急~有數學唔識做??

2010-02-25 6:36 am
1.設f(x)=4x3次方+kx2次方-243,其中k為常數,已知x+3為f(x)的因式
(a)求K的值
(b)因式分解f(x)

2.設f(x)=x3次方-2x2次方-9x+18
(a)求f(2)
(b)因式分解f(x)

要有過程

回答 (2)

2010-02-26 6:05 am
✔ 最佳答案
As follows:


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2010-02-25 1:33 pm




1.設f(x)=4x^3+kx^2-243,其中k為常數,已知x+3為f(x)的因式

(a)求k的值

Sol

x+3=0 =>x=-3

f(-3)=4*(-3)^3+k(-3)^2-243=0

-108+9k-243=0

9k=351

k=39

(b)因式分解f(x)

Sol

4x^3+39x^2-243

=(4x^3+12x^2)+(27x^2+81x)+(-81x-243)

=4x^2(x+3)+27x(x+3)-81(x+3)

=(4x^2+27x-81)(x+3)

=(4x-9)(x+9)(x+3)



2.設f(x)=x^3-2x^2-9x+18

(a)求f(2)

Sol

f(2)=8-8-18+18=0

(b)因式分解f(x)

x^3-2x^2-9x+18

=(x^3-2x^2)+(-9x+18)

=x^2(x-2)-9(x-2)

=(x^2-9)(x-2)

=(x-3)(x+3)(x-2)







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