factor polynomial 100y^2-49?

it is a difference of two squares: x² - y² = (x+y)(x-y)
so it is: (10y + 7)(10y - 7)

100y² - 49 = 0
100y² = 49
10y = 7
y = 7/10

= 10y + 7
= 10y - 7

Answer: Factors (10y + 7)(10y - 7)

Proof:
= (10y + 7)(10y - 7)
= 100² - 70y + 70y - 49
= 100² - 49

Use DOTS (Difference Of Two Squares)

100y^2=(10y)(10y)
49=(7)(7)

Use those square roots to add the two and subtract the two multiplied together.
(10y-7)(10y+7)

(10y-7)(10y+7)

(10y - 7)(10y + 7)

a^2 - b^2 = (a + b)(a - b)

100y^2 - 49
= (10y)^2 - 7^2
= (10y + 7)(10y - 7)

(10y-7)(10y+7)

This problem is a little interesting actually if you look at it...
So we have 100y^2-49 notice that 100 and 49 are perfect squares (10^2 = 100 and 7^2 =49). Thus we have:

100y^2 - 49 = [(10^2)(y^2) - (7^2)]
and we want to factor this polynomial, we notice there is no middle term with a 1st degree coefficient for y, so we can assume that its a (a+y)(a-y) kind of format, just a good place to start. Plus this is format is confirmed by the fact that we have those perfect squares present. So moving on to the Solution:

100y^2 - 49
= (10^2)(y^2) - (7^2)
= (10y-7)(10y+7)

(And you can check this answer by using the F.O.I.L method)
(10y-7)(10y+7) = 100y^2 + 70y - 70y -49 = 100y^2 - 49

100y^2-49

= 100y^2 - 70y + 70y -49 =0
= 10y(10y-7) + 7(10y-7) =0
= (10y+7)(10y-7)=0
therefore (10y+7)=0

and
(10y-7)=0


this implies
y= -10/7
and
y= 10/7