Question

How do you solve (10-5r)^2 using fctoring

Answer 1

Look, that 10 - 5r = 5(2-r). So:

`(10-5r)^2 = \left[ 5(2-r)\right]^2 = 5^2 \cdot (2-r)^2 =25(2-r)^2\\ \hbox{These bracket we will delete using this formula:} \\ (a-b)^2 = a^2 - 2ab + b^2 \\ \hbox{So we've got:} \\ 25 \cdot (r^2 - 2 \cdot 2r + 2^2)=\boxed{25(r^2-4r+4)} \\ \hbox{Or if you want to have it without bracket:} \\ 25r^2-100r+100`

Answer 2

(10-5r)^2=(10-5r)(10-5r) or
25r^2-100r+100
divide the whole thing by 5 and get
5(5r^2-20r+20)
divide the whole thing by 5 again
25(r^2-4r+4)
factor by finding what two numbers add to get -4 and multiply to get 4 (-2 and -2 work)
r^2-4r+4=(r-2)(r-2) or 25((r-2)^2)