x & - & 1) \\
    &&&&&&&0

    \end{array}$
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  4. #4
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    Quote Originally Posted by TheEmptySet View Post
    Your idea is sound but you need to remember to include all powers of x when you do division

    To clarify you started with

    $\displaystyle 2x^3+3x^2-1=2x^3+3x^2+0x-1$

    $\displaystyle \begin{array}{crrrrrrr}
    & 2x^2&+& x &-&1 &&\\ \cline{2-8}
    (x+1) \vline & 2x^3 &+& 3x^2&+&0&-&1 \\
    &-(2x^3 & + & 2x^2) & &&&\\
    &&& x^2 & + & 0x &&\\
    &&&-(x^2 & + & x) &&\\
    &&&&&-x & - & 1 \\
    &&&&&-(-x & - & 1) \\
    &&&&&&&0

    \end{array}$
    Sorry about the picture...I tired to post it form math type into here and it became huge so I did a gif but that didn't work out well. I wanted to show the division in full process but I couldn't figure out how to add the spaces.

    Your right...I forgot about adding the 0x...thank you though. I kind of feel stupid now..lol
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  5. #5
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    Quote Originally Posted by sara213 View Post
    $\displaystyle 2x^3+3x^2-1$

    This is the equation that needs to be fully factored. So, from what I see here it would be impossible to factor it normally, since 1 can't be multiplied by 3 or any other number to get 1 other then 1.

    I figure this is a good idea to use factor theorem. I figured out that p(-1) =0 that would mean that x+1 is a factor of this equation.

    next I did division



    Click image for larger version. 

Name:	math1.gif 
Views:	52 
Size:	1.8 KB 
ID:	21393


    I got stuck half way through the division when I reached the (x^2 -1)(x^2-x) I don't think i can subtract 1-x. Not sure where to go from here. any help appreciated.


    Besides long division you can do this by simple "eye", meaning: just complete the second factor to what you need:

    $\displaystyle 2x^3+3x^2-1=(x+1)(2x^2+x-1)$ ...how did I find the factor $\displaystyle 2x^2+x-1$ ? Well, we need $\displaystyle 2x^3$ , so we must

    multiply $\displaystyle x\mbox { by } 2x^2$ . After this we need $\displaystyle 3x^2$ , and we already have $\displaystyle (x+1)(2x^2...)$ , so we have so

    far only $\displaystyle 2x^2$ , so we need to add $\displaystyle x$ ...and etc.

    Of course, after this we can continue $\displaystyle 2x^2+x-1=(x+1)(2x-1)$ and etc.

    Tonio
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