Polynomials: Adding, Subtracting, Dividing & Multiplying

All math students and many science students encounter polynomials at some stage during their studies, but thankfully they're easy to deal with once you learn the basics. The main operations you'll need to do with polynomial expressions are adding, subtracting, multiplying and dividing, and while division can be complex, most of the time you'll be able to handle the basics with ease.

Polynomial describes an algebraic expression with one or more terms involving a variable (or more than one), with exponents and possibly constants. They can't include division by a variable, can't have negative or fractional exponents and must have a finite number of terms.

This example shows a polynomial:

\(x^3 + 2 x^ 2 – 9 x – 4\)

And this shows another one:

\(xy^2 – 3 x + y\)

There are many ways of classifying polynomials, including by degree (the sum of the exponents on the highest power term, e.g. 3 in the first example) and by the number of terms they contain, such as monomials (one term), binomials (two terms) and trinomials (three terms).

Adding and subtracting polynomials depends on combining "like" terms. A like term is one with the same variables and exponents as another, but the number they're multiplied by (the coefficient) can be different. For example, ​x​² and 4 ​x​ ² are like terms because they have the same variable and exponent, and 2 ​xy​ ⁴ and 6 ​xy​ ⁴ are like terms as well. However, ​x​², ​x​³, ​x​²​y​² and ​y​² are not like terms, because each one contains different combinations of variables and exponents.

Add polynomials by combining like terms in the same way you would with other algebraic terms. For example, look at the problem:

\((x^3 + 3 x ) + (9 x^3 + 2 x + y)\)

Collect the like terms to get:

\((x^3 + 9 x^3) + (3 x + 2 x ) + y\)

And then evaluate by simply adding together the coefficients and combining into a single term:

\(10 x^3 + 5 x + y\)

Note that you can't do anything with ​y​ because it has no like term.

Subtraction works in the same way:

\((4 x^4 + 3 y^2 + 6 y ) – (2 x^4 + 2 y^2 + y)\)

First, note that all the terms in the right hand bracket are subtracted from those in the left hand bracket, so write it as:

\(4 x^4 + 3 y^2 + 6 y – 2 x^4 – 2 y^2- y\)

Combine like terms and evaluate to get:

\((4 x^4 – 2 x^4) + (3 y^2 – 2 y^2) + (6 y – y) = 2 x^4 + y^2 + 5 y\)

For a problem like this:

\((4 xy + x^2) – (6 xy – 3 x^2)\)

Note that the minus sign is applied to the whole expression in the right bracket, so the two negative signs before 3​x​² become an addition sign:

\((4 xy + x^2) – (6 xy – 3 x^2) = 4 xy + x^2 – 6 xy + 3 x^2\)

Then calculate as before.

Multiply polynomial expressions by using the distributive property of multiplication. In short, multiply every term in the first polynomial by every term in the second one. Look at this simple example:

\(4 x × (2 x^2 + y)\)

You solve this using the distributive property, so:

\(\begin{aligned}\)
\(4 x × (2 x^2 + y) &= (4 x × 2 x^2) + (4 x × y)\)
\(&= 8 x^3 + 4 xy\)
\(\end{aligned}\)

Tackle more complicated problems in the same way:

\(\begin{aligned}
(2 y^3 + 3 x ) × &(5 x^2 + 2 x )\)
\(&= (2 y^3 × (5 x^2 + 2 x )) + (3 x × (5 x^2 + 2 x ))\)
\(&= (2 y^3 × 5 x^2) + (2 y^3 × 2 x ) + (3 x × 5 x^2) + (3 x × 2 x )\)
\(&= 10 y^3x^2 + 4 y^3x + 15 x^3 + 6 x^2
\end{aligned}\)

These problems can get complicated for bigger groupings, but the basic process is still the same.

Dividing polynomial expressions takes longer but you can tackle it in steps. Look at the expression:

\(\frac{x^2 – 3 x – 10}{x + 2}\)

First, write the expression like a long division, with the divisor on the left and the dividend on the right:

\(x + 2 )\overline{x^2 – 3 x – 10}\)

Divide the first term in the dividend by the first term in the divisor, and put the result on the line above the division. In this case, ​x​² ÷ ​x​ = ​x​, so:

\(\begin{aligned}
&x \
x + 2 )&\overline{x^2 – 3 x – 10}
\end{aligned}\)

Multiply this result by the whole divisor, so in this case, (​x​ + 2) × ​x​ = ​x​² + 2 ​x​ . Put this result below the division:

\(\begin{aligned}
&x \
x + 2 )&\overline{x^2 – 3 x – 10} \
&x^2 + 2 x
\end{aligned}\)

Subtract the result on the new line from the terms directly above it (note that technically you change the sign, so if you had a negative result you'd add it instead), and put this on a line below it. Move the final term from the original dividend down too.

\(\begin{aligned}
&x \
x + 2 )&\overline{x^2 – 3 x – 10} \
&x^2 + 2 x \
&0 – 5 x – 10
\end{aligned}\)

Now repeat the process with the divisor and the new polynomial on the bottom line. So divide the first term of the divisor (​x​) by the first term of the dividend (−5 ​x​ ) and put this above:

\(\begin{aligned}
&x -5\
x + 2 )&\overline{x^2 – 3 x – 10} \
&x^2 + 2 x \
&0 – 5 x – 10
\end{aligned}\)

Multiply this result (−5 ​x​ ÷ ​x​ = −5) by the original divisor (so (​x​ + 2) × −5 = −5 ​x​ −10) and put the result on a new bottom line:

\(\begin{aligned}
&x -5\
x + 2 )&\overline{x^2 – 3 x – 10} \
&x^2 + 2 x \
&0 – 5 x – 10 \
& -5 x – 10
\end{aligned}\)

Then subtract the bottom line from the next one up (so in this case change the sign and add), and put the result on a new bottom line:

\(\begin{aligned}
&x -5\
x + 2 )&\overline{x^2 – 3 x – 10} \
&x^2 + 2 x \
&0 – 5 x – 10 \
&-5 x – 10 \
& 0 \quad 0
\end{aligned}\)

Since there is now a row of zeros at the bottom, the process is finished. If there were non-zero terms remaining, you would repeat the process again. The result is on the top line, so:

\(\frac{x^2 – 3 x – 10}{x + 2} = x – 5\)

This division and some others can be solved more simply if you can factor the polynomial in the dividend.