How do you use the rational roots theorem to find all possible zeros of #P(x)= x^3 + 3x^2 - 4#?

1 Answer
Aug 9, 2016

Answer:

#P(x)# has zeros #1, -2, -2#

Explanation:

#P(x) = x^3+3x^2-4#

By the rational root theorem, any rational zeros of #P(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#

We find:

#P(1) = 1+3-4 = 0#

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

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

The remaining quadratic is a perfect square trinomial:

#x^2+4x+4 = (x+2)^2#

Hence the remaining zero is #-2# with multiplicity #2#.