Solving a 4th Degree Polynomial: Finding Missing Variables Using Given Roots

[Corkery]

Homework Statement

My friend and I were having trouble with an equation, but I don't have the exact one with me so I will write one that is similar and is the same concept:

Equation: x^4 + 2x^3 - 9x^2 - ax + b

So basically they want you to find the two missing variables in the 4th degree polynomial. They give you two roots to work without of the possible 4.

The roots they give you are: (x-1) and (x-2)

Homework Equations

The Attempt at a Solution

I would think to use synthetic division and use one of the roots, but I'm not sure. Even if I'm coorect to use synthetic division what do I do once I reach the variables. I'm so lost.

 

[rock.freak667]

Well if i am understanding the question correctly then (if a,b,c and d are roots)
x^4 + 2x^3 - 9x^2 - 2x + 8= (x-a)(x-b)(x-c)(x-d) where a and b are known so that you can expand and equate coefficients

 

[Corkery]

the roots they give you are (x-1) and (x-2). Does that make a difference or would I continue to do what you said. Oh actually now that I think about it, I completely messed up on a few things.

 

[Corkery]

ok now its ok to re-read. Sorry for any inconvenience.

 

[Hurkyl]

I would think to use synthetic division and use one of the roots, but I'm not sure. Even if I'm coorect to use synthetic division what do I do once I reach the variables. I'm so lost.

Why can't you do the same thing you normally do? It's just arithmetic, isn't it?

 

[Corkery]

can you please explain into further depth in such a way that I will understand? Please.

 

[Hurkyl]

can you please explain into further depth in such a way that I will understand? Please.

(1) Do synthetic division up until the point where you get stuck.
(2) Imagine what you would do if you had a number instead of a variable.
(3) Do that to the variable.

 

[Corkery]

yes, but that means you can plug in any two sets of numbers that work and they they ARE equal to those varibles when they really arent.

 

[Hurkyl]

yes, but that means you can plug in any two sets of numbers that work and they they ARE equal to those varibles when they really arent.

I can't figure out what you're saying here.

 

[learningphysics]

f(x) = x^4 + 2x^3 - 9x^2 - ax + b

what is f(1) and f(2) ? you know that x=1 and x=2 are roots...

 

[Corkery]

yeah but when you plug that in how do you know what to plug for the variables to get the correct number for the variable. You could be plugging in two numbers that give you remainder of 0 but don't give you what the actual value of the variable unless you get 0 on the other root also. Is there a less tedious and time consuming way to do this?

 

[HallsofIvy]

yeah but when you plug that in how do you know what to plug for the variables to get the correct number for the variable. You could be plugging in two numbers that give you remainder of 0 but don't give you what the actual value of the variable unless you get 0 on the other root also. Is there a less tedious and time consuming way to do this?

?You know that the variables are 1 and 2 because you were told that 1 and 2 are roots! (In your original post, you say that (x-1) and (x-2) are roots. What you really mean are that (x-1) and (x-2) are factors of the polynomial so 1 and 2 are the roots.) You are given two of the roots and, as far as I can see, there is no reason to try to find the other two.

Setting f(1)= 0 and f(2)= 0 give you two linear equations to solve for a and b. Surely solving two linear equations is not "tedious and time consuming"!

 

[sutupidmath]

x^4 + 2x^3 - 9x^2 - ax + b,

what halls and others are saying is that since you are only asked to find the variables a and b and not required to find other two roots of the polynomial than you can simply do this
first let x=1 and since 1 is the root of say p(x)=x^4 + 2x^3 - 9x^2 - ax + b, than p(1)=0
in particular 1+2-9-a+b=0, b-a=6,
and when u plug in x=2, p(2)=0, so you basically have
16+16-36-2a+b=0, so b-2a=4,
all you need to do now, as halls stated is solve this linear system and you will get what you are looking for, which indeed are the values of a and b.

 

[Hurkyl]

Well, I was trying to point out that you can still synthetically divide, since that's where the OP claimed to have problems. For example,

x^4 + 2x^3 - 9x^2 - ax + b
= (x+1)(x^3 + x^2 - 10 x + (10 - a)) + (a + b - 10)

(of course, -1 is not known to be a root. I didn't want to do the OP's problem for him!)

Once he performed the synthetic division, he could invoke knowledge about what result he's supposed to get...