How do you find the zeros of #x^3-x^2-7x+15=0#?

1 Answer
May 19, 2016

Answer:

#x=-3# and #x=2+-i#

Explanation:

#f(x) = x^3-x^2-7x+15#

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 #15# and #q# a divisor of the coefficient #1# of the leading term.

So the only possible rational zeros are:

#+-1#, #+-3#, #+-5#, #+-15#

Trying each of these in turn, we find:

#f(-3) = -27-9+21+15 = 0#

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

#x^3-x^2-7x+15 = (x+3)(x^2-4x+5)#

We can factor #x^2-4x+5# by completing the square and using the difference of squares identity:

#a^2-b^2=(a-b)(a+b)#

with #a=(x-2)# and #b=i# as follows:

#x^2-4x+5 = x^2-4x+4+1 = (x-2)^2-i^2 = (x-2-i)(x-2+i)#

Hence zeros:

#x = 2+-i#