F4 Maths Dividing Polynomials2

1. A=x^4+6x^3-7x^2+3x+9 B=x^2+3x-2

　　　 1+3-14
　　　----------------------------
1+3-2 ) 1+6-7+3+9
　　　 1+3-2
　　　 ----------------
　　　　 3-5+3
　　　　 3+9-6
　　　　--------------------
　　　　　 -14 +9 + 9
　　　　　 -14-42+28
　　　　 --------------------
　　　　　　　 51-19

Quotient = x^2+3x-14
Remainder = 51x-19
(x^4+6x^3-7x^2+3x+9) = (x^2+3x-14) (x^2+3x-2)+51x-19


2. A=x^3+6x^2+6x-6 B=x^2+3x-3

　　　 1+3
　　　---------------------
1+3-3 ) 1+6+6-6
　　　 1+3-3
　　　 --------------
　　　　 3+9-6
　　　　 3+9-9
　　　　-------------
　　　　　　 3

Quotient = x+3
Remainder = 3
x^3+6x^2+6x-6 = (x+3) (x^2+3x-3)+3


Alternative method :
Let x^3+6x^2+6x-6≡(Px+Q) (x^2+3x-3)+Rx+S
x^3+6x^2+6x-6≡Px^3+3Px^2-3Px+Qx^2+3Qx-3Q+Rx+S
x^3+6x^2+6x-6≡Px^3+(3P+Q)x^2-(3P-3Q-R)x-3Q+S

Compare the coefficients :
P=1 . . . . . . . . . . . (1)
3P+Q=6 . . . . . . . . (2)
-(3P-3Q-R)=6
-3P+3Q+R=6 . . . . (3)
-3Q+S=-6 . . . . . . . (4)

Put (1) into (2) : 3+Q=6
Q=3

Put P=1 and Q=3 in to (3) : -3+9+R=6
R=0

Put Q=3 into (4) : -9+S=-6
S=3

Therefore x^3+6x^2+6x-6≡(x+3) (x^2+3x-3)+3


2009-09-28 09:02:12 補充：
It is quite obvious that :


In Q1,

x^4+6x^3-7x^2+3x+9 = (x^2+3x-14) (x^2+3x-2)+51x-19

A = (x^2+3x-14) B + 51x - 19


In Q2,

x^3+6x^2+6x-6 = (x+3) (x^2+3x-3)+3

A = (x+3) B + 3

2009-09-28 09:27:42 補充：
In Q1,

x^4+6x^3-7x^2+3x+9 = (x^2+3x-14) (x^2+3x-2)+51x-19

A = (x^2+3x-14) B + 51x - 19


In Q2,

x^3+6x^2+6x-6 = (x+3) (x^2+3x-3)+3

A = (x+3) B + 3

Is it no problem for Q1 ?

2009-09-27 23:20:52 補充：
wkho28 :

You haven't write the relationship between A and B.

I think Q1 have some word mistake.