What are all the possible rational zeros for #f(x)=x^3-10x^2+16x+15# and how do you find all zeros?

1 Answer
Sep 19, 2016

Answer:

#f(x)# has zeros #x=3# and #x = 7/2+-sqrt(69)/2#

Explanation:

#f(x) = x^3-10x^2+16x+15#

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

That means that the only possible rational zeros are:

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

We find:

#f(3) = 27-90+48+15 = 0#

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

#x^3-10x^2+16x+15 = (x-3)(x^2-7x-5)#

We can find the zeros the remaining quadratic #x^2-7x-5# which is in standard form #ax^2+bx+c# with #a=1#, #b=-7# and #c=-5# using the quadratic formula:

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

#color(white)(x) = (7+-sqrt((-7)^2-4(1)(-5)))/(2*1)#

#color(white)(x) = (7+-sqrt(49+20))/2#

#color(white)(x) = 7/2+-sqrt(69)/2#