22x maybe?
what is the question anyway?
factor it?
put in an equal sign and solve it
find zeros of parabola ?
by the way
x^2 + 22 x + 121 = (x+11)^2
(x+11)^2 -y^2 = (x+11-y)(x+11+y)
because
a^2-b^2 = (a-b)(a+b)
I need help with this problem, and I think it's a perfect square trinomial problem.
X^2+22+121-y^2
I know 11 is the square of 121.
1 answer