How do you solve #12x ^ { 3} + 5x ^ { 2} - 6x + 1= 0#?

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Answer 1 · 2016-12-11T11:36:28

The roots are #-1#, #1/3# and #1/4#

Notice that #12 = 5 + 6 + 1#, so is #+-1# a solution?

Yes, with #x = -1# we find:

#12(-1)^3+5(-1)^2-6(-1)+1 = -12+5+6+1 = 0#

So #(x+1)# must be a factor:

#12x^3+5x^2-6x+1 = (x+1)(12x^2-7x+1)#

Then note that #12 = 3*4# and #7 = 3+4#

So we find:

#12x^2-7x+1 = (3x-1)(4x-1)#

So the other two roots are #x = 1/3# and #x = 1/4#