What are all the possible rational zeros for #y=x^3+x^2-9x+7# and how do you find all zeros?

1 Answer
George C.
Sep 25, 2016

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

Explanation:

#f(x) = x^3+x^2-9x+7#

By the rational roots theorem, any rational zeros of #f(x)# must be expressible in the form #p/q# for integers #p, q# with #p# a divisor of the constant term #7# and #q# a divisor of the coefficient #1# of the leading term.

That means that the only possible rational zeros are:

#+-1, +-7#

Since the sum of the coefficients of #f(x)# is zero (i.e. #1+1-9+7 = 0#) then #f(1) = 0#, #x=1# is a zero and #(x-1)# a factor:

#x^3+x^2-9x+7 = (x-1)(x^2+2x-7)#

We can factor the remaining quadratic by completing the square and using the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

as follows:

#x^2+2x-7 = x^2+2x+1-8#

#color(white)(x^2+2x-7) = (x+1)^2-(2sqrt(2))^2#

#color(white)(x^2+2x-7) = ((x+1)-2sqrt(2))((x+1)+2sqrt(2))#

#color(white)(x^2+2x-7) = (x+1-2sqrt(2))(x+1+2sqrt(2))#

Hence the remaining zeros are:

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

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