Polynomial Arithmetic

We can add, subtract, multiply polynomials. Polynomials can involve many variables raised to whole number powers so for simplicity we will be using polynomials only involving the variable $x$.

Addition and subtraction

When adding/subtracting polynomials, we simply add/subtract the terms with the same power of $x$.

For example, let’s add the polynomials $x^2-3x+4$ and $5x^2+3x+7$. To make visualizing the addition easier, we will be using brackets: $$ \begin{equation}\begin{aligned} &(x^2-3x+4) + (5x^2+3x+7)\\ =&(x^2+5x^2)+(-3x+3x)+(4+7)\\ =&6x^2+0x+11\\ =&6x^2+11\\ \end{aligned}\end{equation} $$

For a regular addition, we do not need to specify the brackets or reorder the terms. These were just done to make emphasis in the demonstration above.

Challenge

Add the following polynomials:

• $x^3+9x-1$ and $-4x^3-19x+5$
• $2x^2-5$ and $x^3+4x$
• $1-x^4$ and $2x^4-x$

The same principles apply for subtraction. For example, consider the difference between the same polynomials given in the example above: $$ \begin{equation}\begin{aligned} &(x^2-3x+4) - (5x^2+3x+7)\\ =&(x^2-5x^2)+(-3x-3x)+(4-7)\\ =&-4x^2-6x-3\\ \end{aligned}\end{equation} $$

Multiplication

When multiplying two polynomials, we need to find the sum of the products of each term in the first with each term in the second. For example, consider two binomials, $x-2$ and $x^2+1$. Each has $2$ terms and thus multiplying the two polynomials would involve $2\times 2$ multiplications - each term in the first, $x$ and $-2$, by each term in the second, $x^2$ and $1$. Again, brackets are used for emphasis:

$$ \begin{equation}\begin{aligned} &(x-2)(x^2+1)\\ =&(x\times x^2)+(x\times 1)+(-2\times x^2)+(-2\times 1)\\ =&x^3+x-2x^2-2\\ =&x^3-2x^2+x-2\\ \end{aligned}\end{equation} $$

Note that the final line was obtained by rearranging the terms according to the commutative and associative properties of multiplication.

Challenge

Multiply the following polynomials:

• $x-1$ and $x+3$ (hint: $4$ products)
• $4x^2+5x+3$ and $x-5$ (hint: $6$ products)
• $5-x^2$ and $x^2+2x+1$ (hint: $6$ products)

Division

Long division of polynomials involves the same steps as long division of integers:

• Divide by the first term in the divisor
• Multiply by the entire divisor
• Subtract
• Bring down the remaining terms

Let’s look at an example. We can divide $x^2+5x+7$ by $x+3$. First, we divide the first term in dividend, $x^2$ by the first term in the divisor, $x$:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ \textcolor{crimson}{x}+3 \overline{) \ \ \textcolor{crimson}{x^2}+5x+7} \end{array} \end{aligned}\end{equation} $$

We then multiply this result, $x$ by the entire divisor, $x+3$:

$$ \begin{equation}\begin{aligned} \begin{array}{r} \textcolor{crimson}{x}\ \ \ \ \ \ \ \\ \textcolor{crimson}{x+3} \overline{) \ \ x^2+5x+7} \\ \underline{-(x^2+3x)}\ \ \ \ \ \end{array} \end{aligned}\end{equation} $$

We then subtract:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x+3 \overline{) \ \ \textcolor{crimson}{x^2+5x}+7} \\ \underline{-(\textcolor{crimson}{x^2+3x})}\ \ \ \ \ \\ 2x\ \ \ \ \ \ \ \end{array} \end{aligned}\end{equation} $$

And bring down the remaining terms:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x+3 \overline{) \ \ x^2+5x+7} \\ \underline{-(x^2+3x)}\ \ \ \ \ \\ 2x\textcolor{crimson}{+7} \end{array} \end{aligned}\end{equation} $$

We then repeat, divide the first term in the dividend, $2x$ by the first term in the divisor, $x$:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x+2\\ \textcolor{crimson}{x}+3 \overline{) \ \ x^2+5x+7} \\ \underline{-(x^2+3x)}\ \ \ \ \ \\ \textcolor{crimson}{2x}+7 \end{array} \end{aligned}\end{equation} $$

Then multiply the result, $2$ by the entire divisor, $x+3$:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x+\textcolor{crimson}{2}\\ \textcolor{crimson}{x+3} \overline{) \ \ x^2+5x+7} \\ \underline{-(x^2+3x)}\ \ \ \ \ \\ 2x+7 \\ \underline{-(2x+6)} \end{array} \end{aligned}\end{equation} $$

Finally, we subtract:

$$ \begin{equation}\begin{aligned} \begin{array}{r} x+2\\ x+3 \overline{) \ \ x^2+5x+7} \\ \underline{-(x^2+3x)}\ \ \ \ \ \\ \textcolor{crimson}{2x+7} \\ \underline{-(\textcolor{crimson}{2x+6})} \\ 1 \end{array} \end{aligned}\end{equation} $$

The quotient is $x+2$ and the remainder is $1$. We can write the dividend in terms of the quotient, divisor and remainder: $$ \begin{equation}\begin{aligned} x^2+5x+7=(x+2)(x+3)+1\\ \end{aligned}\end{equation} $$

Example 2

Show that $x^2-1$ divided by $x-1$ is $x+1$. First, we divide: $$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x-1 \overline{) \ \ x^2+0x-1} \end{array} \end{aligned}\end{equation} $$

Notice how we place a $0x$ to represent the missing power of $x^1$. Then we multiply: $$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \end{array} \end{aligned}\end{equation} $$

Subtract: $$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \\ x\ \ \ \ \ \ \ \end{array} \end{aligned}\end{equation} $$

Bring down: $$ \begin{equation}\begin{aligned} \begin{array}{r} x\ \ \ \ \ \ \ \\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \\ x-1 \end{array} \end{aligned}\end{equation} $$

Divide: $$ \begin{equation}\begin{aligned} \begin{array}{r} x+1\\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \\ x-1 \end{array} \end{aligned}\end{equation} $$

Multiply: $$ \begin{equation}\begin{aligned} \begin{array}{r} x+1\\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \\ x-1 \\ \underline{-(x-1)} \end{array} \end{aligned}\end{equation} $$

Subtract: $$ \begin{equation}\begin{aligned} \begin{array}{r} x+1\\ x-1 \overline{) \ \ x^2+0x-1} \\ \underline{-(x^2-x)}\ \ \ \ \ \ \\ x-1 \\ \underline{-(x-1)} \\ 0 \end{array} \end{aligned}\end{equation} $$

The result is thus: $$ \begin{equation}\begin{aligned} x^2-1=(x+1)(x-1)+0\\ \end{aligned}\end{equation} $$

Challenge

Find the quotient and remainder for the following divisions:

• $x^2+5x+2$ divided by $x+2$
• $3x^2+5x-3$ divided by $3x+1$ (hint: the remainder is $\frac{-13}{3}$ not a linear polynomial)
• $2x^3+3x^2-9$ divided by $x-1$
• $4x^2-25y^2$ divided by $2x+5y$
• $x^2-9y^2+y$ divided by $x+3y$

It should be noted that the degree of the remainder is always less than the degree of the divisor.

For example, if we do a division of a polynomial by a linear divisor (degree of $1$), the remainder will be a constant (degree of $0$). Likewise, if we divide a polynomial by a quadratic divisor (degree of $2$), the remainder will either be a linear polynomial (degree of $1$) or a constant (degree of $0$).