If p (x) = 2x3 + ax2 + 3x – 5 and q (x) = x3 + x2 – 4x + a leave the same remainder when divided by (x – 2), show that
We have,
p (x) = 2x3 + ax2 + 3x – 5
q (x) = x3 + x2 – 4x + a
It is given in the question that, when p (x) and q (x) is divided by (x – 2) it leaves same remainder
∴ p (2) = q (2)
2 (2)3 + a (2)2 + 3 (2) – 5 = (2)3 + (2)2 – 4 (2) + a
2 × 8 + a × 4 + 3 × 2 – 5 = 8 + 4 – 4 × 2 + a
16 + 4a + 6 – 5 = 12 – 8 + a
16 + 4a + 1 = 4 + a
4a – a = 4 – 17
3a = - 13
a =
Hence proved