國中數學題目 因式分解

1.因式分解下列各式
(1)3(2x-3)^2 (5x-1)+2(3-2x)(1-5x)^2 (2)ab(x^2+y^2)-xy(a^2+b^2)

1-1

3(2x-3)^2 (5x-1)+2(3-2x)(1-5x)^2
=3(2x-3)^2 (5x-1)-2(2x-3)(5x-1)^2
= (2x-3)(5x-1) (6x-9-10x+2)
= (2x-3)(5x-1) (-4x-7)

1-2

ab(x^2+y^2) - xy(a^2+b^2)
=abx^2 + aby^2 - xya^2- xyb^2
=ax(bx-ay) - by(bx-ay)
=(bx-ay) (ax-by)

2.已知x-2是兩多項式A=5x^2+ax+2與B=x^2-x+b的公因式，則：
(1)a+b的值為何? (2)將2A-3B的結果做因式分解

2-1

利用餘式定理
將x=2代入多項式A和B

<<代入A>>
20+2a+2=0
2a = -22
a = -11

<<代入B>>
4-2+b=0
b = -2

a+b= -11-2 = -13 <answer>

2-2

2A-3B
=10x^2-22x+4-3x^2+3x+6
=7x^2-19x+10

用7x^2-19x+10除以x-2得:7x-5
所以7x^2-19x+10= (x-2)(7x-5) <answer>

如果有不懂可以再問!^^

factorize
3(2x-3)^2 (5x-1)+2(3-2x)(1-5x)^2
=(2x-3)(5x-1)(6x-9+2-10x)
=-(2x-3)(5x-1)(7+4x)
=-(10x^2-17x+3)(7+4x)
ab(x^2+y^2)-xy(a^2+b^2)
=abx^2+aby^2-xya^2-xyb^2
=ax(bx-ay) +by(ay-bx)
=ax(bx-ay)-by(bx-ay)
=(ax-by)(bx-by)

A=5x^2+ax+2
A=5x^2+7x+2
A=(x+1)(5x+2)
a=7
B=x^2-x+b
B=x^2-x+b
B=x^2-x-2
B=(x+1)(x-2)
b=-2
factorize
2(5x^2+7x+2)-3(x^2-x+b)
=2(x+1)(5x+2)-3(x+1)(x-2)
=-[3(x+1)(x-2)-2(x+1)(5x+2)]
=-[(x+1)(3x+3-10x-4)]
=(x+1)(7x+1)
=7x^2+8x+1

2010-11-28 15:57:00 補充：
so a+b
=7-2
=5
therefore it couldn't be affect that in Q2b

(1)(2X-3)(5X-1)(16X-11)
(2)(bX-aY)(aX-bY)
(3)令X-2=0，X=2帶入多項式A和B
得知a=-11，b=-2
求a+b=-13
2A-3B=(X-2)(7X-5)