How do you factor #x^5 -4x^4 + 8x^3 - 14x^2 +15x -6 #?

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Answer 1 · 2015-11-12T07:38:05

Find factor #(x-1)# by examining the sum of the coefficients, divide and repeat, then factor by grouping to find:

#x^5-4x^4+8x^3-14x^2-14x^2+15x-6#

#= (x-1)(x-1)(x-2)(x^2+3)#

#f(x) = x^5-4x^4+8x^3-14x^2-14x^2+15x-6#

First notice that #f(1) = 0#, since the sum of the coefficients is #0#.

So #(x-1)# is a factor of #f(x)#.

#f(x) = (x-1)(x^4-3x^3+5x^2-9x+6)#

Again, the remaining quartic factor is divisible by #(x-1)# since its coefficients also add up to #0#.

#x^4-3x^3+5x^2-9x+6 = (x-1)(x^3-2x^2+3x-6)#

The remaining cubic factor is factorizable by grouping:

#x^3-2x^2+3x-6 = (x^3-2x^2)+(3x-6) = x^2(x-2)+3(x-2)#

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

The remaining quadratic factor has no linear factors with Real coefficients since #x^2+3 >= 3 > 0# for all #x in RR#

Putting this together we get:

#x^5-4x^4+8x^3-14x^2-14x^2+15x-6 = (x-1)(x-1)(x-2)(x^2+3)#

If you allow Complex coefficients then you can factor further using:

#x^2+3 = (x-sqrt(3)i)(x+sqrt(3)i)#

So:

#x^5-4x^4+8x^3-14x^2-14x^2+15x-6#

#= (x-1)(x-1)(x-2)(x-sqrt(3)i)(x+sqrt(3)i)#