Question

How do I do (5y-4)^2 with steps? Thank you!

Answer 1

$(5y-4)^2 \\ \\ (5y)^2 - 2 \times 5y \times 4 + 4^2 \ \textless \ -- square \ of \ difference \\ \\ 5^2y^2-2 \times 5y \times 4 + 4^2 \ \textless \ -- distributive \ property \\ \\ 25y^2 - 2 \times 5y \times 4 + 4^2 \ \textless \ -- simplify \\ \\ 25y^2 - 2 \times 5y \times 4 + 16 \ \textless \ -- simplify \\ \\ 25y^2 - 40y + 16 \ \textless \ -- simplify \\ \\ Answer: 25y^2 - 40y + 16$

Answer 2

first, you have to turn (5y-4)^2 into (5y-4)(5y-4).
From there, you would FOIL, or multilpy the 5y by the 5y, which is 25y^2, the multiply the 5y by the -4, which is -20y, then multiply the -4 by the 5y which is -20y, and then multiply the -4 by the -4, which is 16. It would then look like this: 25y^2 - 20y - 20y +16. Then you would add (or in this case, subtract) like terms. You would add -20y and -20y, which is -40y. The end expression would be 25y^2 - 20y +16. Hope this helps!