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