How do you factor completely: #3x^2 + 12x + 7#?

1 Answer
Aug 15, 2015

Answer:

It depends on what kinds of numbers we are using.

Explanation:

#3x^2 + 12x + 7#

Using integers (or rational numbers) this quadratic cannot be factored any further. It is already factored completely.
(The only possibilities are #(3x+1)(x+7)# and #(3x+7)(x+1)#, neither of which works.)

If we are using Real numbers or Complex numbers, we can factor the quadratic by finding its zeros.

Solve
#3x^2 + 12x + 7 = 0# using your choice of completing the square of the quadratic formula.

#x = (-(12) +- sqrt ((12)^2-4(3)(7)))/(2(3))#

# = (-12 +- sqrt (144-84))/6 = (-12 +- sqrt 60)/6 = (-12 +- 2sqrt 15)/6#

# = -2 +- sqrt 15/3#

The zeros are:

#z_1 = -2 + sqrt 15/3# and z_2 = -2 - sqrt 15/3#

The factors are:

#(x-z_1)(x-z_2) = (x -(-2 + sqrt 15/3))(x- (-2 - sqrt 15/3))#

# = (x + 2 - sqrt 15/3))(x + 2 + sqrt 15/3))#