What are all the possible rational zeros for #y=4x^3+12x^2+x+3# and how do you find all zeros?

1 Answer
Nov 8, 2016

Answer:

Use the rational root theorem to find "possible" rational zeros:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-3#

Factor by grouping to find the actual zeros: #-3#, #+-1/2i#

Explanation:

#f(x) = 4x^3+12x^2+x+3#

By the rational root 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 #3# and #q# a divisor of the coefficient #4# of the leading term.

That means that the only possible rational zeros are:

#+-1/4, +-1/2, +-3/4, +-1, +-3/2, +-3#

In fact, since all of the coefficients of #f(x)# are positive, any Real zeros are negative, leaving us with possible rational zeros:

#-1/4, -1/2, -3/4, -1, -3/2, -3#

We could try each of these in turn, but there's something else we can notice about this particular cubic: The ratio between the first and second terms is the same as that between the third and fourth terms. As a result, this cubic will factor by grouping:

#4x^3+12x^2+x+3 = (4x^3+12x^2)+(x+3)#

#color(white)(4x^3+12x^2+x+3) = 4x^2(x+3)+1(x+3)#

#color(white)(4x^3+12x^2+x+3) = (4x^2+1)(x+3)#

Hence #f(x)# has one Real zero #x=-3#

The other two zeros are non-Real Complex since #4x^2+1 > 0# for all Real values of #x#. We can factor this expression using the difference of squares identity:

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

with #a=2x# and #b = i# as follows:

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

Hence the two remaining zeros are #x = +-1/2i#