In previous posts, we have discussed how to solve systems of linear equations. Now, we will talk about a bigger family of equations: the polynomials. Formally, the polynomials are all the equations of the form:

    \[a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 = 0\]

where a_i is some constant (for i = 0, 1, ..., n) and x is a variable. Basically, polynomials are combinations of constants and the positive powers of x multiplied by any coefficient.

We say that the highest exponent of a variable in a polynomial is its degree. So for example,

    \[x^{90} + 2 = 0 \textrm{ has degree 90}\]

    \[x^2 + 2x - 8 = 0 \textrm{ has degree 2}\]

    \[0x^{90} + x^2 + 2x - 8 = 0 \textrm{ has degree 2}\]

since if we simplify the last equation, the x^{90} term goes away and we just get x^2 + 2x - 8 = 0 which we earlier noted has degree 2.

We say equations with just constants have degree 0. So for example,

    \[10 - 8 = 2 \textrm{ has degree 0}\]

Now, some polynomials play a distinctive role in mathematics and so they get special names. Polynomials with degree 2 (i.e. 2 is the highest power of any variable) are called quadratic equations. Here are some examples of quadratic equations:

    \[x^2 + x = 0\]

    \[3x^2 - 2x + 5 = 0\]

    \[\frac{x^2}{9} - x + 3 = x^2 - 2x + 8\]

And while the last equation may not look like our definition of a polynomial, we can see how by subtracting x^2 -2x + 8 from both sides, we can put it into that form:

    \[\frac{-17x^2}{9} + 3x - 5 = 0\]

Polynomials with degree 3 are called cubic. Here are some examples of those:

    \[x^3  = 0\]

    \[9x^3 + x^2 - 2x + 1 = 0\]

    \[\frac{x^3}{9} - x + 3 = x^2 - 2x + 8\]

In the next few posts, we will discuss two strategies for solving quadratic equations: via the quadratic formula or via factoring.

Practice Problems

Find the degree of the following equations:

  1. \frac{9x^3}{x} - 9x  = 0, where x \neq 0

    Answer

  2. 9x^4 - 20x^9 = 0

    Answer

  3. x - 2x + 3x^2 - 4x + x^5 = 0

    Answer


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