HELP!!INTERSTING QUESTION (Form 4)

只是用符號的問題，大致如下：
let g(x) be 5x2-x-4.
g(x)= (x-1)(5x+4)

because g(x) is a factor of f(x),
f(g(x))=0
f(1)=0
f(-4/5)=0

f(1)= 5+p+q-12=0
p+q-7=0 ....(1)

以上都可以。
f(-4/5) = -2.56+0.64p-0.8q-12= 0
是-0.8q而不是+0.8q。
p-1.25q-22.75=0..... (2)
(1)-(2),
2.25q+15.75=0
q=-7
p=14

Of course, you can use the method of factor theorem, but I would like to solve the problem in the following way.

f(x)
= 5x^3 + px^2 + qx - 12
= 5x^3 - x^2 - 4x + x^2 + 4x + px^2 + qx - 12
= x(5x^2 - x - 4) + (p+1)x^2 + (q+4)x - 12
= x(5x^2 - x - 4) + 3{[(p+1)/3]x^2 + [(q+4)/3]x - 4}

As x(5x^2 - x - 4) is divisible by 5x^2 - x - 4,
for f(x) being divisible by 5x^2 - x - 4,
3{[(p+1)/3]x^2 + [(q+4)/3]x - 4} must be divisible by 5x^2 - x - 4.

Thus, we have
[(p+1)/3]x^2 + [(q+4)/3]x - 4 = 5x^2 - x - 4.

By comparing coefficients, we have
(p+1)/3 = 5, (q+4)/3 = -1.

Therefore, we have, p = 14, q = -7.

let g(x) be 5x^2-x-4.
g(x)= (x-1)(5x+4)

because g(x) is a factor of f(x),
f(g(x))=0
f(1)=0
f(-4/5)=0

f(1)= 5+p+q-12=0
p+q-7=0 ....(1)

f(-4/5) = -2.56+0.64p+0.8q-12=0
p+1.25q-22.75=0..... (2)
(2)-(1),
0.25q= 15.75
q=63
p=-56

2007-02-22 23:54:53 補充：
多謝下面位兄台幫我執漏~