數學題[F.4程度]

f(x) = Q(x)(6x^2 + x - 2) + (Ax + B)

f(x) = Q(x)(2x - 1)(3x + 2) + (Ax + B)

f(1/2) = A/2 + B = 4...(1)

f(-2/3) = -2A/3 + B = 11...(2)

Solving (1) and (2)

7A/6 = -7

A = -6

B = 7

So, the remainder is -6x + 7

first, you have to know the relation between divident, divisor, quotient, remainder...etc
Dividend = divisor x quotient + remainder
for example, 7 divide by 3, the remainder is 1
7 can be written as 2x3 + 1
for any f(x) = D(x) Q(x) + R(x) ..... general equation

second, you have to remember the degree of remainder should be less than the degree of divisor .... just like the remainder 1 should be less than the divisor 3 as example above.

for example, if divisor is (x^3 + x^2) which the degree is 3, the remainder should be with degree of 2 and else ....

f(x) = (2x-1) A(x) + 4 => f(1/2) = 4
f(x) = (3x+2) B(x) + 11 => f(-2/3) = 11

let f(x) = (6x^2 + x - 2) Q(x) + ax + b
f(x) = (2x-1)(3x+2) Q(x) + ax + b
put x = 1/2 => a/2 + b = 4 .... (1)
put x = -2/3 => -2a/3 + b = 11 .... (2)
(1) - (2) => 7a/6 = -7 => a = -6, then b = 7

我認為你應該學左 remainder theorem,否則點會做呢條數?
用remainder theorem就會出到f(x) = Q(x)(6x^2 + x - 2) + (Ax + B)
Q(x)係商,唔使理
001依家當個remainder 係 (Ax + B)
f(x) = Q(x)(2x - 1)(3x + 2) + (Ax + B) 係將上面條式既(6x^2 + x - 2) factorize