How do you factor completely #y=2x^3 - 11x^2 + 14x - 3#?

1 Answer
Apr 24, 2016

Answer:

#2x^3-11x^2+14x-3 = (2x-3)(x-2-sqrt(3))(x-2+sqrt(3))#

Explanation:

#f(x) = 2x^3-11x^2+14x-3#

By the rational root theorem, any zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-3# and #q# a divisor of the coefficient #2# of the leading term.

That means that the only possible rational zeros are:

#+-1/2#, #+-1#, #+-3/2#, #+-3#

Trying each in turn, we eventually find:

#f(3/2) = 27/4-99/4+21-3 = (27-99+84-12)/4 = 0#

So #x=3/2# is a zero and #(2x-3)# a factor:

#2x^3-11x^2+14x-3 = (2x-3)(x^2-4x+1)#

The remaining quadratic factor can be factored by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a=(x-2)# and #b=sqrt(3)# as follows:

#x^2-4x+1 = x^2-4x+4-3#

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

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

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

Putting it all together:

#2x^3-11x^2+14x-3 = (2x-3)(x-2-sqrt(3))(x-2+sqrt(3))#