How do you find all the zeros of #f(x)=x^3+x^2-7x+2#?

1 Answer
Apr 16, 2016

Answer:

#f(x)# has zeros #x=2# and #x=-3/2+-sqrt(13)/2#

Explanation:

By the rational root theorem, any rational zeros of #f(x)# must be 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#

We find:

#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 factor the remaining quadratic expression by completing the square. I will multiply by #4# first to cut down on the fractions involved:

#4(x^2+3x-1)#

#=4x^2+12x-4#

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

#=(2x+3)^2-(sqrt(13))^2#

#=((2x+3)-sqrt(13))((2x+3)+sqrt(13))#

#=(2x+3-sqrt(13))(2x+3+sqrt(13))#

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