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

2 Answers
Alan P.
May 11, 2016

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)

Nghi N.
May 11, 2016

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)

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