Question

If p(x) = x⁴ + ax³ + bx² + cx + d = 0 and p(1) = 1, p(2) = 2, p(3) = 3. Find p(-2) + p(6).

Answer 1

P(x) = x⁴ + ax³ + bx² + cx + d

\(\because\) P(1) = 1

⇒ 1 + a + b + c + d = 1

⇒ a + b + c + d = 0---(1)

∵ P(2) = 2

⇒ 16 + 8a + 4b + 2c + d = 1

⇒ 8a + 4b + 2c + d = -15---(2)

∵ P(3) = 3

⇒ 81 + 27a + 9b + 3c + d = 3

⇒ 27a + 9b + 3c + d = -78--(3)

P(-2) + P(6) = (16 - 8a + 4b - 2c + d) + (1296 + 216a + 36b + 6c + d)

 = 1312 + 208a + 40b + 4c + 2d

 = 1312 + 16(27a + 9b + 3c + d) - 30(8a + 4b + 2c + d) + 16(a + b + c + d)

 = 1312 + 16 x -78 - 30 x -15 + 16 x 0 (From equations (1), (2), (3))

 = 1312 - 1248 + 450 + 0

 = 514