# Is it possible to factor #y=x^2/2+4x+8 #? If so, what are the factors?

##### 1 Answer

May 31, 2018

#### Answer:

#### 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#