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

##### 1 Answer

#### 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:

#### Explanation:

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

By the rational root theorem, any *rational* zeros of

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 *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

The other two zeros are non-Real Complex since

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

with

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

Hence the two remaining zeros are