Third degree equation, one answer give, how do I find the others?

Question

The equation is 12x(^3) - 23x(^2) -3x + 2 = 0.

One of the solutions is x=2, how can I find the rest? I've tried several methods, but I can't seem to factor it in any way.

Thanks in advance!

Answer 1

I agree with norman.  Use synthetic division which drops the degree down to two, then either factor the quadratic equation or use quadratic formula.  Good luck to you !

Answer 2

2]....12 -23 -3 +2

........... 24..2..-2

......12....1.-1...0

   12x^3 - 23x^2 - 3x + 2

=(x-2)(12x^2 + x -1)

=(x-2)(4x-1)(3x+1)

Answer 3

12x³ - 23x² - 3x + 2 = 0

As long as you got that one solution, you'll find it easier:

First, divide the polynomial by (x - 2):

(12x³ - 23x² - 3x + 2) / (x - 2)

I'm going to use synthetic division (I hope you can do that because I can't show you here). You should get:

12x² + x - 1

Therefore, your third degree equation now becomes:

(x - 2)(12x² + x - 1) = 0

Now solve for that quadratic to get your other 2 solutions:

x = [-b ± √(b² - 4ac)] / 2a

x = [-1 ± √(1² - 4*12*-1)] / 2*12

x = [-1 ± √(1 + 48)] / 24

x = (-1 ± √49) / 24

x = (-1 ± 7) / 24

x = 6/24 = 1/4

x = -8/24 = -1/3

So your third degree polynomial now becomes:

(x - 2)(x + 1/3)(x - 1/4) = 0

===========================

Your solutions are:

x = 2

x = -1/3

x = 1/4