A F. 2 Maths question - Expansion by using identities

A F. 2 Maths question - Expansion by using identities

Expand (-x/4-y/2)(8y-4x).

Please show your working steps clearly and explain your answer in details.


(-x/4 - y/2)(8y - 4x)

= [(-1/4)(x + 2y)][4(2y - x)] 【Extracting common constants】

= (-1/4)(4)(2y + x)(2y - x)

= (-1)[(2y)2 - x2] 【Using identity a2-b2=(a-b)(a+b)】

= -(4y2 - x2)

= x2 - 4y2

2007-01-09 22:48:56 補充：
小小補充：factorizaton (因式分解) 常用到的恆等式有a²-b² = (a-b)(a+b) ..... 最常用(a-b)² = a²-2ab+b²(a+b)² = a²+2ab+b²a³-b³ = (a-b)(a²+ab+b²)a³+b³ = (a+b)(a²-ab+b²)只是代不同的項至 a 和 b。

正確答案
(-x/4-y/2)(8y-4x).
=(-x/4-y/2)*4(2y-x)
=4*(-x/4-y/2)*(2y-x)
=(-x-2y)*(2y-x)
=-(x+2y)*(2y-x)
=-(2y+x)(2y-x)
=-(2y+x)^2

(-x/4-y/2)(8y-4x).
=(-x/4-y/2)*4(2y-x)
=4*(-x/4-y/2)*(2y-x)
=(-x-2y)*(2y-x)
=-(x+2y)*(2y-x)
=-(2y+x)(2y-x)
=-(2y+x)^2