Polynomials are special type of functions.
They are expressions consisting of variables and coefficients. The general form of a polynomial with a leading exponent $n$ (degree $n$ polynomial) is:
$$p(x) = a_n x^n + a_{n-1} x^{n-1} + … + a_1 x + a_0$$
where $a_0, a_1, … , a_{n-1}, a_n \in \mathbb{C}$ (polynomial with complex coefficients) or $\mathbb{R}$ (polynomial with real coefficients) and $a_n \neq 0$.
First degree polynomial
$$ a x + b, a \neq 0 $$
is sometimes called linear polynomial, while second degree polynomial
$$a x^2 + b x + c, a \ne 0$$
is called quadratic polynomial (or quadratic function).
More about linear polynomials you can find in lessons One-step equations, Two-step equations and Multi-step equations.
More about quadratic polynomials you can find in lesson Quadratic equations.
Often we want to find roots of polynomial.
Root (or zero) of a function $f$ is number $x$ such that $f(x) = 0$.
As you probably know, every first degree polynomial has exactly one root
$$x = – \frac{b}{a} $$
and every second degree polynomial has exactly two roots (real or complex) which are given by formula
$$x_{1,2} =\frac{-b \pm \sqrt[]{b^2-4ac}}{2a}.$$
Example 1 Find all roots of the function $ f(x) = -5 x + 10$.
$$-5 x + 10 = 0$$
$$-5x = -10$$
$$ x = 2 $$
Example 2 Find all roots of the function $ g(x) = x^2 – x – 6$.
Using the formula for roots of a quadratic polynomial with $a = 1$, $b = -1$, $c = -6$ we can calculate
$$x_{1,2}=\frac{-(-1) \pm \sqrt[]{(-1)^2-4\cdot1\cdot(-6)}}{2(-1)}$$
$$x_{1,2}=\frac{1 \pm 5}{-2}$$
$$x_1 = -3, x_2 = 2.$$
Another way to find roots of polynomial is to factor it, that is to write it as a product of linear polynomials
$$(x-x_n)(x-x_{n-1})\cdot…\cdot(x-x_1).$$
Then $x_1, x_2, … , x_{n-1}, x_n$ are obviously roots (zeros) of that polynomial.
To learn more about factorization see lesson Meaning of term factoring polynomials.
Example 3 Find all roots of the function $ h(x) = x^3 – 4 x^2 – x + 4$.
First we are going to factor given polynomial:
$$ x^3 – 4 x^2 – x + 4 = (x^3 – 4 x^2) – (x – 4) = x^2 (x – 4) – (x – 4) = (x – 4) (x^2 – 1) = (x – 4) (x – 1) (x + 1) $$
where we used formula for difference of squares to write $x^2 – 1$ as $(x – 1) (x + 1)$ in last equality. Now we immediately see that roots of $h$ are $-1, 1$ and $4$.
So third degree polynomial in Example 3. has three roots.
Example 4 Find all roots of the function $ p(x) = x^4+ 5 x^3+ 3 x^2 – 9x$.
Again, first step is to factor polynomial:
$$ x^4+ 5 x^3+ 3 x^2 – 9x = x(x^3 + 5x^2 + 3x – 9) = x(x^3 + 3x^2 + 2x^2 + 3x – 9) =$$
$$x(x^2 (x+3)+2x^2+6x-3x-9=x^2(x+3)+2x(x+3)-3(x+3))=x((x+3)(x^2+2x-3))=$$
$$x((x+3)(x^2+3x-x-3))=x((x+3)(x(x+3)-(x+3)))=x((x+3)((x+3)(x-1)))=$$
$$x(x+3)^2(x-1)=(x-0)(x+3)^2(x-1)$$
We see that roots are $0,1$ and $-3$. Note that factor $(x+3)$ which corresponds to root $-3$ is to the power of $2$. We say that root $-3$ has multiplicity of $2$, while roots $0$ and $1$ have multiplicity of $1$.
Multiplicity of a root is power of corresponding linear factor in full factorization of polynomial.
Counting in multiplicity, we see that fourth degree polynomial $p$ in Example 4. has four roots.
This general result holds:
Theorem 1 (Fundamental Theorem of Algebra) Every polynomial with degree $n \geq 1$ has, counted with multiplicity, exactly $n$ roots (real or complex) .
Remark 1 Note that theorem gives existence of exactly $n$ roots, but roots don’t have to be real numbers – even if polynomial coefficients are real numbers.
Example 5 Find all roots of the function $ q(x) = x^2+1$.
Using formula for roots of quadratic polynomial we obtain $x_{1,2} =\frac{0 \pm \sqrt[]{0^2-4\cdot1\cdot1}}{2\cdot1}=\pm i$.
We see that even simple polynomial like $q$ has $2$ complex roots.
As we have seen in previous examples, finding roots of polynomial is much easier if we have factored it (that is, written it as a product of linear factors). Closed formulas for roots (like formula for roots of quadratic polynomial) only exist for polynomials with degree up to four. Formulas for roots of polynomials of degree three and four are relatively complex and most of the times it is easier to factor this polynomials and read their roots from that factorization.
Another (equivalent) way to express Theorem 1 is:
Theorem 1′ Every polynomial with degree $n \geq 1$ can be, in a unique way, expressed as the product of $n$ linear polynomials (with complex coefficients).
Remark 2 Even if polynomial has real coefficients its factorization can contain complex numbers.
Fundamental theorem of algebra has many important consequences and applications. Some of them are:
- it shows us the need for complex numbers: as we have seen in Example 5 polynomial $x^2 + 1$ does not have real roots (or in other words equation $x^2 + 1 = 0$ does not have real solutions) so if we want to be able to solve such equations we need concept of complex numbers
- the field of complex numbers is algebraically closed (contains a root for every non-constant polynomial with complex coefficients), so it follows that any theorem concerning algebraically closed fields applies to the field of complex numbers
- it can be used for finding roots of polynomials and factorization of polynomials to linear factors (what we have already seen)
- in calculus integrating rational functions (quotients of polynomials) is done with factorization of numerator and denominator to linear factors and using technique called partial fractions
- in linear algebra it guarantees existence of eigenvalues from characteristic polynomial and so on.
