I'm doing this question as part of some AS Maths revision, and keep getting -10 although apparently the answer is -3.
The remainder when 2x^3 + ax^2 + 4 is divided by x-2 is five more than when it is divided by x-1. Find a.
I'm pretty confident on the remainder theorem so it's probably just a silly mistake I'm making that I'm too tired to spot for myself!
My working is as follows anyway:
(x-2) --> f(2)
16 + 4a + 4 = 5r
20 + 4a = 5r
(x-1) --> f(1)
2 + a + 4 = r
6 + a = r
Solve simultaneously:
20 + 4a = 5(6 + a)
20 + 4a = 30 + 5a
4a = 10 + 5a
0 = 10 + a
a = -10
Help!
The remainder when 2x^3 + ax^2 + 4 is divided by x-2 is five more than when it is divided by x-1. Find a.
I'm pretty confident on the remainder theorem so it's probably just a silly mistake I'm making that I'm too tired to spot for myself!
My working is as follows anyway:
(x-2) --> f(2)
16 + 4a + 4 = 5r
20 + 4a = 5r
(x-1) --> f(1)
2 + a + 4 = r
6 + a = r
Solve simultaneously:
20 + 4a = 5(6 + a)
20 + 4a = 30 + 5a
4a = 10 + 5a
0 = 10 + a
a = -10
Help!
-
5 more means add 5 not multiply by 5 . f(2)=f(1)+5 , 20+4a=6+a+5 , 3a=-9 , a=-3 .