# Help needed please

• Jan 20th 2009, 06:26 PM
MathMack
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!
• Jan 20th 2009, 06:31 PM
Mush
Quote:

Originally Posted by MathMack
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 $\displaystyle (y-a)$ is a factor of the polynomial. You would then divide the original polynomial by $\displaystyle (y-a)$ using long division/synthetic division. This will enable you to write the expression in the form:

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

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

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

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

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

4.) $\displaystyle 2x^3 + x^2 -32x -16=0$. Another cubic, so apply the rules I laid out for question 1! Hint: x = 4 is a root.
• Jan 20th 2009, 06:34 PM
MathMack
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.
• Jan 20th 2009, 06:42 PM
Mush
Quote:

Originally Posted by MathMack
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.
• Jan 20th 2009, 06:49 PM
MathMack
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.
• Jan 20th 2009, 07:00 PM
DeMath
Quote:

Originally Posted by MathMack
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!

$\displaystyle \left. 4 \right){\text{ }}2{x^3} + {x^2} = 32x + 16 \Rightarrow$

$\displaystyle \Rightarrow {x^2}\left( {2x + 1} \right) = 16\left( {2x + 1} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 16, \hfill \\2x + 1 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} = \pm 4, \hfill \\{x_3} = - \frac{1}{2}. \hfill \\ \end{gathered} \right.$
• Jan 20th 2009, 07:06 PM
DeMath
For number 2

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

$\displaystyle \Rightarrow {x^2}\left( {2x + 3} \right) = 4\left( {2x + 3} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 4, \hfill \\2x + 3 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} = \pm 2, \hfill \\{x_3} = - \frac{3}{2}. \hfill \\ \end{gathered} \right.$
• Jan 20th 2009, 07:10 PM
MathMack
Thank you Dmath, on the 2a (a+1) = a^2 +8, I'm lost to how I solve that one.
• Jan 20th 2009, 07:14 PM
DeMath
$\displaystyle \left. 1 \right){\text{ }}{y^3} + 3{y^2} - 4y - 12 = 0 \Rightarrow$

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

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

$\displaystyle \Rightarrow \left[ \begin{gathered}y + 3 = 0, \hfill \\{y^2} - 4 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_1} = - 3, \hfill \\{x_3} = \pm 2. \hfill \\ \end{gathered} \right.$
• Jan 20th 2009, 07:25 PM
DeMath
$\displaystyle \left. 3 \right){\text{ 2}}a\left( {a + 1} \right) = {a^2} + 8 \Rightarrow {\text{2}}{a^2} + 2a = {a^2} + 8$

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

$\displaystyle \Rightarrow \left[ \begin{gathered}a + 1 = - 3, \hfill \\a + 1 = 3; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{a_1} = - 4, \hfill \\{a_2} = 2. \hfill \\ \end{gathered} \right.$
• Jan 20th 2009, 07:40 PM
MathMack
http://www.mathhelpforum.com/math-he...964f9408-1.gif

http://www.mathhelpforum.com/math-he...ae940207-1.gif

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 ?