Question

How do you find the number of roots for `f(x) = x^4 + 5x^3 +10x^2 + 20x + 24` using the fundamental theorem of algebra?

Answer 1

The FTOA tells us that there are four zeros counting multiplicity.

Other methods help us find them: `-2`, `-3`, `2i`, `-2i`.

Explanation:

`f(x) = x^4+5x^3+10x^2+20x+24`

The fundamental theorem of algebra tells you that any non-constant polynomial in one variable with Real or Complex coefficients has a zero in the field of Complex numbers, `CC`.

A corollary of this, often stated as part of the FTOA, is that a polynomial with Complex coefficients of degree `n > 0` has exactly `n` Complex (possibly Real) zeros counting multiplicity.

In our example, `f(x)` is of degree `4`, so it has exactly four zeros counting multiplicity.

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What else can we find out about these zeros?

First note that all of the coefficients of `f(x)` are positive, so `f(x)` has no positive zeros.

If we reverse the signs on the terms of odd degree, then the pattern of signs of coefficients is: `+ - + - +`, which has `4` changes of sign. Since the coefficients of `f(x)` are Real, that means that it has `0`, `2` or `4` negative Real zeros.

Using the rational root theorem, the only possible rational zeros are expressible in the form `p/q` for integers `p, q` with `p` a divisor of the constant term `24` and `q` a divisor of the coefficient `1` of the leading term.

That means that the only possible rational zeros are:

`-1, -2, -3, -4, -6, -8, -12, -24`

Trying each of these in turn, we find:

`f(-2) = 16-40+40-40+24 = 0`

So `-2` is a zero and `(x+2)` a factor:

`x^4+5x^3+10x^2+20x+24 = (x+2)(x^3+3x^2+4x+12)`

The remaining cubic factor factors by grouping as follows:

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

`= x^2(x+3)+4(x+3)`

`= (x^2+4)(x+3)`

`= (x-2i)(x+2i)(x+3)`

So the zeros are: `-2`, `-3`, `2i`, `-2i`.