How do you find the product #(2/5y-4)^2#?

1 Answer
Jun 16, 2017

#(2/5y-4)=color(magenta)(4/25y^2-8/5y+16)#

Explanation:

Use the general relation #(a-b)^2=a^2-2ab+b^2#
with #a=2/5y# and #b=4#

or work it through in detail using the distributive property:
#(2/5y-4)^2#
#color(white)("XXX")=color(red)(""(2/5y-4))xxcolor(blue)(""(2/5y-4))#

#color(white)("XXX")=color(red)(2/5y)xxcolor(blue)(""(2/5y-4))color(red)(-4)xxcolor(blue)(""(2/5y-4))#

#color(white)("XXX")=(color(red)(2/5y)xxcolor(blue)(2/5y)-color(red)(2/5y)xxcolor(blue)4)-(color(red)4xxcolor(blue)(2/5y)-color(red)4xxcolor(blue)(""(-4)))#

#color(white)("XXX")=(4/25y^2-8/5y)-(8/5y-16)#

#color(white)("XXX")=4/25y^2-16/5y+16#