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Math Help - Help needed please

  1. #1
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    Help needed please

    I need some help on figuring out how to solve these equations, the directions say the problems represent a mix of linear, quadratic, and third degree polynomial equations. Here they are:

    1.) y^3 +3y^2 - 4y - 12= 0

    2.) 2x^3 + 3x^2 = 8x +12

    3.) 2a (a+1) = a^2 +8

    4.) 2x^3 + x^2 = 32x +16

    Thanks for your help! It's appreciated!
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  2. #2
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    Quote Originally Posted by MathMack View Post
    I need some help on figuring out how to solve these equations, the directions say the problems represent a mix of linear, quadratic, and third degree polynomial equations. Here they are:

    1.) y^3 +3y^2 - 4y - 12= 0

    2.) 2x^3 + 3x^2 = 8x +12

    3.) 2a (a+1) = a^2 +8

    4.) 2x^3 + x^2 = 32x +16

    Thanks for your help! It's appreciated!
    For cubics (third degree polynomials!), the first step requires guessing a root! Basically you have to guess a value of x, then apply it and see if it gives you zero! If it doesn't, then it's not a root. However, if it does, then it is a root.

    In your first example, imagine you let y = a, and you found out that the result was indeed zero! Then you could say that  (y-a) is a factor of the polynomial. You would then divide the original polynomial by  (y-a) using long division/synthetic division. This will enable you to write the expression in the form:

     (y-a)(f(y)) , where  f(y) is a quadratic that can be solved using the quadratic equation/factorisation!

    In question 1, guessing y = 2 gives you a root, hence  (y-2) is a factor. If you divide y^3 +3y^2 - 4y - 12, by  (y-2) , you get y^2+5y+6 . Hence:

    y^3 +3y^2 - 4y - 12 = (y^2+5y+6)(y-2) = 0

    You can factorise that quadratic easily to find the other roots!

    For the last 3 problems you have rearrange them to get 0 on the RHS, and you also need the expland the LHS of 3).

    2.) 2x^3 + 3x^2 - 8x -12 = 0 Which is a cubic, so apply the rules I used before! Hint: x = 2 is a root.

    3.) a^2+2a -8 = 0 Which is a quadratic, easily solveable using factorisation.

    4.) 2x^3 + x^2 -32x -16=0. Another cubic, so apply the rules I laid out for question 1! Hint: x = 4 is a root.
    Last edited by Mush; January 20th 2009 at 07:41 PM.
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  3. #3
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    I'm sorry, i'm really confused, my book has the answer for the 1st one as -3, -2, 2. I'm not sure how to get everything on the left and then how to solve it. THanks for your help.
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  4. #4
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    Quote Originally Posted by MathMack View Post
    I'm sorry, i'm really confused, my book has the answer for the 1st one as -3, -2, 2. I'm not sure how to get everything on the left and then how to solve it. THanks for your help.
    Have a look at my post. I've editted it with extra information.
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  5. #5
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    Okay, I have that far figured out, I have a question on number 3, what happened to the a+1?? How do I figure out the rest of them? I'm lost to when I get everything on that side how to factor it or figure it out.
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  6. #6
    Senior Member DeMath's Avatar
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    Quote Originally Posted by MathMack View Post
    I need some help on figuring out how to solve these equations, the directions say the problems represent a mix of linear, quadratic, and third degree polynomial equations. Here they are:

    1.) y^3 +3y^2 - 4y - 12= 0

    2.) 2x^3 + 3x^2 = 8x +12

    3.) 2a (a+1) = a^2 +8

    4.) 2x^3 + x^2 = 32x +16

    Thanks for your help! It's appreciated!
    \left. 4 \right){\text{ }}2{x^3} + {x^2} = 32x + 16 \Rightarrow

    \Rightarrow {x^2}\left( {2x + 1} \right) = 16\left( {2x + 1} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 16, \hfill \\2x + 1 = 0; \hfill \\ \end{gathered}  \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} =  \pm 4, \hfill \\{x_3} =  - \frac{1}{2}. \hfill \\ \end{gathered}  \right.
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  7. #7
    Senior Member DeMath's Avatar
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    For number 2

    \left. 2 \right){\text{ }}2{x^3} + 3{x^2} = 8x + 12 \Rightarrow

    \Rightarrow {x^2}\left( {2x + 3} \right) = 4\left( {2x + 3} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 4, \hfill \\2x + 3 = 0; \hfill \\ <br />
\end{gathered}  \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} =  \pm 2, \hfill \\{x_3} =  - \frac{3}{2}. \hfill \\ \end{gathered}  \right.
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  8. #8
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    Thank you Dmath, on the 2a (a+1) = a^2 +8, I'm lost to how I solve that one.
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  9. #9
    Senior Member DeMath's Avatar
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    \left. 1 \right){\text{ }}{y^3} + 3{y^2} - 4y - 12 = 0 \Rightarrow

    \Rightarrow {y^2}\left( {y + 3} \right) - 4\left( {y + 3} \right) = 0

    \Rightarrow \left( {y + 3} \right)\left( {{y^2} - 4} \right) = 0

    \Rightarrow \left[ \begin{gathered}y + 3 = 0, \hfill \\{y^2} - 4 = 0; \hfill \\ \end{gathered}  \right. \Rightarrow \left[ \begin{gathered}{x_1} =  - 3, \hfill \\{x_3} =  \pm 2. \hfill \\ \end{gathered}  \right.
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  10. #10
    Senior Member DeMath's Avatar
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    \left. 3 \right){\text{ 2}}a\left( {a + 1} \right) = {a^2} + 8 \Rightarrow {\text{2}}{a^2} + 2a = {a^2} + 8

    \Rightarrow {a^2} + {\text{2}}a + 1 = 9 \Rightarrow {\left( {a + 1} \right)^2} = 9

    \Rightarrow \left[ \begin{gathered}a + 1 =  - 3, \hfill \\a + 1 = 3; \hfill \\ \end{gathered}  \right. \Rightarrow \left[ \begin{gathered}{a_1} =  - 4, \hfill \\{a_2} = 2. \hfill \\ \end{gathered}  \right.
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  11. #11
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    I got the first part down, the 2a^2 +2a = a^2 +8, how do you go about getting it to be a^2 +2a +1 = 9 ?
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