Question

The polynomial p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7 when divided by x + 1 leaves the remainder 19. Find the values of a. Also find the remainder when p(x) is divided by x + 2.

Answer 1

Given, p(x) = x⁴ – 2x³ + 3x² – ax + 3a – 7

When we divide p(x) by x + 1, then we get the remainder p(–1).

Now, p(–1) = (–1)⁴ – 2(–1)³ + 3(–1)² – a(–1) + 3a – 7

= 1 + 2 + 3 + a + 3a – 7

= 4a – 1

p(–1) = 19

⇒ 4a – 1 = 19

⇒ 4a = 20

∴ a = 5

∴ Required polynomial = x⁴ – 2x³ + 3x² – 5x + 3(5) – 7  ...[Put a = 5 on p(x)]

= x⁴ – 2x³ + 3x² – 5x + 15 – 7

= x⁴ – 2x³ + 3x² – 5x + 8

When we divide p(x) by x + 2, then we get the remainder p(–2)

Now, p(–2) = (–2)⁴ – 2(–2)³ + 3(–2)² – 5(–2) + 8

= 16 + 16 + 12 + 10 + 8

= 62

Hence, the value of a is 5 and remainder is 62.