What are the zeros in x^3+3x^2-3x-5?

1 Answer
Jan 7, 2016

Answer:

#{-1, -1+sqrt(6), -1-sqrt(6)}#

Explanation:

Applying the rational roots theorem, we find that any rational roots of the given expression will be of the form #p/q# where #p# is a divisor of #-5# and #q# is a positive divisor of #1#. Then, the possible rational roots are #+-1# and #+-5#.

Trying these out, we find that #(-1)^3+3(-1)^2-3(-1)-5 = 0# Thus #x-(-1) = x+1# is a factor of #x^3+3x^2-3x-5#

Dividing, we get
#x^3+3^2-3x-5 = (x + 1)(x^2 + 2x -5)#

To find the remaining roots, we can simply apply the quadratic formula
#ax^2+bx+c = 0 => x = (-b+-sqrt(b^2-4ac))/(2a)#
to obtain the roots of #x^2+2x-5# as

#(-2+-sqrt(2^2-4(1)(-5)))/(2(1)) = -1 +- sqrt(6)#

The the final set of roots (zeros) of the expression is

#{-1, -1+sqrt(6), -1-sqrt(6)}#