# Can someone give me the solution of that trinomial

1. May 24, 2006

### pendulum

Can I please have the solution of the trinomial

x^3+(11+(8/3))*x^2+(8/3)*(m+10)*x+(160/3)(m-1)=0

in terms of the unknown variable "t"?

I' ve heard that it can be done in 'mathematica'.

2. May 24, 2006

### daveb

There is no unknown variable "t". There is an "x" and there is an "m".

3. May 24, 2006

### pendulum

Sorry I meant "m"

4. May 24, 2006

### daveb

What have you done so far?

5. May 24, 2006

### pendulum

I don't understand what you mean. I' ve done nothing.
I don't have a mathematica I that's what you are asking.
I don't remember the Horner 'thing' or the polynomial division, and I was wondering if someone could help me.

6. May 24, 2006

### Curious3141

I've attached the maple output, you'll be able to see it as soon as one of the mentors approves it. I just didn't have the energy to format it into tex, anyway, this way there's no chance of transcription error.

#### Attached Files:

• ###### math.JPG
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7. May 28, 2006

### pendulum

Thanks Curious

8. Jun 29, 2006

### nn987

solution

with mathematica
Solve[x^3+(11+(8/3))*x^2+(8/3)*(m+10)*x+(160/3)(m-1)Š0,m]

m -> {160 - 80x - 41x^2 - 3x^3 / 8 (20+x) }

9. Jun 29, 2006

### VietDao29

We are solving the equation for x kin terms of m, not for m in terms of x. :)

10. Jul 2, 2006

### nn987

cubic equation

ok :
solved the equation for x in terms of m

#### Attached Files:

• ###### cubic.pdf
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66
11. Jul 7, 2006

### Robokapp

x^3+(11+(8/3))*x^2+(8/3)*(m+10)*x+(160/3)(m-1)=0

x^3+(11+(8/3))*x^2 all this can be written as...a

a+8mx/3+80x/3+160m/3-160/3=0
160/3-80x/3-a=8mx/3+160m/3
160-80x-3a=m(8x+160)

m={160-80x-3[x^3+(11+(8/3))*x^2]}/(8x+160)

and you can simplify it however you please. i don't see the difficulty since the cubic part didn't have an m in it lol.