# Thread: Solving a polynomial problem without guess & check!

1. ## Solving a polynomial problem without guess & check!

Hey everyone,

I have the following problem and I know I could solve with guess & check, but how would I solve it to be able to prove my answer?

Problem:

Find the range of values of 'c' such that the function f(x) = 2x3 - x2 - 6x + c has a zero between x = 0 and x = 1

Thanks everyone!

2. Originally Posted by qcom
Hey everyone,

I have the following problem and I know I could solve with guess & check, but how would I solve it to be able to prove my answer?

Problem:

Find the range of values of 'c' such that the function f(x) = 2x^3 - x^2 - 6x + c has a zero between x = 0 and x = 1
in future, use the caret symbol ^ to indicate exponents, or learn Latex.

$f(x) = 2x^3 - x^2 - 6x + c$

$f(0) = c$

$f(1) = 2 - 1 - 6 + c = c - 5$

for a zero to exist between $x = 0$ and $x = 1$, there has to be a sign change of the function values

if $f(0) = c > 0$ , then $f(1) = c-5 < 0$

$0 < c < 5$

looking at the other possibility ...

if $f(0) = c < 0$ , then $f(1) = c-5 > 0$

looks like that possibility can't happen ... do you see why?

3. Sorry for the messy exponents, what is Latex by the way?

Now, I understood how you logically proceeded with the problem, now I think it's not possible because the question calls for the zeros to be between 0 and 1 and that would require a sign change, yet a sign change is impossible between those two integers?

4. Originally Posted by qcom
Sorry for the messy exponents, what is Latex by the way?

Now, I understood how you logically proceeded with the problem, now I think it's not possible because the question calls for the zeros to be between 0 and 1 and that would require a sign change, yet a sign change is impossible between those two integers?
I'm afraid you don't understand ... I checked two possible cases.

Case #1 : f(0) positive and f(1) negative. this case worked as shown below ...

$f(x) = 2x^3 - x^2 - 6x + c$

$f(0) = c$

$f(1) = 2 - 1 - 6 + c = c - 5$

for a zero to exist between $x = 0$ and $x = 1$, there has to be a sign change of the function values ...

if $f(0) = c > 0$ , then $f(1) = c-5 < 0$

$0 < c < 5$

Case #2 : f(0) negative and f(1) positive does not work.

I suggest you graph f(x) using values of c between 0 and 5 and confirm for yourself that a zero for f(x) occurs between x = 0 and x = 1.

5. OK, I see where I was confused, thanks for your help and patience!