To multiply two polynomials, we have to carry out the multiplication process term-by-term. Consider a simple example of multiplying two binomials *f* and *g*:

\[\begin{array}{l}\left\{ \begin{array}{l}f\left( x \right) = 1 + {x^2}\\g\left( x \right) = 2x + 3{x^3}\end{array} \right.\\ \Rightarrow f\left( x \right)g\left( x \right) = \left( {1 + {x^2}} \right)\left( {2x + 3{x^3}} \right)\\ \,\,\,\,\,\,\;\;\;\; = \left\{ \begin{array}{l}1 \times 2x + 1 \times 3{x^3}\\\,\,\,\,\,\,\,\,\,\,\,\, + \\{x^2} \times 2x + {x^2} \times 3{x^3}\end{array} \right.\\ \,\,\,\,\,\,\;\;\;\; = 2x + 3{x^3} + 2{x^3} + 3{x^5}\\\;\;\;\;\,\,\,\,\,\, = 2x + 5{x^3} + 3{x^5} & \end{array}\]

In the first bracket, there were two terms. And in the second bracket, there were two terms as well. Multiplying the two brackets term-by-term yielded four terms. However, two of those terms turned out to have the same degree, and so eventually, we were left with only three terms (with distinct powers of the variable *x*).

In general, suppose that two polynomials *f* and *g* have *p* and *q* terms respectively. Multiplying the two polynomials term-by-term will yield \(p \times q\) terms. However, some of those terms might have the same degree, and thus the *actual* number of terms in the product polynomial will be less than this.

Consider the example above once again. In this, *f* was a quadratic polynomial (its degree was 2), whereas *g* was a cubic polynomial (its degree was 3). The degree of the product polynomial was 5 (the sum of the two degrees). This is a general result. If the degrees of two polynomials are *m* and *n* respectively, then the degree of their product polynomial will be \(m + n\).