How do you factor #3x^2 - 12x - 4#?

2 Answers
May 11, 2016

Answer:

Note that there are no "pretty" factors for this expression.
The best I could come up with is
#color(white)("XXX")3(x-2+(2sqrt(12))/3)(x-2-(2sqrt(12))/3)#

Explanation:

The rational root theorem reveals no rational roots
so we can apply the quadratic formula to find zeros for this expression.

For a quadratic of the form #ax^2+bx+c#
zeros occur at #x=(-b+-sqrt(b^2+4ac))/(2a)#

Applying this to the given equation results in
#color(white)("XXX")x=2+-(2sqrt(12))/3#
So
#color(white)("XXX")(x-(2+(2sqrt(12))/3)) and (x-(2-(2sqrt(12))/3))# are both factors of the given expression.

Dividing the original expression by these factors leaves an additional factor of #3#
So the complete factorization is
#color(white)("XXX")3(x-2+(2sqrt(12))/3)(x-2-(2sqrt(12))/3)#

May 11, 2016

Answer:

#3(x - 2 - (4sqrt3)/3)(x - 2 + (4sqrt3)/3)#

Explanation:

Use the improved quadratic formula (Socratic Search)
#D = d^2 = b^2 - 4ac = 144 + 48 = 192 = 64(3)# --> #d = +-8sqrt3#
There are 2 real roots:
#x = -b/(2a) +- d/(2a) = 12/6 +- (8sqrt3)/6 = 2 +- (4sqrt3)/3#
#x1 = 2 + (4sqrt3)/3# and #x2 = 2 - (4sqrt3)/3#
Factored form:
#y = a(x - x1)(x - x2) = 3(x - 2 - 4sqrt3/3)(x - 2 + 4sqrt3/3)#