數學恒等式:(2u-5w)^2 =??

Remember the formulas:
(a士b)^2=a^2士2ab+b^2
(a+b)(c+d)=ac+ad+bc+cd

(1)
(2u-5w)^2
=(2u)^2-2(2u)(5w)+(5w)^2
=4u^2-20uw+25w^2

(2)
(2x+3)^2
=(2x)^2+2(2x)(3)+3^2
=4x^2+12x+9

(3)
(2x+1)(3x-2)
=(2x)(3x)+(2x)(-2)+(1)(3x)+(1)(-2)
=6x^2-4x+3x-2
=6x^2-x-2

(4)
(3p+2q)(2p-q)
=(3p)(2p)+(3p)(-q)+(2q)(2p)+(2q)(-q)
=6p^2-3pq+4pq-2q^2
=6p^2+pq-2q^2

(5)
(4p-3s)(5p+2s)
=(4p)(5p)+(4p)(2s)+(-3s)(5p)+(-3s)(2s)
=20p^2+8ps-15ps-6s^2
=20p^2-7ps-6s^2

(2u - 5w)^2=4u^2 -20uw + 25w^2

(2x + 3)^2=4x^2 + 12x +9

(2x + 1)(3x - 2)=6x^2 + x -2

(3p + 2q)(2p - q)=6q^2 + q - 2q^2

(4p-3s)(5p+2s)=20p^2-7ps -6s^2

1.(2u - 5w)^2 = 4u^2-20uw+25w^2

2.(2x + 3)^2 = 4x^2+12x+9

3.(2x + 1)(3x - 2) = 6x^2-x-2

4.(3p + 2q)(2p - q) = 6p^2+pq-2q^2

5.(4p-3s)(5p+2s) = 20p^2-7ps-6s^2