Polynomial Division

You do polynomial division the way you do long division of numbers. It's difficult to describe the general procedure in words, so I'll work through some examples step-by-step.

Example. Find the quotient and remainder when $x^2 + 2 x - 3$ is divided by $x - 2$ .

You may also see this kind of problem written like this: "Perform the division $\dfrac{x^2 + 2 x - 3}{x - 2}$ ."

Set up the division the way you'd set up division of numbers. To start, look at the first term (x) in $x - 2$ and the first term ($x^2$ ) in $x^2 +
   2 x - 3$ .

$$\hbox{\epsfysize=0.35in \epsffile{polynomial-division-1a.eps}}$$

Ask yourself: "What times x gives $x^2$ ?"

$$(\hbox{?}) \cdot x = x^2.$$

You can see that x works, so put it on top:

$$\hbox{\epsfysize=0.3in \epsffile{polynomial-division-1b.eps}}$$

Next, multiply the $x - 2$ by the x on top, and put the result $x^2 - 2 x$ under $x^2 + 2 x - 3$ . Line up terms with the same power of x:

$$\hbox{\epsfysize=0.7in \epsffile{polynomial-division-1c.eps}}$$

Subtract: $(x^2 + 2 x) -
   (x^2 - 2 x) = 4 x$ .

$$\hbox{\epsfysize=0.7in \epsffile{polynomial-division-1d.eps}}$$

(When you subtract, be careful of the signs! In this case, the $x^2$ terms cancel, but $2 x - (-2 x) = 2 x + 2 x = 4
   x$ .)

Next, bring down the -3 and put it next to the $4 x$ :

$$\hbox{\epsfysize=0.7in \epsffile{polynomial-division-1e.eps}}$$

Look at the first term (x) in $x - 2$ and the first term ($4 x$ ) in $4 x^2
   - 3$ .

$$\hbox{\epsfysize=0.85in \epsffile{polynomial-division-1f.eps}}$$

Ask yourself: "What times x gives $4 x$ ?"

$$(\hbox{?}) \cdot x = 4 x.$$

You can see that 4 works, so put it on top:

$$\hbox{\epsfysize=0.75in \epsffile{polynomial-division-1g.eps}}$$

Multiply the $x - 2$ by the 4 on top, and put the result $4 x - 8$ under $4 x - 3$ . Line up terms with the same power of x:

$$\hbox{\epsfysize=1.15in \epsffile{polynomial-division-1h.eps}}$$

(You don't multiply $x -
   2$ by the $x + 4$ on top, just the 4; you already multiplied by the "x" in $x + 4$ in an earlier step.)

Subtract: $(4 x - 3) - (4 x
   - 8) = 5$ .

$$\hbox{\epsfysize=1.15in \epsffile{polynomial-division-1i.eps}}$$

At this point, the "x" of "$x - 2$ " doesn't go into 5, so the division is finished.

The quotient is $x + 4$ , the expression on the top. And the remainder is 5.

If you're just asked for the quotient and remainder, you're done.

If the problem asked you to do the division $\dfrac{x^2 + 2 x - 3}{x - 2}$ , then you'd write it this way:

$$\dfrac{x^2 + 2 x - 3}{x - 2} = (x + 4) + \dfrac{5}{x - 2}.$$

The quotient $x + 4$ goes in front. The remainder 5 goes on top of the fraction. The expression $x - 2$ , which was on the bottom of the original fraction, goes on the bottom of the new fraction.

If you have trouble remembering where everything goes, think about how you convert improper fractions to mixed numbers. Let's say you want to convert $\dfrac{38}{7}$ to a mixed number. To do this, you divide 38 by 7:

$$\hbox{\epsfysize=0.65in \epsffile{polynomial-division-1j.eps}}$$

Then $\dfrac{38}{7} =
   5\frac{3}{7}$ .

However, if you think about it, $5\frac{3}{7}$ --- which you read as "5 and $\dfrac{3}{7}$ " --- means "$5 + \dfrac{3}{7}$ ", so

$$\dfrac{38}{7} = 5 + \dfrac{3}{7}.$$

This is the same as what I did with the polynomials: The quotient 5 goes in front. The remainder 3 goes on top of the fraction. The expression 7, which was on the bottom of the original fraction $\dfrac{38}{7}$ , goes on the bottom of the new fraction.

In other words, as often happens when you're doing algebra, you can figure out "what to do with variables" by thinking about "what you know to do with numbers".


Example. Find the quotient and remainder when $2 x^3 - 2 x + 1$ is divided by $x + 1$ .

Since $2 x^3 - 2 x + 1$ is missing an $x^2$ term, I'll write "$0\,x^2$ " as a placeholder. It is optional, but this makes it less likely that you'll make a mistake when you do the subtraction.

Look at the first term (x) in $x + 1$ and the first term ($2 x^3$ ) in $2 x^3 - 2 x + 1$ :

$$\hbox{\epsfysize=0.35in \epsffile{polynomial-division-2a.eps}}$$

Ask yourself: "What times x gives $2 x^3$ ?"

$$(\hbox{?}) \cdot x = 2 x^3.$$

You can see that $2 x^2$ works, so put it on top:

$$\hbox{\epsfysize=0.4in \epsffile{polynomial-division-2b.eps}}$$

Next, multiply $x + 1$ by the $2 x^2$ on top, and put the result $2 x^3 + 2 x^2$ under $2 x^3 - 2 x + 1$ . Line up terms with the same power of x:

$$\hbox{\epsfysize=0.6in \epsffile{polynomial-division-2c.eps}}$$

Note that the $2 x^2$ goes under the $0 x^2$ I put in as a placeholder.

Subtract: $(2 x^3 + 0 x^2)
   - (2 x^3 + 2 x^2) = -2 x^2$ .

$$\hbox{\epsfysize=0.85in \epsffile{polynomial-division-2d.eps}}$$

Next, bring down the $-2
   x$ and put it next to the $-2 x^2$ :

$$\hbox{\epsfysize=0.8in \epsffile{polynomial-division-2e.eps}}$$

Look at the first term (x) in $x + 1$ and the first term ($-2 x^2$ ) in $-2 x^2 - 2 x$ :

$$\hbox{\epsfysize=1in \epsffile{polynomial-division-2f.eps}}$$

Ask yourself: "What times x gives $-2 x^2$ ?"

$$(\hbox{?}) \cdot x = -2 x^2.$$

You can see that $-2 x$ works, so put it on top:

$$\hbox{\epsfysize=0.8in \epsffile{polynomial-division-2g.eps}}$$

Multiply $x + 1$ by the $-2 x$ on top (just the "$-2 x$ " --- you multiplied by the "$2 x^2$ "), and put the result $-2 x^2 - 2 x$ under $-2 x^2 - 2 x$ :

$$\hbox{\epsfysize=1.1in \epsffile{polynomial-division-2h.eps}}$$

Subtract: $(-2 x^2 - 2 x) -
   (2 x^2 - 2 x) = 0$ .

$$\hbox{\epsfysize=1.2in \epsffile{polynomial-division-2i.eps}}$$

Bring down the 1 --- since $0 + 1 = 1$ , I just write the 1:

$$\hbox{\epsfysize=1.2in \epsffile{polynomial-division-2j.eps}}$$

Since $x + 1$ does not go into 1, the division is finished:

$$\hbox{\epsfysize=1.2in \epsffile{polynomial-division-2k.eps}}$$

The quotient is $2 x^2 - 2
   x$ and the remainder is 1.

In fraction form, this would be written as

$$\dfrac{2 x^3 - 2 x + 1}{x + 1} = (2 x^2 - 2 x) + \dfrac{1}{x + 1}.\quad\halmos$$


Example. Find the quotient and remainder when $3 x^3 - x^2 - 7 x +
   12$ is divided by $x^2 - 2 x + 1$ .

Look at the first term ($x^2$ ) in $x^2 - 2 x + 1$ and the first term ($3 x^3$ ) in $3
   x^3 - x^2 - 7 x + 12$ :

$$\hbox{\epsfysize=0.42in \epsffile{polynomial-division-3a.eps}}$$

Ask yourself: "What times $x^2$ gives $3 x^3$ ?"

$$(\hbox{?}) \cdot x^2 = 3 x^3.$$

You can see that $3 x$ works, so put it on top:

$$\hbox{\epsfysize=0.35in \epsffile{polynomial-division-3b.eps}}$$

Next, multiply $x^2 - 2 x
   + 1$ by the $3 x$ on top, and put the result $3 x^2 - 6 x^2 + 3 x$ under $3 x^3 - x^2 - 7 x + 12$ . Line up terms with the same power of x:

$$\hbox{\epsfysize=0.68in \epsffile{polynomial-division-3c.eps}}$$

Subtract: $(3 x^3 - x^2 -
   7 x + 12) - (3 x^2 - 6 x^2 + 3 x) = 5 x^2 - 10 x$ .

$$\hbox{\epsfysize=0.8in \epsffile{polynomial-division-3d.eps}}$$

Next, bring down the 12 and put it next to the $5 x^2 - 10 x$ :

$$\hbox{\epsfysize=0.8in \epsffile{polynomial-division-3e.eps}}$$

Look at the first term ($x^2$ ) in $x^2 - 2 x + 1$ and the first term ($5 x^2$ ) in $5
   x^2 - 10 x$ :

$$\hbox{\epsfysize=0.94in \epsffile{polynomial-division-3f.eps}}$$

Ask yourself: "What times $x^2$ gives $5 x^2$ ?"

$$(\hbox{?}) \cdot x^2 = 5 x^2.$$

You can see that 5 works, so put it on top:

$$\hbox{\epsfysize=0.8in \epsffile{polynomial-division-3g.eps}}$$

Multiply $x^2 - 2 x + 1$ by the 5 on top (just the "5" --- you already multiplied by the "$3 x$ "), and put the result $5 x^2 - 10 x + 5$ under $5 x^2 - 10 x + 12$ :

$$\hbox{\epsfysize=1.32in \epsffile{polynomial-division-3h.eps}}$$

Subtract: $(5 x^2 - 10 x +
   12) - (5 x^2 - 10 x + 5) = 7$ .

$$\hbox{\epsfysize=1.2in \epsffile{polynomial-division-3i.eps}}$$

Since $x^2 - 2 x + 1$ does not go into 7, the division is finished.

The quotient is $3 x + 5$ and the remainder is 7. In fraction form, this would be written as

$$\dfrac{3 x^3 - x^2 - 7 x + 12}{x^2 - 2 x + 1} = 3 x + 5 + \dfrac{7}{x^2 - 2 x + 1}.\quad\halmos$$


Example.

$$\dfrac{x^4 - 4x^2 - 3}{x^2 - 1} = x^2 - 3 - \dfrac{6}{x^2 - 1}, \quad\hbox{or}\quad x^4 - 4x^2 - 3 = (x^2 - 3)(x^2 - 1) - 6. \quad\halmos$$


Example.

$$\dfrac{x^3 + 4x^2 - 5x - 20}{x + 4} = x^2 - 5, \quad\hbox{or}\quad x^3 + 4x^2 - 5x - 20 = (x^2 - 5)(x + 4).$$

Sine the remainder is 0, this shows that $x^3 + 4x^2 - 5x - 20$ factors as $(x + 4)(x^2 - 5)$ .


The Remainder Theorem. If $p(x)$ is a polynomial, the remainder when $p(x)$ is divided by $x - a$ is $p(a)$ .


Example. Divide $p(x) = 3x^4 - 2x + 3$ by $x - 2$ :

$$3x^4 - 2x + 3 = (x - 2)(3x^3 + 6x^2 + 12x + 22) + 47.$$

The remainder is 47. On the other hand,

$$p(2) = 3\cdot 16 - 4 + 3 = 47.$$

The remainder is the same as $p(2)$ .


The Root Theorem. $x - a$ divides $p(x)$ evenly if and only if $p(a) = 0$ (i.e. $x = a$ is a root of $p(x)$ ).


Example. Let $p(x) = 2x^4 + 3x^3 + 1$ .

$$p(-1) = 2 - 3 + 1 = 0.$$

Therefore, $x - 1$ should divide $2x^4 + 3x^3 + 1$ . In fact,

$$2x^4 + 3x^3 + 1 = (x - 1)(2x^3 + x^2 - x + 1).\quad\halmos$$


Example. Let $p(x) = x^2 - 6x + 8$ . Then

$$p(x) = (x - 2)(x - 4).$$

Since $x - 2$ and $x - 4$ are factors, $x = 2$ and $x =
   4$ should be roots --- and they are:

$$p(2) = 4 - 12 + 8 = 0, \quad p(4) = 16 - 24 + 8 = 0.\quad\halmos$$


Example. If you know or can guess a factor, you can sometimes complete a factorization by long division. For example $x^3 - 27$ has $x = 3$ as a root. By the fact I stated earlier, this means that $x - 3$ is a factor. Divide $x - 3$ into $x^3
   - 27$ to get $x^2 + 3x + 9$ .

Therefore,

$$x^3 - 27 = (x - 3)(x^2 + 3x + 9).\quad\halmos$$


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