The polynomials are a very simple and useful family of functions. For example quadratic polynomials of the form $f(x) = ax^2 + bx +c$ often arise in the description of physics phenomena.

• $x$: the variable
• $f(x)$: the polynomial. We sometimes sometimes denote polynomials $P(x)$ to

distinguish them from generic function $f(x)$.

• degree of $f(x)$: the largest power of $x$ that appears in the polynomial
• roots of $f(x)$: the values of $x$ for which $f(x)=0$

The most general polynomial of the first degree is a line $f(x) = mx + b$, where $m$ and $b$ are arbitrary constants.

The most general polynomial of second degree is $f(x) = a_2 x^2 + a_1 x + a_0$, where again $a_0$, $a_1$ and $a_2$ are arbitrary constants. We call $a_k$ the coefficient of $x^k$ since this is the number that appears in front of it.

By now you should be able to guess that a third degree polynomial will look like $f(x) = a_3 x^3 + a_2 x^2 + a_1 x + a_0$.

In general, a polynomial of degree $n$ has equation: \[ f(x) = a_n x^n + a_{n-1}x^{n-1} + \cdots + a_2 x^2 + a_1 x + a_0. \] or if you want to use the sum notation we can write it as: \[ f(x) = \sum_{k=0}^n a_kx^k, \] where $\Sigma$ (the capital Greek letter sigma) stands for summation.

Very often you will have to solve a polynomial equations of the form: \[ A(x) = B(x), \] where $A(x)$ and $B(x)$ are both polynomials. Remember that solving means to find the value of $x$ which makes the equality true.

For example, say the revenue of your company, as function of the number of products sold $x$ is given by $R(x)=2x^2 + 2x$ and the costs you incur to produce $x$ objects is $C(x)=x^2+5x+10$. A very natural question to ask is the amount of product you need to produce to break even, i.e., to make your revenue equal your costs $R(x)=C(x)$. To find the break-even $x$, you will have to solve the following equation: \[ 2x^2 + 2x = x^2+5x+10. \]

This may seem complicated since there are $x$s all over the place and it is not clear how to find the value of $x$ that makes this equation true. No worries though, we can turn this equation into the “standard form” and then use the quadratic equation. To do this, we will move all the terms to one side until we have just zero on the other side: \[ \begin{align} 2x^2 + 2x \ \ \ -x^2 &= x^2+5x+10 \ \ \ -x^2 \nl x^2 + 2x \ \ \ -5x &= 5x+10 \ \ \ -5x \nl x^2 - 3x \ \ \ -10 &= 10 \ \ \ -10 \nl x^2 - 3x -10 &= 0. \end{align} \]

Remember that if we do the same thing on both sides of the equation, it remains true. Therefore, the values of $x$ that satisfy \[ x^2 - 3x -10 = 0, \] namely $x=-2$ and $x=5$, will also satisfy \[ 2x^2 + 2x = x^2+5x+10, \] which was the original problem that we were trying to solve.

This “shuffling of terms” approach will work for any polynomial equation $A(x)=B(x)$. We can always rewrite it as some $C(x)=0$, where $C(x)$ is a new polynomial that has as coefficients the difference of the coefficients of $A$ and $B$. Don't worry about which side you move all the coefficients to because $C(x)=0$ and $0=-C(x)$ have exactly the same solutions. Furthermore, the degree of the polynomial $C$ can be no greater than that of $A$ or $B$.

The form $C(x)=0$ is the standard form of a polynomial and, as you will see shortly, there are formulas which you can use to find the solution(s).

The formula for solving the polynomial equation $P(x)=0$ depend on the degree of the polynomial in question.

For first degree: \[ P_1(x) = mx + b = 0, \] the solution is $x=b/m$. Just move $b$ to the other side and divide by $m$.

For second degree: \[ P_2(x) = ax^2 + bx + c = 0, \] the solutions are $x_1=\frac{-b + \sqrt{ b^2 -4ac}}{2a}$ and $x_2=\frac{-b - \sqrt{b^2-4ac}}{2a}$.

Note that if $b^2-4ac < 0$, the solutions involve taking the square root of a negative number. In those cases, we say that no real solutions exist.

The solutions to the cubic polynomial equation \[ P_3(x) = x^3 + ax^2 + bx + c = 0, \] are given by \[ x_1 = \sqrt[3]{ q + \sqrt{p} } \ \ + \ \sqrt[3]{ q - \sqrt{p} } \ -\ \frac{a}{3}, \] and \[ x_{2,3} = \left( \frac{ -1 \pm \sqrt{3}i }{2} \right)\sqrt[3]{ q + \sqrt{p} } \ \ + \ \left( \frac{ -1 \pm \sqrt{3}i }{2} \right) \sqrt[3]{ q - \sqrt{p} } \ - \ \frac{a}{3}, \] where $q \equiv \frac{-a^3}{27}+ \frac{ab}{6} - \frac{c}{2}$ and $p \equiv q^2 + \left(\frac{b}{3}-\frac{a^2}{9}\right)^3$.

Note that, in my entire career as an engineer, physicist and computer scientist, I have never used the cubic equation to solve a problem by hand. In math homework problems and exams you will not be asked to solve equations of higher than second degree, so don't bother memorizing the solutions of the cubic equation. I included the formula here just for completeness.

There is also a formula for polynomials of degree $4$, but it is complicated. For polynomials with order $\geq 5$, there does not exist a general analytical solution.

When solving real world problems, you will often run into much more complicated equations. For anything more complicated than the quadratic equation, I recommend that you use a computer algebra system like sympy to find the solutions. Go to http://live.sympy.org and type in:

 >>> solve( x**2 - 3*x +2, x)      [ shift + Enter ]
 [1, 2]

Indeed $x^2-3x+2=(x-1)(x-2)$ so $x=1$ and $x=2$ are the two solutions.

Sometimes you can solve polynomials of fourth degree by using the quadratic formula. Say you are asked to solve for $x$ in \[ g(x) = x^4 - 3x^2 -10 = 0. \] Imagine this comes up on your exam. Clearly you can't just type it into a computer, since you are not allowed the use of a computer, yet the teacher expects you to solve this. The trick is to substitute $y=x^2$ and rewrite the same equation as: \[ g(y) = y^2 - 3y -10 = 0, \] which you can now solve by the quadratic formula. If you obtain the solutions $y=\alpha$ and $y=\beta$, then the solutions to the original fourth degree polynomial are $x=\sqrt{\alpha}$ and $x=\sqrt{\beta}$ since $y=x^2$.

Of course, I am not on an exam, so I am allowed to use a computer:

 >>> solve(y**2 - 3*y -10, y)
 [-2, 5]
 >>> solve(x**4 - 3*x**2 -10 , x)
 [sqrt(2)i, -sqrt(2)i, -sqrt(5) , sqrt(5) ]

Note how a 2nd degree polynomial has two roots and a fourth degree polynomial has four roots, two of which are imaginary, since we had to take the square root of a negative number to obtain them. We write $i=\sqrt{-1}$. If this was asked on an exam though, you should probably just report the two real solutions: $\sqrt{5}$ and $-\sqrt{5}$ and not talk about the imaginary solutions since you are not supposed to know about them yet. If you feel impatient though, and you want to know about the complex numbers right now you can skip ahead to the section on complex numbers.