If polynomials ax3 + 3x2 ? 3 and 2x3 ? 5x + a when divided by (x - 4) leave the remainders as R1 and R2 respectively. Find the values of a in each of the following cases, if 1. R1 = R2 2. R1 + R2 = 0 3. 2R1 ? R2 = 0

Let the remainder be R1 and R2 as given :

R1 = ax^3 + 3x^2 – 3

Now, x – 4 => x = 4

R1 = a(4)^3 + 3(4)^2 – 3
R1 = 64a + 48 – 3
R1 = 64a + 45

R2 = 2x^2 – 5x + a
R2 = 2(4)^2 – 5(4) + a
R2 = 32 – 20 + a
R2 = 12 + a

R1 + R2 = 0

64a + 45 + 12 + a = 0
65a = – 57
a = – 57/65

2R1 – R2 = 0

2( 64a + 45 ) – 12 – a = 0
128a + 90 – 12 – a = 0
127a = – 78
a = – 78/127