How do you find all the real and complex roots of #x^3 - x^2 - 7x + 15=0#?

1 Answer
George C.
Jan 29, 2016

Use the rational root theorem to help find one root, then factor that out to find the remaining two Complex roots:

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

Explanation:

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

By the rational root theorem, any rational roots of #f(x) = 0# must be expressible in the form #p/q# for some integers #p# and #q# where #p# is a factor of the constant term #15# and #q# is a factor of the coefficient #1# of the leading term.

That means that the only possible rational roots are:

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

Let's try the first few...

#f(1) = 1-1-7+15 = 8#

#f(-1) = -1-1+7+15 = 20#

#f(3) = 27-9-21+15 = 12#

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

So #x=-3# is a root of #f(x) = 0# and #(x+3)# is a factor of #f(x)#

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

The remaining quadratic factor has negative discriminant, so no Real roots, but we can find its Complex roots using the quadratic formula:

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

#= (4+-sqrt((-4)^2-(4*1*5)))/(2*1)#

#=(4+-sqrt(-4))/2 = 2+-i#

Related questions
See all questions in Complex Conjugate Zeros
Impact of this question
1893 views around the world
You can reuse this answer
Creative Commons License