Is it possible to factor #y=x^2-12x+6 #? If so, what are the factors?

1 Answer
Apr 27, 2016

Answer:

Factors of #y=x^2-12x+6# are #y=(x-6+sqrt30)(x-6-sqrt30)#

Explanation:

In #y=x^2-12x+6# discriminant is #(-12)^2-4xx1xx6=144-24=120#, which is not a perfect square of a rational number and hence we cannot factorise it easily by splitting middle term.

Hence we can use quadratic formula to first obtain zeros of function, which are #(-(-12)+-sqrt120)/(2xx1)# or #(12+-2sqrt30)/2=6+-sqrt30# and hence factors of #y=x^2-12x+6# are #y=(x-6+sqrt30)(x-6-sqrt30)#