What are all the possible rational zeros for #f(x)=x^3+4x^2+5x+2# and how do you find all zeros?

1 Answer
Sep 5, 2016

Answer:

#f(x)# has zeros #-1#, #-1# and #-2#.

Explanation:

#f(x) = x^3+4x^2+5x+2#

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

So the only possible rational zeros are:

#+-1, +-2#

Note also that #f(x)# only has positive coefficients, so it has no positive Real zeros, which only leaves #-1# and #-2# as possible rational zeros.

We find:

#f(-1) = -1+4-5+2 = 0#

So #x=-1# is a zero and #(x+1)# a factor:

#x^3+4x^2+5x+2 = (x+1)(x^2+3x+2)#

The only possible rational zeros of the remaining quadratic are also #-1# and #-2#, with corresponding factors #(x+1)# and #(x+2)# and indeed:

#x^2+3x+2 = (x+1)(x+2)#

So all the zeros of #f(x)# are rational, being #-1# with multiplicity #2# and #-2# with multiplicity #1#.