Question

please help.

for the given polynomial p(x) and the given c use the remainder theorem to find p(c) P(x)=8x^4-5x^2-2,1/3

Answer 1

can you help?

Answer 2

@DanJS

Answer 3

is there difference between p and P?

Answer 4

Also you don't need remainder theorem to find P(1/3)

Answer 5

thats how it was written...

Answer 6

oh then how are p and P related?

Answer 7

i think its the same just a typo on the teachers part because its only in the written part once.

Answer 8

P(1/3) can be found by replacing the x's with 1/3

Answer 9

Evaluate P(1/3)

Answer 10

I think the remainder is 1/3 when you substitute x = c

Answer 11

so replace all x's with 1/3?

Answer 12

\[P(\frac{ 1 }{ 3 }) = 8\left( \frac{ 1 }{ 3 } \right)^4 -5\left( \frac{ 1 }{ 3 } \right)^2-2\]

Answer 13

is the question find P(c) given c is 1/3
or is the question find c such that P(c) is 1/3 ?*

Answer 14

use the remainder theorem to find P(C) so i believe its using 1/3 as a remainder

Answer 15

if it is find c such that P(c) is 1/3
then @campbell_st is right

Answer 16

if I'm right, then its a tough question

Answer 17

lol yep

Answer 18

i think its using the remainder

Answer 19

c is given as 1/3 in the question. P(x) is given. You are asked to find P(c) using the remainder theorem. You may be overthinking it. The asker may just not have been consistent with using the upper case P.

Answer 20

they're the same. Its a typo on the question.

Answer 21

so do i just replace all X with 1/3?

Answer 22

That's right.

Answer 23

I don't see what this question has to do with remainder theorem

like you can say use remainder theorem to find the remainder when
P(x)/(x-c) by replacing the x in P(x) with the value for c...


Like just evaluating P(c) alone is not using the remainder theorem

Answer 24

according to remainder theorem
\[f(x) = (x-c)g(x) + a\]

now you can easily find the solution
8x^4-5x^2-2 = (x-c)g(x) + a
p(c) = 0 + 1/3
p(c) = 1/3

Answer 25

The remainder theorem states that when dividing a polynomial P(x) by a binomial (x-c), the remainder is simply P(c). I believe that's what this question is asking. But I admit it's written in a very unusual way since there is no mention of the binomial x-c

Answer 26

so maybe they want us to do synthetic division or long division

Answer 27

\[\frac{P(x)}{x-\frac{1}{3}}\]