Question

A polynomial p of degree 4 leaves a remainder of 2 when divided by x-1 and a remainder of 42 when divided by x+1. Find the polynomial p given that x-2 is a factor of p, p(-2)=128 and p(0)=16

Answer 1

let's right what we are given...
p is degree 4
\[p(x)=ax^4+bx^3+cx^2+d \\ \]
p(1)=2 <---p divided by (x-1) leaves remainder 2
p(-1)=42 <---p divided by (x+1) leaves remainder 42
p(2)=0 <---p divided by (x-2) leaves remainder 0
p(-2)=128
p(0)=16

Answer 2

oops missed the x term in my 4th degree polynomial

Answer 3

\[p(x)=ax^4+bx^3+cx^2+dx+e \]*

Answer 4

use p(0)=16 to find e

Answer 5

then use the others to give you a system of equations

Answer 6

I have to be right back

Answer 7

in the meantime show me what you have after using all of the conditions I mentioned above

Answer 8

I got it.... thank you soooo much.. that was helpful.....

Answer 9

np

Answer 10

I don't know how to solve it......... plz help

Answer 11

you got e right? e is the easiest to find..
the others one will probably cause more difficulty

Answer 12

p(0)=16 was given
replace x with 0

Answer 13

\[p(x)=ax^4+bx^3+cx^2+dx+e \\ p(0)=a(0)^4+b(0)^3+c(0)^2+d(0)+e \\ \text{ but } p(0)=16 \\ 16=0+0+0+0+e \\ 16=e\]

Answer 14

Let me see the other equations you get by pluggin in the conditions given