How can you factorize #x^2 + 4x -4# ?

I found #(x+2+2sqrt2)(x+2-2sqrt2)# by using delta but isn't there sth better? Thanks

2 Answers
Jan 2, 2018

Explanation:

substitute: a=1
b=4
c=-4

Jan 2, 2018

#(x+2+2sqrt2)(x+2-2sqrt2)#

Explanation:

#"to factorise find the roots using the "color(blue)"quadratic formula"#

#"given a quadratic in standard form ";ax^2+bx+c=0#

#"then the roots can be found using the quadratic formula"#

#•color(white)(x)x=(-b+-sqrt(b^2-4ac))/(2a)#

#x^2+4x-4=0" is in standard form"#

#"with "a=1,b=4" and "c=-4#

#rArrx=(-4+-sqrt(16+16))/2=(-4+-sqrt32)/2=(-4+-4sqrt2)/2#

#rArrx=-2+-2sqrt2larrcolor(blue)"roots(zeros)"#

# rArr(x-(-2-2sqrt2))(x-(-2+2sqrt2))" factors"#

#rArrx^2+4x-4=(x+2-2sqrt2)(x+2-2sqrt2)#