What are all the possible rational zeros for #f(x)=x^3-4x^2-3x+14# and how do you find all zeros?

1 Answer
Aug 18, 2016

Answer:

#f(x)# has zeros #2# and #1+-2sqrt(2)#

Explanation:

#f(x) = x^3-4x^2-3x+14#

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

That means that the only possible rational zeros are:

#+-1, +-2, +-7, +-14#

Trying each in turn, we find:

#f(2) = 8-4(4)-3(2)+14 = 8-16-6+14 = 0#

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

#x^3-4x^2-3x+14 = (x-2)(x^2-2x-7)#

We can factor the remaining quadratic by completing the square:

#x^2-2x-7#

#=x^2-2x+1-8#

#=(x-1)^2-(2sqrt(2))^2#

#=((x-1)-2sqrt(2))((x-1)+2sqrt(2))#

#=(x-(1+2sqrt(2))(x-(1-2sqrt(2)))#

Hence zeros:

#x = 1+-2sqrt(2)#