F.2因式分解

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

4(2x+y)^2+4(2x+y)+1
=4(4x^2+y^2)+4(2x+y)+1
=16x^2+4y^2+8x+4y+1
=8x(2x+1)+4y(y+1)+1

4(2x+y)^2+4(2x+y)+1

= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2----------- using the identies

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

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

4(2x+y)^2+4(2x+y)+1


= [2(2x + y)]^2 + 2(2)(2x + y) + 1^2-----------------應用恆等式:a^2+2ab+b^2


= [2(2x + y) + 1]^2----------------------------------------十字相乘


= (4x + 4y + 1)^2 -----------------------------------------答案


十字相乘的過程:

2(2x+y)↘1
2(2x+y)↗1
－－－－－－－－
2(2x+y)+2(2x+y)=4(2x+y)=2(2)(2x+y)

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

因為 a^2+2ab+b^2 = (a+b)^2