If the polynomials ax3+3x2-3x and 2x3-5x+a when divided by (x-4) leave the remainder R1 and R2 respectively. Find the value of a in each of the following cases, if
(i) R1 = R2 (ii) R1 + R2=0
(iii) 2R1-R2 = 0.
Let, p (x) = ax3+3x2-3 and q (x) = 2x3-5x+a be the given polynomials.
Now,
R1 = Remainder when p (x) is divided by (x – 4)
= p (4)
= a (4)3 + 3 (4)2 – 3 [Therefore, p (x) = ax3+3x2-3]
= 64a + 48 – 3
R1 = 64a + 45
And,
R2 = Remainder when q (x) is divided by (x – 4)
= q (4)
= 2 (4)3 – 5 (4) + a [Therefore, q (x) = 2x3-5x+a]
= 128 – 20 + a
R2 = 108 + a
(i) Given condition is,
R1 = R2
64a + 45 = 108 + a
63a – 63 = 0
63a = 63
a = 1
(ii) Given condition is R1 + R2 = 0
64a + 45 + 108 + a = 0
65a + 153 = 0
65a = -153
a = 
(iii) Given condition is 2R1 – R2 = 0
2 (64a + 45) – (108 + a) = 0
128a + 90 – 108 – a
127a – 18 = 0
127a = 18
a = 