# F(x) question

1. May 25, 2004

### chris_tams

If i have f(x) = 2x^3 + 7x -3

And when f(x) is divided by (2x-k) I have -8

How do i find values for k?

2. May 25, 2004

### TALewis

I might be misunderstanding your question, but are you saying that

$$\frac{2x^3+7x-3}{2x-k}=-8$$

and you wish to solve for k as a function of x?

3. May 25, 2004

### Gokul43201

Staff Emeritus
How do you actually divide 2 polynomials ? Do the long division ...you'll find the remaider to be R = f(k), where f is a cubic in k. Equate this to -8 and you have the 3 values of k.

4. May 25, 2004

### matt grime

Yes and no there's no need to solve as a function of x: he's saying that if we write

2x^3+7x-3 = (2x-k)(x^2+ax+b)-8

what is k?

It's a division algorithm for polynomials question; by comparing coeffs on each side you ought to be able to solve a set of simulatneous equations to find k (and a and b)

5. May 25, 2004

### Gokul43201

Staff Emeritus
I'm assuming you mean the remainder is -8. It's not clear, and evidently, TALewis has interpreted differently.

And Matt's way is much easier.

Last edited: May 25, 2004
6. May 25, 2004

### TALewis

Ah, if you mean the remainder is -8, that makes much more sense.

7. May 25, 2004

### chris_tams

yes, what i mean is that when, 2x^3+7x-3/(2x-k) This equals some polynomial funtion with a remainder of -8. How do i work out what the polynomial is?????????

Last edited: May 25, 2004
8. May 25, 2004

### Gokul43201

Staff Emeritus
Look at matt's post again.

Expand the RHS and equate coefficients of the different powers of x, between the LHS and RHS. For instance, "Is there an x^2 term in the LHS ?...No !" So you equate the coefficient in the RHS to zero...and so on. Solving from these equations gives you k.

9. May 26, 2004

### chris_tams

Yes youth. Cheers !