Let g(x)=px^3+qx^2-28x+15, where p and q are constants. 2x^2-11x+5 is a factor of g(x)
(A) find the values of p and q
(B) solve the equation g(x)=0
F.4 POLYNOMIALS
2010-11-23 2:01 am
回答 (2)
2010-11-23 2:15 am
✔ 最佳答案
(a) px^3+qx^2-28x+15 =Q(x)(2x^2-11x+5)=Q(x)(2x-1)(x-5)Sub. x=1/2=>p/8+q/4+1=0 ; Sub. x=5=>125p+25q-125=0
Solve it, we have p+2q+8=0 ; 5p+q-5=0
p+2(5-5p)+8=0 => p=2, q=-5
(b) g(x)=0
2x^3-5x^2-28x+15=0
(x+3)(2x-1)(x-5)=0
x=-3,1/2,5
2010-11-23 2:17 am
(A). Because (px^3+qx^2-28x+15)=(2x^2-11x+5)(x+3)
(px^3+qx^2-28x+15)=(2x^3-5x^2-28x+15)
Therefore, p=2, q=-5
(B). For g(x)=0
(2x^3-5x^2-28x+15)=0
(2x^2-11x+5)(x+3)=0
x+3=0 or 2x-1=0 or x-5=0
x=-3 or x=0.5 or x=5
(px^3+qx^2-28x+15)=(2x^3-5x^2-28x+15)
Therefore, p=2, q=-5
(B). For g(x)=0
(2x^3-5x^2-28x+15)=0
(2x^2-11x+5)(x+3)=0
x+3=0 or 2x-1=0 or x-5=0
x=-3 or x=0.5 or x=5
收錄日期: 2021-04-26 14:00:52
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101122000051KK02464