What are all the possible rational zeros for #f(x)=x^3-6x^2+7x+4# and how do you find all zeros?

1 Answer
Dec 25, 2016

Answer:

The "possible" rational zeros are: #+-1, +-2, +-4#

The actual zeros are: #4#, #-1+sqrt(2)#, #-1-sqrt(2)#

Explanation:

#f(x) = x^3-6x^2+7x+4#

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

That means that the only possible rational zeros are:

#+-1, +-2, +-4#

Trying each in turn, we eventually find:

#f(4) = 64-96+28+4 = 0#

So #x=4# is a zero and #(x-4)# a factor:

#x^3-6x^2+7x+4 = (x-4)(x^2-2x-1)#

The remaining quadratic can be factored by completing the square:

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

#color(white)(x^2-2x-1) = (x+1)^2-sqrt(2)^2#

#color(white)(x^2-2x-1) = ((x+1)-sqrt(2))((x+1)+sqrt(2))#

#color(white)(x^2-2x-1) = (x+1-sqrt(2))(x+1+sqrt(2))#

Hence the other two zeros are:

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