maths f.4 (4)

2008-02-29 7:38 am
EXPAnd (1+px+X^2)^2 and express the ANS in descending powers of x

if the coefficient of x in the expansion of (1+px+X^2)^2 is -6
find the coefficient of x^2

consider 2 polynomials p(x) = x^3 -13x +12 and Q(x)= x^2 -4X +3
simplify
A) p(x) -(x-4)Q(X)
B) P(x)/x-3 -2Q(x)

suppose that 2x^3 +3x^2 -5X+c = (2x+A)(x+b) (X+2)
find the values of ABC

hence solve the equation 2x ^3 +3x^2 -5X+c=0

回答 (2)

2008-02-29 8:11 am
✔ 最佳答案
(1+px+x^2)^2
= 1+px+x^2 + px + (px)^2 + px^3 + x^2 + px^3 + x^4
= 1+2px+(p^2 +2)x^2+2px^3+x^4
= x^4 + 2px^3 + (p^2+2)x^2 + 2px + 1

the coefficient of x in the expansion of (1+px+X^2)^2 is -6
Therefore
2p=-6
p=-3

the coefficient of x^2
=p^2 + 2
=9+2
=11


A)
x^3-13x+12 - (x-4)(x^2-4x+3)
=x^3-13x+12 - (x^3-4x^2+3x-4x^2+16x-12)
=8x^2-32x+24

B)
(x^3-13x+12)/(x-3) - 2(x^2-4x+3)
=(x-3)(x^2+3x-4)/(x-3) - 2(x^2-4x+3)
=(x^2+3x-4) - 2(x^2-4x+3)
= -x^2+11x-10

2x^3 +3x^2 -5x+c = (2x+A)(x+B) (x+2)
2x^3 +3x^2 -5x+c = (2x+A)[x^2+(B+2)x+2B]
2x^3 +3x^2 -5x+c = 2x^3 + 2(B+2)x^2 + 4Bx + Ax^2 + A(B+2)x + 2AB

Compare the coefficient of x^2 , x and constant, we have

2(B+2)+A=3--------------(1)
A(B+2)+4B=-5------------(2)
C=2AB--------------------(3)
By (1) ,
A=3-2(B+2) --------(4)

Put (4) into (2)
[3-2(B+2)](B+2) + 4B = -5
3B+6 - 2B^2 - 8B - 8 + 4B = -5
2B^2 + B - 3= 0
(2B+3)(B-1)=0
B=1 OR -3/2
By (4),
When B=1 , A=-3
When B=-3/2 , A=2
Therefore C=2AB= -6

2x ^3 +3x^2 -5X+c=0
2x ^3 +3x^2 -5X-6=0
(2x+2)(x-3/2)(x+2)=0
Therefore , x=-1 or 3/2 or -2
2008-02-29 8:08 am
(1+px+x^2)^2
= (1+px+x^2)(1+px+x^2)
= (1+px+X^2) + (px+p^2x^2+px^3) + (x^2+px^3+x^4)
= x^4 + 2px^3 + (2 + p^2)x^2 + 2px + 1

Coefficient of x = -6
2p = -6
p = -3

Coefficient of x^2 = 2 + (-3)^2 = 11

A) p(x) -(x-4)Q(X)
= x^3 -13x +12 -(x - 4)(x^2 - 4x + 3)
= x^3 - 13x + 12 - x^3 + 4x^2 - 3x + 4x^2 - 16x + 12
= 8x^2 - 32x + 24

B) By long division, p(x) = (x - 3)(x^2 + 3x - 4)
P(x)/x-3 -2Q(x)
= x^2 + 3x - 4 - 2x^2 + 8x - 6
= -x^2 + 11x - 10

L.H.S.
= (2x+A)(x+B) (X+2)
= [2x^2 + (2B + A)x + AB] (x + 2)
= 2x^3 + 4x^2 + (2B + A)x^2 + 2(2B + A)x + ABx + 2AB
= 2x^3 + (2B + A + 4)x^2 + (4B + 2A + AB)x + 2AB

Comparing the like terms in L.H.S. and R.H.S.,
2B + A + 4 = 3 ... (1)
4B + 2A + AB = -5 ... (2)
2AB = C ... (3)

From (1), A = -1 - 2B ... (4)
Substitute (4) into (1),
4B - 2 - 4B + B(-1 - 2B) = -5
3 - B - 2B^2 = 0
2B^2 + B - 3 = 0
(2B + 3)(B - 1) = 0
B = -3/2 or 1
A = -1 - 2(-3/2) = 2
or A = -1 - 2(1) = -3
C = 2(2)(-3/2) = -6
or C = 2(1)(-3) = -6

The values of A,B and C are 2, -3/2 and -6 or -3,1,-6 respectively.

2x ^3 +3x^2 -5x + c = 0
2x ^3 +3x^2 -5x - 6 = 0
(2x+2)(x- 3/2)(x+2) = 0
x = -1 or x = 3/2 or x = -2


收錄日期: 2021-04-13 15:13:57
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080228000051KK03419

檢視 Wayback Machine 備份