try to keep up here:


let f(x) = kx³ + Ax² + Bx + C

(x-3) & (x+2) are factors

f(3) = 27k + 9A + 3B + C = 0 . . . . . . [1]

f(-2) = -8k + 4A - 2B + C = 0 . . . . . . [2]

[1] - [2],
=> 35k + 5A + 5B = 0
=> 7k + A + B = 0 . . . . . . [3]

f(-1) = -k + A - B + C
f(2) = 8k + 4A + 2B + C


-k + A - B + C = 4 x (8k + 4A + 2B + C )
=>-k + A - B + C = 32k + 16A + 8B + 4C
=>33k + 15A + 9B + 3C = 0
=> 11k + 5A + 3B + C = 0 . . . . . . [4]

-8k + 4A - 2B + C = 0
[4] - [2]
=> 19k + A + 5B = 0
=> A = -19k - 5B . . . . . . [5]

subst. [5] into [3] ;
=> 7k + (-19k - 5B) + B = 0
=> -12k - 4B = 0
=> B = -3k . . . . . . [6]



passes thru' (4, 12) ;
=> 64k + 16A + 4B + C = 12 . . . . . . [7]
27k + 9A + 3B + C = 0

[7] - [1] ;
=> 37k + 7A + B = 12 . . . . . . [8]

subst. [5] , [6] into [8] ;
=> 37k + 7[ -19k - 5(-3k) ] - 3k = 12
=> 34k + 7[ -4k ] = 12
=> k = 2

B = -6

A = -34 +30 = -8

-8k + 4A - 2B + C = 0
C = 8k - 4A + 2B
C = 16 + 32 - 12 = 36

f(x) = 2x³ - 8x² - 6x + 36

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