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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