How do you solve #3x^3 -2x^2 -12x + 8 = 0#?

2 Answers
Jul 6, 2017

Answer:

With a large polynomial like this, you'll need to factor in order to isolate the x and solve for it.

Explanation:

This is the equation you're given: #3x^3-2x^2-12x+8=0#

Separate the polynomial into two parts, and factor that way.
#(3x^3-2x^2)+(-12x+8)=0#
#x^2(3x-2)-4(3x-2)#
#(3x-2)(x^2-4)=0#

You can factor out #(x^2-4)# to #(x+2)(x-2)#.

The fully factored polynomial is #(3x-2)(x+2)(x-2)#.

But remember... the question asks you to solve the equation.
Set the factored polynomial to zero and solve for x (you will end up with three values of x):
#(3x-2)(x+2)(x-2)=0#

#3x-2=0# ---> #x=2/3#
#x+2=0# ---> #x=-2#
#x-2=0# ---> #x=2#

Your final answers are #x=2/3,-2,2#.

Jul 6, 2017

Answer:

#x=-2#

#x=2#

#x=2/3#

Explanation:

#3x^3 -2x^2 -12x +8 =0#

Common factors

#3x(x+2)(x-2)-2(x+2)(x-2)=0#

#(3x-2)(x+2)(x-2) = 0#

Three solutions (B.S#=#both sides)

(1) #" "x+2=0" "# (#-2# B.S)

#x=-2#

(2) #" "x-2=0" "# (#+2# B.S)

#x=2#

(3) #" "3x-2=0" "# (#+2# B.S)

#3x=2" "# (#div3# B.S)

#x=2/3#

So, the problem has 3 solutions, #x=-2#, #x=2#, and #x=2/3#.

-Fish