代數......（急．．．）

圖片參考：http://hk.yimg.com/i/icon/16/18.gif

４（ｙ－ｚ)（２ｚ＋３ｙ）
=(4y-4z)(2z+3y)
=8yz+12y^2-8z^2-12zy
=12y^2-8z^2-4yz
=4(3y^2-2z^2-yz)
（４ｗ^2　－ｗ）（１－５ｗ＋７ｗ^2）
w(4w-1)(7w^2-5w+1)
=(28w^3-7w^2-20w^2+5w+4w-1)w
=(28w^3-27w^2+9w-1)w
=28w^4-27w^3+9w^2-w
３（２ｃ－５ｄ）（２ｃ^3－５ｃｄ－３ｄ）
=３(4c^4-10c^2d-6cd-10c^3d+25cd^2+15d^2)
=３(4c^4-10c^2d-6cd-10c^3d+25cd^2+15d^2)
=12c^4-30c^2d-18cd-30c^3d+75cd^2+45d^2

(1)
4(y-z)(2z+3y)
=4[(y-z)(2z)+(y-z)(3y)]
=4[2yz-2z^2+3y^2-3yz]
=4[-2z^2+3y^2-yz]
=-8z^2+12y^2-4yz

(2)
(4w^2-w)(1-5w+7w^2)
=(4w^2-w)(1)-(4w^2-w)(5w)+(4w^2-w)(7w^2)
=4w^2-w-20w^3+5w^2+28w^4-7w^3
=9w^2-27w^3+28w^4-w

(3)
3(2c-5d)(2c^3-5cd-3d)
=3[(2c-5d)(2c^3)-(2c-5d)(5cd)-(2c-5d)(3d)]
=3[4c^4-10c^3 d-10c^2 d+25cd^2-6cd+15d^2]
=12c^4-30c^3 d-30c^2 d+75cd^2-18cd+45d^2

the first one is 20yz+4(3y２次-2z２次)
the second one is w(28w3次+7w２次-w-1)
the third one is i don't know la!