急~有數學唔識做??
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)
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)
收錄日期: 2021-04-23 20:58:29
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