[難題] 因式分解

You can refer to your math book solutions
Formula
a(b+c)+d(b+c)
=(b+c)(a+d)

Q1

ab(3x-1)-bc(3x-1)
=(3x-1)(ab-bc)
=b(3x-1)(a-c)

(3x-1) is the common factor

Q2

5(3m-n)^2-(n-3m)
=5(3m-n)^2+(3m-n)
=(3m-n)(5+3m-n)

Replace the n and 3m of behind first, and 5 will not be multipled

Q3

9mx-96ny+144my-6nx
=9m(x+16y)-6n(16y+x)
=9m(x+16y)-6n(x+16y)
=(x+16y)(9m-6n)
=3(x+16y)(3m-2n)

If the numbers replace in (+), the (-) will not be changed

Q4

16m^2y-6m^2x-3mx+8my
=2m^2(8y-3x)-m(3x-8y)
=-2m^2(3x-8y)-m(3x-8y)
OR 2m^2(8y-3x)+m(8y-3x)
=-(3x-8y)(2m^2+m)
=-m(3x-8y)(2m+1)

When you change the (8y-3x) to (3x-8y), it turns to minus

2006-11-05 23:50:14 補充：
Q2A mistake5(3m-n)^2-(n-3m)=5(3m-n)^2+(3m-n)=(3m-n)(5+3m-n)Is not rightShould be=(3m-n)[5(3m-n)+(3m-n)]=(3m-n)(15m-5n+3m-n)=(3m-n)(18m-6n)=6(3m-n)(3m-n)=6(3m-n)^2

2006-11-05 23:54:58 補充：
Q2 has misake again, sorry...=(3m-n)[5(3m-n) (3m-n)]=(3m-n)(15m-5n 3m-n)=(3m-n)(18m-6n)=6(3m-n)(3m-n)=6(3m-n)^2Is not rightThe behind of (3m-n) turned to 1=(3m-n)[5(3m-n) 1]=(3m-n)(15m-5n 1)The 5 is not the common factor in (15m-5n 1), because of the 1

1. ab(3x - 1) - bc(3x - 1)
= (3x - 1) (ab - bc)

2. 5(3m - n)^2 - (n - 3m)
= 5(3m - n)^2 + (3m - n)
= (3m - n) [5(3m-n) + 1]
= (3m - n) (15m - 5n + 5)

3. 9mx - 96ny + 144my - 6nx
= 9mx - 6nx + 144my - 96ny
= 3x (3m - 2n) + 48y (3m - 2n)
= (3m - 2n) ( 3x + 48y)

4. 16m^2y - 6m^2x - 3mx + 8my
= 16m^2y - 6m^2x + 8my - 3mx
= 2m^2 (8y - 3x) + m(8y - 3x)
= 2m2^ + m (8y - 3x)