How do you find all zeros with multiplicities of #f(x)=3x^3+3x^2-11x-10#?

1 Answer
Apr 10, 2017

Answer:

The zeros are #-2#, #1/2+sqrt(69)/6#, #1/2-sqrt(69)/6# all with multiplicity #1#.

Explanation:

Given:

#f(x) = 3x^3+3x^2-11x-10#

By the rational roots theorem, any rational zeros of #f(x)# are expressible in the form #p/q# for integers #p#, #q# with #p# a divisor of the constant term #-10# and #q# a divisor of the coefficient #3# of the leading term.

That means that the only possible rational zeros are:

#+-1/3, +-2/3, +-1, +-5/3, +-2, +-10/3, +-5, +-10#

Trying each in turn, we eventually find:

#f(-2) = 3(color(blue)(-2))^3+3(color(blue)(-2))^2-11(color(blue)(-2))-10#

#color(white)(f(-2)) = -24+12+22-10#

#color(white)(f(-2)) = 0#

So #-2# is a zero and #(x+2)# a factor:

#3x^3+3x^2-11x-10 = (x+2)(3x^2-3x-5)#

The remaining quadratic factor is in the form:

#ax^2+bx+c#

with #a=3#, #b=-3# and #c=-5#

We can find its zeros using the quadratic formula:

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

#color(white)(x) = (3+-sqrt((-3)^2-4(3)(-5)))/(2*3)#

#color(white)(x) = (3+-sqrt(9+60))/(2*3)#

#color(white)(x) = 1/2+-sqrt(69)/6#