F.2 math factorize quick,,plz help

1) (2x-y)^2 -x^2
=4x^2-4xy+y^2-x^2
=3x^2-4xy+y^2
=(x-y)(3x-y)
2) (2x+y)^2 - (x+2y)^2
=4x^2+4xy+y^2-x^2-4xy-4y^2
=3x^2-3y^2
=3(x+y)(x-y)
3)1st way:
4(2x+y)^2+4(2x+y)+1
=[2(2x+y)+1]^2
=(4x+2y+1)^2
2nd way:
let (2x+y) be A
4(2x+y)^2+4(2x+y)+1
=4A^2+4A+1
=(2A+1)^2
=[2(2x+y)+1]^2
=(4x+2y+1)^2

As you see, in Q3, the 1st way and the 2nd way are actually the same. But if you cannot really understand what's going on in the 1st way, the 2nd way is easier to understand.

Hope I can help you.

1)2(2x-y)^2-X^2
=(2x-y)(2x-y)-x^2
=(2x-y)(2x-y-x^2)

2)(2X+y)^2-(x+2y)^2
=(2x+y)(2x+y)-(x+2y)(x+2y)
=(2x+y)(x+2y-1)

3)4(2x+y)^2+4(2x+y)+1
=4(2x+1)(2x+y)+4(2x+y)+1
=(2x+y)(2x+y)[4(2x+y)+1]
=(2x+y)(8x+4y+1)

1) (2x-y)^2 -x^2
= (2x-y+x)(2x-y-x)
= (3x-y(x-y)

2) (2x+y)^2 - (x+2y)^2
= (2x+y+x+2y)(2x+y-x-2y)
= (3x+3y)(x-y)
= 3(x+y)(x-y)

3) 4(2x+y)^2+4(2x+y)+1
=[2(2x+y)]^2 + 2[2(2x+y)] + 1^2
=[2(2x+y)+1]^2
=(4x+2y+1)^2

個個答案都一樣

1)
(2x-y)^2 -x^2
= [(2x-y)+x] [(2x-y)-x]
= (2x-y+x)(2x-y-x)
= (3x-y)(x-y)

2)
(2x+y)^2 - (x+2y)^2
= [(2x+y)+(x+2y)][(2x+y)-(x+2y)]
= (2x+y+x+2y)(2x+y-x-2y)
= (3x+3y)(x-y)
= 3(x+y)(x-y)

3)
4(2x+y)^2+4(2x+y)+1
= [2(2x+y)+1]^2
= (4x+2y+1)^2

有錯請指正!

1) (2x-y)^2 -x^2
= (2x - y + x)(2x - y - x)
= (3x - y)(x - y)

2) (2x+y)^2 - (x+2y)^2
= (2x + y + (x + 2y))(2x + y - (x + 2y))
= (3x + 3y)(x - y)
= 3(x + y)(x - y)

3) 4(2x+y)^2+4(2x+y)+1
= [2(2x + y)]^2 + 2[2(2x + y)][1] + 1^2
= [2(2x + y) + 1]^2
= (4x + 2y + 1)^2

1) (2x-y)^2 -x^2
= (2x-y+x)(2x-y-x)
= (3x-y(x-y)

2) (2x+y)^2 - (x+2y)^2
= (2x+y+x+2y)(2x+y-x-2y)
= (3x+3y)(x-y)
= 3(x+y)(x-y)

3) 4(2x+y)^2+4(2x+y)+1
=[2(2x+y)]^2 + 2[2(2x+y)] + 1^2
=[2(2x+y)+1]^2
=(4x+2y+1)^2

1) (2x-y)^2 -x^2
= (2x-y+x)(2x-y-x)
= (3x-y(x-y)

2) (2x+y)^2 - (x+2y)^2
= (2x+y+x+2y)(2x+y-x-2y)
= (3x+3y)(x-y)
= 3(x+y)(x-y)

3) 4(2x+y)^2+4(2x+y)+1
=[2(2x+y)]^2 + 2[2(2x+y)] + 1^2
=[2(2x+y)+1]^2
=(4x+2y+1)^2