factorization

1. x^2(y+z)+y^2(z+x)+z^2(x+y)+2xyz= yx^2 + zx^2 + zy^2 + xy^2 + xz^2 + yz^2 + 2xyz= (yx^2 + xy^2) + (xz^2 + yz^2) + ( zx^2 + zy^2 + 2xyz)= xy(x + y) + (x + y)z^2 + z(x + y)^2= (x + y) (xy + z^2 + z(x + y))= (x + y) (z + x)(z + y)
2. (x+y+z)(yz+zx+xy)-xyz= xyz + zx^2 + yx^2 + zy^2 + xyz + xy^2 + yz^2 + xz^2 + xyz - xyz= zx^2 + yx^2 + zy^2 + xy^2 + yz^2 + xz^2 + 2xyz= (z + y)x^2 + (z + x)y^2 + (x + y)z^2 + 2xyzBy question 1) := (x + y) (z + x)(z + y)
3. a^2b-a^2c-ac^2-ab^2-b^2c+bc^2+2abc= (ba^2 - ab^2) + (bc^2 - ac^2) - ca^2 - cb^2 + 2abc= ab(a - b) - (a - b)c^2 - c(a^2 - 2ab + b^2)= (a - b) (ab - c^2 - c(a - b))= (a - b)(b - c)(a + c)
4.x^2-y^2+2zx+2yz+2y-2z-1(arrange in powers of x)= x^2 + 2zx + (2yz+2y-2z-1 - y^2)= x^2 + 2zx + (- y^2 + 2(z+1)y - (2z+1))= x^2 + 2zx + ( - y + 1)(y - 2z - 1)I think the question should be = x^2 - 2zx + ( - y + 1)(y - 2z - 1)= (x - y + 1)(x + y - 2z - 1)
5. (xy-1)(x-1)(y+1)-xy= (xy-1) (xy-1 - x + y) - xy= (xy - 1)^2 + (y - x)(xy - 1) - xy= (xy - 1 + y)(xy - 1 - x)
arrange in powers of a and b6. a^2+2ab+b^2-x^2-6x-9= (a+b)^2 - (x^2 + 6x + 9)= (a+b)^2 - (x+3)^2= (a+b-x-3)(a+b+x+3)
7. I think the question is ab+2ac+3b^2+6bc-5a-13b+4c-10 = a(b+2c) + 3b(b+2c) - 5a - 13b + 4c - 10= (b+2c)(a+3b) + [2(b+2c) - 5(a+3b)] + (2)(-5)= (b+2c - 5)(a+3b + 2)

2010-07-22 17:52:47 補充：
Sorry Q4 is no problem :

4.x^2-y^2+2zx+2yz+2y-2z-1(arrange in powers of x)
= x^2 + 2zx + (2yz+2y-2z-1 - y^2)
= x^2 + 2zx + (- y^2 + 2(z+1)y - (2z+1))
= x^2 + 2zx + ( y - 1)(- y + 2z + 1)

= (x + y - 1)(2zx - y + 2z + 1)

2010-07-22 18:07:57 補充：
typing error : (x + y - 1)(2zx - y + 2z + 1)


should be = (x + y - 1)(x - y + 2z + 1)

2010-07-22 18:34:46 補充：
Corr.

5. (xy-1)(x-1)(y+1)-xy
= (xy-1) (xy-1 + x - y) - xy
= (xy - 1)^2 + (x - y)(xy - 1) - xy
= (xy - 1 + x)(xy - 1 - y)