f.3 math 因式分解　講解釋

1. ax + ay + 2by + 2bx
= a (x + y) + 2b (y + x) --- Take out the common factor
= (x + y) (a + 2b)

2. 25 + 60n + 36b^2
= 5^2 + 2 (5) (6b) + (6b)^2 --- Identity
= (5 + 6b)^2

3. -343x^3 + 64y^3
= - (343x^3 - 64y^3)
= - [ (7x)^3 - (4y)^3 ]
= - (7x - 4y) [ (7x)^2 + (7x) (4y) + (4y)^2 ] --- Identity
= - (7x - 4y) (49x^2 + 28xy + 16y^2)
= (4y - 7x) (49x^2 + 28xy + 16y^2)

4. b^2 - 6b + 8
= (b - 2) (b - 4) --- Cross Method

我唔知點講比你listen!但我know how to計1. 2. 4.(我只是f.2)

1. ax+ay+2by+2bx
=a(x+y)+2b(y+x)
=a(x+y)+2b(x+y)
=(a+2b)(x+y)

2. 25+60b+36b^2
=5^2+60b+(6b)^2
=5^2+(6b)^2+60b
=(5+6b)(5-6b)+60b

4. b^2-6b+8
=b(b-6)+8

1. ax+ay+2by+2bx
= a(x+y)+2b(y+x)
= (x+y)(a+2b)

2. 25+60b+36b^2
=(5+6b^2)

3. -343x^3+64y^3
= -64(64x^3+y^3)
= -64(8x+y)(64x^2-128xy+y^2)

4. b^2-6b+8
= (b-4)(b-2)

1) ax+ay+ 2by+2bx

= (ax+ay) + ( 2by+2bx )

= a ( x+y )+ 2b (x+y )

= ( a + 2b ) ( x +y )

抽common factor


2) 25+60b+36b^2

= 36b^2 + 60b +25

= ( 6b )^2 + (2)(6b)(5) + (5)^2

= ( 6b + 5 )^2

跟據 ( x + y )^2 = x^2+2xy+y^2


3) -343x^3+64y^3

= - ( 343x^3 + 64 y^3)


= - ( 7x + 4y) [ ( 7x )^2 - (7x)(4y) + (4y)^2 ]

= - ( 7x + 4y ) ( 49x^2 - 28xy + 16y^2 )

跟據 x^3+y^3 = ( x+y) ( x^2 -xy +y^2)


4.b^2-6b+8

這是 x 相乘

b - 2

b - 4
______
- 6b

所以 = (b-8)(b-2)

2007-10-31 18:28:39 補充：
第3題= - ( 7x 4y ) ( 49x^2 - 28xy 16y^2 )= ( 4y - 7x) ( 49x^2 - 28xy 16y^2 )

1.ax+ay+2by+2bx
=a(x+y)+2b(x+y)
=(x+y)(a+2b)
2.25+60b+36b^2
=(6b+5)^2
3.-343x^3+64y^3
=(4y-7x)(16y^2+28xy+49y^2)
4.b^2-6b+8
=(b-4)(b-2)