What are the possible rational roots of #x^3-4x^2+x+2=0# and then determine the rational roots?

1 Answer
Apr 2, 2017

Answer:

"Possible" rational roots: #+-1#, #+-2#

Rational root: #1#

Irrational roots: #3/2+-sqrt(17)/2#

Explanation:

Given:

#x^3-4x^2+x+2 = 0#

By the rational roots theorem, any rational roots of this cubic are expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #2# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

#+-1#, #+-2#

In addition, notice that the sum of the coefficients is zero. That is:

#1-4+1+2 = 0#

Hence we can tell that #x=1# is a zero and #(x-1)# a factor:

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

#color(white)(0) = (x-1)(x^2-3x-2)#

The remaining roots can be found using the quadratic formula.

#x = (3+-sqrt((-3)^2-4(1)(-2)))/(2*1) = 3/2+-sqrt(17)/2#