How do you use the rational root theorem to find the roots of #f(x)=x^3+x^2-8x-6 #?

1 Answer
Mar 23, 2018

Answer:

The zeros of #f(x)# are #-3#, #1+sqrt(3)# and #1-sqrt(3)#

Explanation:

Given:

#f(x) = x^3+x^2-8x-6#

The rational root theorem tells us that any rational zero of #f(x)# is expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #-6# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-2, +-3, +-6#

Trying each of these in turn, we eventually find:

#f(-3) = (color(blue)(-3))^3+(color(blue)(-3))^2-8(color(blue)(-3))-6 = -27+9+24-6 = 0#

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

#x^3+x^2-8x-6 = (x+3)(x^2-2x-2)#

We can find the zeros of the remaining quadratic by completing the square and using the difference of squares identity:

#A^2-B^2 = (A-B)(A+B)#

with# A=(x-1)# and #B=sqrt(3)# as follows:

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

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

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

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

Hence the other two zeros are:

#x = 1+-sqrt(3)#