Is it possible to factor #y=x^2/2+4x+8 #? If so, what are the factors?
1 Answer
May 31, 2018
Explanation:
#"take out a "color(blue)"common factor "1/2#
#y=1/2(x^2+8x+16)#
#x^2+8x+16" is a "color(blue)"perfect square"#
#=(x+4)(x+4)=(x+4)^2#
#y=1/2x^2+4x+8=1/2(x+4)^2#
