Problema Solution
The polynomial ax^3+3x^2-3 and 2x^3-5x+a when divided by x-4 leaves remainder R1 & R2. find a of 2R1-R2=0.
Answer provided by our tutors
x-4) ax^3+3x^2+0x-3(ax^2+(4a+3)x + 4(4a+3)
ax^3-4ax^2
-----------------
(4a+3)x^2-0x
(4a+3)x^2-4(4a+3)x
-----------------------
4(4a+3)x-3
4(4a+3)-16(4a+3)
-----------------------
16(4a+3)-3
x-4)2x^3+0x^2-5x+a(2x^2+8x+27
2x^3-8x^2
-------------------
8x^2-5x
8x^2-32x
--------------
27x+a
27x-108
---------------
108+a
R1 = 16(4a+3)-3
R2 = 108+a
2R1 -R2 = 32(4a+3)-6-108-a = 0
128a+96-114-a = 0
127a = 18
a = 18/127