Question

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

Answer 1

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`