Question 1170553
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x^2 = x squared


p(x) = ax^2 + bx + c


When we divide p(x) over (x-k), the remainder is p(k) according to the remainder theorem.


p(x) is divisible by (x-1), so the remainder is 0
So p(1) = 0
This means p(x) = (x-1)*q(x) for some polynomial q(x) 



When we divide p(x) over (x+1), the remainder is 2
So p(-1) = 2, since k = -1 here.


When we divide p(x) over (x-2), the remainder is 8. 
So p(2) = 8



We have these function statements
p(1) = 0
p(-1) = 2
p(2) = 8


This is another way of saying we know the three points 
(1,0)
(-1,2)
(2,8)
are on the p(x) curve.


Plug in (x,y) = (1,0) into p(x) to get
p(x) = ax^2 + bx + c
p(1) = a(1)^2 + b(1) + c
0 = a(1)^2 + b(1) + c
a+b+c = 0
Solve for c
c = -a-b


Plug (-1,2) into p(x)
p(x) = ax^2 + bx + c
p(-1) = a(-1)^2 + b(-1) + c
2 = a - b + c
a - b + c = 2
Plug in c = -a-b
a - b + c = 2
a - b + (-a-b) = 2
-2b = 2
b = -1


Use that b value to get
c = -a-b
c = -a-(-1)
c = -a+1


Now plug (x,y) = (2,8) into p(x)
p(x) = ax^2 + bx + c
p(2) = a(2)^2 + b(2) + c
p(2) = 4a + 2b + c
8 = 4a + 2b + c
4a + 2b + c = 8


Plug in c = -a+1 and b = -1. Solve for 'a'
4a + 2b + c = 8
4a + 2(-1) + (-a+1) = 8
4a - 2 - a + 1 = 8
3a - 1 = 8
3a = 8+1
3a = 9
a = 9/3
a = 3


So,
c = -a+1
c = -3+1
c = -2


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Answers:
a = 3 
b = -1
c = -2
The polynomial is p(x) = 3x^2-x-2

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