Ray
Lv 6

# F4 Maths Dividing Polynomials2

In each question, divide A by B, and write the relationship between the two.

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

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

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• wkho28
Lv 7

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