中二factorization and identity

1.5(x+y)^2-20(x+y)(x-y)+20(x-y)^2
Let x+y = a
x-y = b
5a^2 - 20ab + 20b^2
= 5(a^2 - 4ab + 4b^2)
= 5(a - 2b)^2
= 5[(x+y - 2(x-y)]^2
= 5(3y - x)^2
2.(ax+2)(x^2+bx+c)=ax^3+dx^2+e
ax^3 + 2x^2 + abx^2 + 2bx + acx + 2c = ax^3+dx^2+e
ax^3 + (2 + ab)x^2 + (2b + ac)x + 2c = ax^3 + dx^2 + 0x + e
2 + ab = d...(1)
2b + ac = 0...(2)

2c = e...(3)
let a = 1 , c = 2 , from(2) : b = - 1 , from(3) e = 4
from (1) : d = 1
let a = 3 , c = 4 , from(2) : b = - 6 , from(3) : e = 8
from (1) : d = 14



2009-10-08 23:56:02 補充：
Corrections

The last line d = - 16

2009-10-08 23:56:58 補充：
Corrections

The last line d = - 16

1).
Let (x+y)=a, (x-y)=b,

5(x+y)^2 - 20(x+y)(x-y) + 20(x-y)^2
=5a^2 - 20ab + 20b^2
=5[a^2 - 4ab + 4b^2]
=5(a-2b)^2

Sub (x+y)=a, (x-y)=b back, the equation becomes
5[(x+y)-2(x-y)]^2
=5(3y-x)^2



2).
LHS
=(ax+2)(x^2+bx+c)
=ax^3 + (ab+2)x^2 + (ac+2b)x + 2c

RHS
=ax^3 + dx^2 + e

For the identity to hold for any arbitrary value of x, LHS has to be equal to RHS. That means the coefficients has to be equal correspondingly, i.e.

x^3: a=a
x^2: (ab+2)=d
x^1: (ac+2b)=0
x^0: 2c=e

Any value of a,b,c,d & e which satisfies the above constraints is a set of answers.

1st set of answers: a=0, b=0, c=0, d=0, e=0
2nd set of answers: a=0, b=0, c=1, d=0, e=0

My 2 answer sets may look a bit tricky, you may think of some others.

2009-10-09 00:02:25 補充：
Sorry. My previous 2 answer sets for question 2 is incorrect.

1st set of answers: a=0, b=0, c=0, d=2, e=0
2nd set of answers: a=0, b=0, c=1, d=2, e=2