How do you find all the zeros of #x^3+2x^2-2x-3#?

1 Answer
George C.
Feb 27, 2016

Use the rational root theorem to help find the first zero #x=-1#, then divide #f(x)# by #(x+1)# to find a quadratic and hence the other two zeros:

#x = (-1+-sqrt(13))/2#

Explanation:

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

By the rational root theorem, any rational zeros of #f(x)# will be expressible in the form #p/q# for some integers #p# and #q# with #p# a divisor of the constant term #-3# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1#, #+-3#

Trying each in turn, we find:

#f(1) = 1+2-2-3 = -2#

#f(-1) = -1+2+2-3 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

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

The remaining quadratic factor is of the form #ax^2+bx+c# with #a=1#, #b=1# and #c = -3#. This has zeros given by the quadratic formula:

#x = (-b+-sqrt(b^2-4ac))/(2a)#

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

#=(-1+-sqrt(13))/2#

Related questions
See all questions in Zeros
Impact of this question
1942 views around the world
You can reuse this answer
Creative Commons License