View Full Version : F(x) question
chris_tams
May25-04, 10:50 AM
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?
TALewis
May25-04, 11:23 AM
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?
Gokul43201
May25-04, 11:29 AM
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.
matt grime
May25-04, 11:30 AM
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)
Gokul43201
May25-04, 11:31 AM
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.
TALewis
May25-04, 11:45 AM
Ah, if you mean the remainder is -8, that makes much more sense.
chris_tams
May25-04, 11:49 AM
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?????????
Gokul43201
May25-04, 07:43 PM
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.
chris_tams
May26-04, 01:54 AM
Yes youth. Cheers !
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