Question

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

Answer 1

`(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]

Answer 2

`(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)`