How do you factor the trinomial x^2+12x+12x+11x2+12x+12x+11?

2 Answers
Oct 26, 2017

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

Explanation:

x^2+12x+12x+11 = x^2+24x+11x2+12x+12x+11=x2+24x+11 or

x^2+24x+144 -144 +11x2+24x+144144+11 or

(x+12)^2 -133 or (x+12)^2 - (sqrt133)^2(x+12)2133or(x+12)2(133)2 or

(x+12+sqrt133)(x+12-sqrt133)(x+12+133)(x+12133) or

(x+12+11.53)(x+12-11.53)(x+12+11.53)(x+1211.53) or

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

Oct 26, 2017

(x+12+sqrt133)(x+12-sqrt133)(x+12+133)(x+12133)

Explanation:

"simplify by collecting like terms"simplify by collecting like terms

rArrx^2+12x+12x+11x2+12x+12x+11

=x^2+24x+11=x2+24x+11

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

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

x^2+24x+11=0x2+24x+11=0

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

rArrx=(-24+-sqrt(24^2-(4xx1xx11)))/2x=24±242(4×1×11)2

color(white)(rArrx)=(-24+-sqrt(576-44))/2x=24±576442

color(white)(rArrx)=(-24+-sqrt532)/2x=24±5322

color(white)(rArrx)=(-24+-2sqrt133)/2=-12+-sqrt133x=24±21332=12±133

rArrx^2+24x+11x2+24x+11

=(x-(-12+sqrt133))(x-(-12-sqrt133))=(x(12+133))(x(12133))

=(x+12-sqrt133)(x+12+sqrt133)=(x+12133)(x+12+133)