Reminder Theorum?

【Question】
(a)
Let f(x) = ax⁴ + bx³ + cx² + dx + e.
(i) If a + c + e = b + d, prove that f(x) is divisible by x + 1.
(ii) Hence, show that 69487 is divisible by 11.

(b)
Let g(x) = 4x⁴ + 9x³ + x² + 9x + 2.
(i) Find the remainder when g(x) is divided by x + 1.
(ii) Hence, show that 49192 is divisible by 11.

【Solution】
(a) (i)
When f(x) is divided by x + 1, the remainder is
f(-1)
= a - b + c - d + e
= (a + c + e) - (b + d)
= 0
Therefore, f(x) is divisible by x + 1.

(a) (ii)
Consider a = 6, b = 9, c = 4, d = 8, e = 7.
Check that a + c + e = 6 + 4 + 7 = 17 = 9 + 8 = b + d.
By (a) (i), f(x) is divisible by x + 1.
For x = 10, that is, f(10) is divisible by 11.
That is, 6(10000) + 9(1000) + 4(100) + 8(10) + 7(1) = 69487 is divisible by 11.

(b) (i)
When g(x) is divided by x + 1, the remainder is
g(-1)
= 4 - 9 + 1 - 9 + 2
= 7 - 18
= -11

(b) (ii)
From (b) (i), it is known that
g(x) = (x + 1)Q(x) - 11 where Q(x) is a polynomial with integer coefficients.
Put x = 10,
g(10) = 11Q(10) - 11 = 11[Q(10) - 1].
Since g(10) and Q(10) - 1 are integers, g(10) is divisible by 11.
That is, 4(10000) + 9(1000) + 100 + 9(10) + 2 = 49192 is divisible by 11.