How do you factor #z^2-20z+100#?

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Answer 1 · 2018-03-29T12:36:49

See process below

There are several ways to solve

1.- Using cuadratic formula

#az^2+bz+c=0# and his solutions #z=(-b+-sqrt(b^2-4ac))/(2a)#

In our case

#a=1; b=-20 and c=100#

#z=(20+-sqrt(400-400))/2=20/2=10# that's double root, so

#(z-10)(z-10)=(z-10)^2#

2.- Using notable identities. In this case

#(a-b)^2=a^2-2ab+b^2# we have #a=z# and #b=10#

#(z-10)^2=(z-10)(z-10)=z^2-2·10·z+10^2=z^2-20z+100#

Answer 2 · 2018-03-29T12:38:18

#(z-10)(z-10)#

By sum and product Means that two numbers whose sum is -#20# and product is 100

=#z^2-10z-10z+100#

=#z(z-10)-10(z-10)#

=#(z-10)(z-10)#

Hope this helps!