1. ## Cubic Equations

2x^3 + x^2 - 18x - 9 = 0

2. Originally Posted by JonathanEyoon

2x^3 + x^2 - 18x - 9 = 0

Use the rational roots theorem to suggest factors, in this case (x+1/2) appears to be a factor, then divide through to get:

2x^3 + x^2 - 18x - 9 = 2(x+1/2)(x^2-9) = 2(x+1/2)(x-3)(x+3)=0

so the roots are x=-1/2, x=3, x=-3.

RonL

3. Originally Posted by JonathanEyoon

2x^3 + x^2 - 18x - 9 = 0

this is a nice one, we can factor by groups.

$2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)$

now factor out the common $(2x + 1)$, we get:

$(2x + 1) \left( x^2 - 9 \right) = 0$

can you take it from here?

EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff

4. Originally Posted by Jhevon
this is a nice one, we can factor by groups.

$2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)$

now factor out the common $(2x + 1)$, we get:

$(2x + 1) \left( x^2 - 9 \right) = 0$

can you take it from here?

EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff
Welcome back slow coach

RonL

5. Originally Posted by Jhevon
this is a nice one, we can factor by groups.

$2x^3 + x^2 - 18x - 9 = x^2 (2x + 1) - 9(2x + 1)$

now factor out the common $(2x + 1)$, we get:

$(2x + 1) \left( x^2 - 9 \right) = 0$

can you take it from here?

EDIT: beaten by CaptainBlack. The long vacation made me even slower with this stuff

{ -3 , -1/2 , 3} correct?

Another question is, usually in a cubic equation, using the method you used is always gonna be the more common method to solving these right?

6. Originally Posted by JonathanEyoon
{ -3 , -1/2 , 3} correct?
yes (as CaptainBlack said)

7. Originally Posted by Jhevon
yes (as CaptainBlack said)

Thanks alot to the both of you

8. Originally Posted by JonathanEyoon
Another question is, usually in a cubic equation, using the method you used is always gonna be the more common method to solving these right?
Most cubic equations cannot be solved using the methods that CaptainBlack
and Jhevon used, though probably the majority that you will meet on low
level HS and University maths courses will be. The usual way of solving them
is numerically, and if we are feeling adventurous using Ferro-Tartaglia-Cardano method.

Note a cubic with real coefficients always has at least one real root/factor

RonL

9. Originally Posted by CaptainBlack
Most cubic equations cannot be solved using the methods that CaptainBlack
and Jhevon used, though probably the majority that you will meet on low
level HS and University maths courses will be. The usual way of solving them
is numerically, and if we are feeling adventurous using Ferro-Tartaglia-Cardano method.

Note a cubic with real coefficients always has at least one real root/factor

RonL
Must cubics with real coefficients always have an odd number of real roots? If so, is it generally true that for all polynomials of odd degree with real coefficients there are an odd number of real roots and for those of even degree an even number of real roots?

10. Originally Posted by rualin
Must cubics with real coefficients always have an odd number of real roots? If so, is it generally true that for all polynomials of odd degree with real coefficients there are an odd number of real roots and for those of even degree an even number of real roots?
Yes. The reason is that for a polynomial with real coefficients any complex roots show up as complex conjugates. So the complex roots always come in pairs. (I have forgotten the name for this theorem, sorry!)

-Dan

11. Originally Posted by topsquark
Yes. The reason is that for a polynomial with real coefficients any complex roots show up as complex conjugates. So the complex roots always come in pairs. (I have forgotten the name for this theorem, sorry!)

-Dan
Like halfwits

RonL