急急急急急!!!! 20分數學polynomial

Q.1
(a)
f(x) = x³ - 6x² + 11x - 6
f(1) = (1)³ - 6(1)² + 11(1) - 6 = 0
(x - 1) is a factor of f(x).

By long division, f(x) ÷ (x - 1) = (x² - 5x + 6)
f(x) = (x - 1)(x² - 5x + 6)
f(x)= (x - 1)(x - 2)(x - 3)

g(x) = x³ - 9x² + 26x - 24
g(2) = (2)³ - 9(2)² + 26(2) - 24 = 0
(x - 2) is a factor of g(x).

By long division, g(x) ÷ (x - 2) = x² - 7x + 12
g(x) = (x - 2)(x² - 7x + 12)
g(x)= (x - 2)(x - 3)(x - 4)

(b)(i)
[f(x)/g(x)] [f(x - 1) / g(x + 1)]
= [(x - 1)(x - 2)(x - 3) / (x - 2)(x - 3)(x - 4)] [(x - 2)(x - 3)(x - 4) / (x -1)(x - 2)(x - 3)]
= 1

(b)(ii)
[f(x)/g(x)] - [f(x - 1) / g(x + 1)]
= [(x - 1)(x - 2)(x - 3) / (x - 2)(x - 3)(x - 4)] - [(x - 2)(x - 3)(x - 4) / (x- 1)(x - 2)(x - 3)]
= [(x - 1) / (x - 4)] - [(x - 4) / (x - 1)]
= [(x - 1)² - (x - 4)²] / (x - 1)(x - 4)
= 3(2x- 5) / (x - 1)(x - 4)


Q.2
(a)(i)
The question should be :
"If a + c + e = b + d, prove that f(x) is divisible by (x + 1)."

f(x) = ax⁴ + bx³ + cx² + dx + e

f(-1)
= a(-1)⁴ + b(-1)³ + c(-1)² + d(-1) +e
= (a + c + d) - (b + d)
= (b + d) - (b + d)
= 0

Since f(-1) = 0, then f(x) is divisible by (x + 1).

(a)(ii)
6 + 4 + 7 = 9 + 8
Put a = 6, b = 9, c = 4, d = 8 and e = 7
Then f(x) = 6x⁴ + 9x³ + 4x² + 8x + 7

f(-1) = 0
(6x⁴ + 9x³ + 4x² + 8x + 7) is divisible by(x + 1).

Put x = 10 :
[6(10)⁴ + 9(10)³ + 4(10)² + 8(10) + 7] is divisibleby [(10) + 1].
Hence, 69487is divisible by 11.

(b)(i)
g(x) = 4x⁴ + 9x³ + x² + 9x + 2
g(-1) = 4(-1)⁴ + 9(-1)³ + (-1)² + 9(-1) + 2 = -11
The required remainder = -11

g(x) + 11 is divisible by (x + 1).

Put x = 10 :
{[4(10)⁴ + 9(10)³ + (10)² + 9(10) + 2] + 11} isdivisible by [(10) + 1].
(49192 + 11) is divisible by 11.
Hence, 49192is divisible by 11.