# 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?

May 27, 2016

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

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

#### Explanation:

$f \left(x\right) = {x}^{4} + 5 {x}^{3} + 10 {x}^{2} + 20 x + 24$

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, $\mathbb{C}$.

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, $f \left(x\right)$ is of degree $4$, so it has exactly four zeros counting multiplicity.

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Bonus

What else can we find out about these zeros?

First note that all of the coefficients of $f \left(x\right)$ are positive, so $f \left(x\right)$ 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 $f \left(x\right)$ 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 $\frac{p}{q}$ 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:

$f \left(- 2\right) = 16 - 40 + 40 - 40 + 24 = 0$

So $- 2$ is a zero and $\left(x + 2\right)$ a factor:

${x}^{4} + 5 {x}^{3} + 10 {x}^{2} + 20 x + 24 = \left(x + 2\right) \left({x}^{3} + 3 {x}^{2} + 4 x + 12\right)$

The remaining cubic factor factors by grouping as follows:

${x}^{3} + 3 {x}^{2} + 4 x + 12 = \left({x}^{3} + 3 {x}^{2}\right) + \left(4 x + 12\right)$

$= {x}^{2} \left(x + 3\right) + 4 \left(x + 3\right)$

$= \left({x}^{2} + 4\right) \left(x + 3\right)$

$= \left(x - 2 i\right) \left(x + 2 i\right) \left(x + 3\right)$

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