How do you find the roots of #x^3+x^2-7x+2=0#?

1 Answer
Nov 11, 2016

Answer:

The roots are #2# and #-3/2 +- sqrt(13)/2#

Explanation:

Given:

#f(x) = x^3+x^2-7x+2#

By the rational roots theorem, any rational zeros of #f(x)# 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 zeros are:

#+-1, +-2#

Note that:

#f(2) = 8+4-14+2 = 0#

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

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

We can find the zeros of the remaining quadratic factor using the quadratic formula with #a=1#, #b=3# and #c=-1# :

#x = (-color(blue)(3) +-sqrt (color(blue)(3)^2 - 4(color(blue)(1))(color(blue)(-1))))/(2(color(blue)(1)))#

#color(white)(x) = (-3+-sqrt(9+4))/2#

#color(white)(x) = -3/2+-sqrt(13)/2#