How do you factor #2x^4-9x^3+19x^2-15x#?

1 Answer
Aug 19, 2016

Answer:

#2x^4-9x^3+19x^2-15x#

#=x(2x-3)(x^2-3x+5)#

#=x(2x-3)(x-3/2-sqrt(11)/2i)(x-3/2+sqrt(11)/2i)#

Explanation:

#f(x) = 2x^4-9x^3+19x^2-15x = x(2x^3-9x^2+19x-15)#

Let #g(x) = 2x^3-9x^2+19x-15#

By the rational roots theorem, any rational zeros of #g(x)# (and therefore of #f(x)#) are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-15# 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, +-5/2, +-3, +-5, +-15#

In addition note using Descartes' rule of signs, we find that this cubic has #1# or #3# positive Real zeros and no negative Real zeros.

So the only possible rational zeros are:

#1/2, 1, 3/2, 5/2, 3, 5, 15#

Trying each in turn, we find:

#g(3/2) = 2(3/2)^3-9(3/2)^2+19(3/2)-15#

#=27/4-81/4+114/4-60/4 = 0#

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

#2x^3-9x^2+19x-15 = (2x-3)(x^2-3x+5)#

The zeros of the remaining quadratic can be found using the quadratic formula:

#x = (3+-sqrt((-3)^2-4(1)(5)))/(2*1)#

#=(3+-sqrt(9-20))/2#

#=3/2+-sqrt(11)/2i#

Hence factors:

#(x-3/2-sqrt(11)/2i)(x-3/2+sqrt(11)/2i)#