1.Let f(x)=x^3+g(x),where g(x) is a quadratic polynomial with real coefficients.When f(x) is divided by (x-1)(x-4) and when f(x) is divided by (x-4)^2,the remainders are -x+k and kx-10 respectively,where k is real number.Find
(a).k,
(b).g(x),
(c).the remainder when(g(x))^3 is divided by x+1.
2.Let f(x) be a polynomial of degree 4 with real coefficients.When f(x) is divided by x-2,the remainder is 4.When f(x) is divided by x+3,the remainder is -6.Let r(x) be the remainder when f(x) is divided by (x-2)(x=3).
(a).Find r(x).
(b).Let g(x)=f(x)-r(x).It is known that g(x) is divisible by x^2+1 and g(1)=-16.Find g(x)
2條pure maths問題
2010-09-19 7:54 am
回答 (2)
2010-09-19 6:35 pm
✔ 最佳答案
1a)f(x)=(x-1)(x-4)p(x)-x+kf(x)=(x-4)^2q(x)+kx-10-4+k=f(4)=4k-10k=2b)let g(x)ax^2+bx+cby long division, we haveb-16+8(a+8)=2b-4+5(a+5)=-1c-4(a+5)=2So, a=-8 b=18 c=-10g(x)=-8x^2+18x-10c)let h(x)=[g(x)]^3h(-1=[g(-1)]^3=-46656the remainder when(g(x))^3 is divided by x+1 is -46656
2010-09-19 10:54:59 補充：
2a)
f(x)=(x-2)p(x)+4
f(x)=(x+3)q(x)-6
f(x)=(x-2)(x+3)h(x)+r(x)
let r(x)=ax+b
f(2)=r(2)=4
f(-3)=r(-3)=-6
So,2a+b-4=0 and 3a-b-6=0
Thus, a=2 b=0
r(x)=2x
2010-09-19 10:55:06 補充：
b)
g(x)=f(x)-r(x)=(x-2)(x+3)h(x)
g(x)=(x^2+1)s(x)
Since h(x) and s(x) are a polynomial of degree 2 with real coefficients
So g(x)=k(x-2)(x+3)(x^2+1)
g(1)=-16
-8k=-16
k=2
2010-09-19 8:48 pm
太陽, can you explain in details on how to get:
b-16+8(a+8)=2
b-4+5(a+5)=-1
c-4(a+5)=2
b-16+8(a+8)=2
b-4+5(a+5)=-1
c-4(a+5)=2
收錄日期: 2021-04-13 17:31:53
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