How do you factor the trinomial #x^2+12x+12x+11#?

2 Answers
Binayaka C.
Oct 26, 2017

#(x+23.53)(x+0.47)#

Explanation:

# x^2+12x+12x+11 = x^2+24x+11# or

# x^2+24x+144 -144 +11# or

#(x+12)^2 -133 or (x+12)^2 - (sqrt133)^2# or

#(x+12+sqrt133)(x+12-sqrt133)# or

# (x+12+11.53)(x+12-11.53)# or

#(x+23.53)(x+0.47)# [Ans]

Jim G.
Oct 26, 2017

#(x+12+sqrt133)(x+12-sqrt133)#

Explanation:

#"simplify by collecting like terms"#

#rArrx^2+12x+12x+11#

#=x^2+24x+11#

#"this does not factor with integer coefficients so"#

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

#x^2+24x+11=0#

#"with "a=1,b=24,c=11#

#rArrx=(-24+-sqrt(24^2-(4xx1xx11)))/2#

#color(white)(rArrx)=(-24+-sqrt(576-44))/2#

#color(white)(rArrx)=(-24+-sqrt532)/2#

#color(white)(rArrx)=(-24+-2sqrt133)/2=-12+-sqrt133#

#rArrx^2+24x+11#

#=(x-(-12+sqrt133))(x-(-12-sqrt133))#

#=(x+12-sqrt133)(x+12+sqrt133)#

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