Answer :
to factor first multily 100 and 1 and get 100
then find what 2 numbers multiply to get 100 and add to get 20
the numbers are 10 and 10
so
split the center term up
100x^2+10x+10x+1
group
(100x^2+10x)+(10x+1)
undistribute
(10x)(10x+1)+(1)(10x+1)
undistribute/reverse distributive property
(10x+1)(10x+1)
(10x+1)^2
then find what 2 numbers multiply to get 100 and add to get 20
the numbers are 10 and 10
so
split the center term up
100x^2+10x+10x+1
group
(100x^2+10x)+(10x+1)
undistribute
(10x)(10x+1)+(1)(10x+1)
undistribute/reverse distributive property
(10x+1)(10x+1)
(10x+1)^2
since
this is a perfect square problem, the answer will be in the form (ax +
b) ^ 2. what can you tell from looking at the original equation? x^2
does not have a coefficient in front of it, so a = 1. the second term is
negative, so b must be negative. finally, the last number 100 is the
square of + or - 10, so b must be +10 or -10, but we already figured b
has to be negative, so b = -10