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!

2. 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.

3. 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.

4. 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.

5. 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.

6. 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!
$\displaystyle \left. 4 \right){\text{ }}2{x^3} + {x^2} = 32x + 16 \Rightarrow$

$\displaystyle \Rightarrow {x^2}\left( {2x + 1} \right) = 16\left( {2x + 1} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 16, \hfill \\2x + 1 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} = \pm 4, \hfill \\{x_3} = - \frac{1}{2}. \hfill \\ \end{gathered} \right.$

7. For number 2

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

$\displaystyle \Rightarrow {x^2}\left( {2x + 3} \right) = 4\left( {2x + 3} \right) \Rightarrow \left[ \begin{gathered}{x^2} = 4, \hfill \\2x + 3 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_{1,2}} = \pm 2, \hfill \\{x_3} = - \frac{3}{2}. \hfill \\ \end{gathered} \right.$

8. Thank you Dmath, on the 2a (a+1) = a^2 +8, I'm lost to how I solve that one.

9. $\displaystyle \left. 1 \right){\text{ }}{y^3} + 3{y^2} - 4y - 12 = 0 \Rightarrow$

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

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

$\displaystyle \Rightarrow \left[ \begin{gathered}y + 3 = 0, \hfill \\{y^2} - 4 = 0; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{x_1} = - 3, \hfill \\{x_3} = \pm 2. \hfill \\ \end{gathered} \right.$

10. $\displaystyle \left. 3 \right){\text{ 2}}a\left( {a + 1} \right) = {a^2} + 8 \Rightarrow {\text{2}}{a^2} + 2a = {a^2} + 8$

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

$\displaystyle \Rightarrow \left[ \begin{gathered}a + 1 = - 3, \hfill \\a + 1 = 3; \hfill \\ \end{gathered} \right. \Rightarrow \left[ \begin{gathered}{a_1} = - 4, \hfill \\{a_2} = 2. \hfill \\ \end{gathered} \right.$

11. 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 ?