<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />
You are viewing an older version of this Concept. Go to the latest version.

# Division of Polynomials

## Using long division to divide polynomials

0%
Progress
Practice Division of Polynomials
Progress
0%
Polynomial Long Division and Synthetic Division

While you may be experienced in factoring, there will always be polynomials that do not readily factor using basic or advanced techniques.  How can you identify the roots of these polynomials?

#### Watch This

http://www.youtube.com/watch?v=brpNxPAkv1c James Sousa: Dividing Polynomials-Long Division

#### Guidance

There are numerous theorems that point out relationships between polynomials and their factors.  For example there is a theorem that a polynomial of degree n\begin{align*}n\end{align*} must have exactly n\begin{align*}n\end{align*} solutions/factors that may or may not be real numbers.  The Rational Root Theorem and the Remainder Theorem are two theorems that are particularly useful starting places when manipulating polynomials.

• The Rational Root Theorem states that in a polynomial, every rational solution can be written as a reduced fraction (x=pq)\begin{align*}\left(x=\frac{p}{q}\right)\end{align*}, where p\begin{align*}p\end{align*} is an integer factor of the constant term and q\begin{align*}q\end{align*} is an integer factor of the leading coefficient. Example A shows how all the possible rational solutions can be listed using the Rational Root Theorem.
• The Remainder Theorem states that the remainder of a polynomial f(x)\begin{align*}f(x)\end{align*} divided by a linear divisor (xa)\begin{align*}(x-a)\end{align*} is equal to f(a)\begin{align*}f(a)\end{align*}. The Remainder Theorem is only useful after you have performed polynomial long division because you are usually never given the divisor and the remainder to start. The main purpose of the Remainder Theorem in this setting is a means of double checking your application of polynomial long division. Example B shows how the Remainder Theorem is used.

Polynomial long division is identical to regular long division.  Synthetic division is a condensed version of regular long division where only the coefficients are kept track of.  In Example B polynomial long division is used and in Example C synthetic long division is used.

Example A

Identify all possible rational solutions of the following polynomial using the Rational Root Theorem.

12x1891x17+x16++2x214x+5=0\begin{align*}12x^{18}-91x^{17}+x^{16}+\cdots+2x^2-14x+5=0\end{align*}

Solution:   The integer factors of 5 are 1, 5.  The integer factors of 12 are 1, 2, 3, 4, 6 and 12.  Since pairs of factors could both be negative, remember to include ±\begin{align*}\pm\end{align*}

±pq=11,12,13,14,16,112,51,52,53,54,56,512\begin{align*}\pm \frac{p}{q}=\frac{1}{1}, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \frac{1}{6}, \frac{1}{12}, \frac{5}{1}, \frac{5}{2}, \frac{5}{3}, \frac{5}{4}, \frac{5}{6}, \frac{5}{12}\end{align*}

While narrowing the possible solutions down to 24 possible rational answers may not seem like a big improvement, it surely is.  This is especially true considering there are only a handful of integer solutions.  If this question required you to find a solution, then the Rational Root Theorem would give you a great starting place.

Example B

Use Polynomial Long Division to divide:

x3+2x25x+7x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}

Solution:  First note that it is clear that 3 is not a root of the polynomial because of the Rational Root Theorem and so there will definitely be a remainder.  Start a polynomial long division question by writing the problem like a long division problem with regular numbers:

¯x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}\overline{}{x-3 \overline { ) {x^3+2x^2-5x+7}}}\end{align*}

Just like with regular numbers ask yourself “how many times does x\begin{align*}x\end{align*} go into x3\begin{align*}x^3\end{align*}?” which in this case is x2\begin{align*}x^2\end{align*}.

x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2\begin{align*}\overset{x^2}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\end{align*}

Now multiply the x2\begin{align*}x^2\end{align*} by x3\begin{align*}x-3\end{align*} and copy below.  Remember to subtract the entire quantity.

x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2 (x33x2)

Combine the rows, bring down the next number and repeat.

x3)x3+2x25x+7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯x2+5x+10 (x33x2)  5x25x (5x215x) 10x+7(10x30)37

The number 37 is the remainder.  There are two things to think about at this point.  First, interpret in an equation:

x3+2x25x+7x3=(x2+5x+10)+37x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}=(x^2+5x+10)+\frac{37}{x-3}\end{align*}

Second, check your result with the Remainder Theorem which states that the original function evaluated at 3 must be 37.  Notice the notation indicating to substitute 3 in for x\begin{align*}x\end{align*}

(x3+2x25x+7)|x=3=33+23253+7=27+1815+7=37\begin{align*}(x^3+2x^2-5x+7)|_{x=3}=3^3+2 \cdot 3^2-5 \cdot 3+7=27+18-15+7=37\end{align*}

Example C

Use Synthetic Division to divide the same rational expression as the previous example.

Solution:  Synthetic division is mostly used when the leading coefficients of the numerator and denominator are equal to 1 and the divisor is a first degree binomial.

x3+2x25x+7x3\begin{align*}\frac{x^3+2x^2-5x+7}{x-3}\end{align*}

Instead of continually writing and rewriting the x\begin{align*}x\end{align*} symbols, synthetic division relies on an ordered spacing.

+3)1  2 5  7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯\begin{align*}\overset{}{+3 \overline{ ) {1 \ \ 2 \ -5 \ \ 7}}} \end{align*}

Notice how only the coefficients for the denominator are used and the divisor includes a positive three rather than a negative three.  The first coefficient is brought down and then multiplied by the three to produce the value which goes beneath the 2.

+3)1  2 5  7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯ 31

Next the new column is added. 2+3=5\begin{align*}2+3=5\end{align*}, which goes beneath the 2nd\begin{align*}2^{nd}\end{align*} column.  Now, multiply 5+3=15\begin{align*}5 \cdot +3=15\end{align*}, which goes underneath the -5 in the 3rd\begin{align*}3^{rd}\end{align*} column.   And the process repeats…

+3)1  2 5  7¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯   315  30  1  510  37

The last number, 37, is the remainder.  The three other numbers represent the quadratic that is identical to the solution to Example B.

1x2+5x+10\begin{align*}1x^2+5x+10\end{align*}

Concept Problem Revisited

Identifying roots of polynomials by hand can be tricky business.  The best way to identify roots is to use the rational root theorem to quickly identify likely candidates for solutions and then use synthetic or polynomial long division to quickly and effectively test them to see if their remainders are truly zero.

#### Vocabulary

Polynomial long division is a procedure with rules identical to regular long division.  The only difference is the dividend and divisor are polynomials.

Synthetic division is an abbreviated version of polynomial long division where only coefficients are used.

#### Guided Practice

1. Divide the following polynomials.

x3+2x24x+8x2\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}\end{align*}

2. Completely factor the following polynomial.

x4+6x3+3x226x24\begin{align*}x^4+6x^3+3x^2-26x-24\end{align*}

3. Divide the following polynomials.

3x52x2+10x5x1\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}\end{align*}

1. x3+2x24x+8x2=x2+4x+4+16x2\begin{align*}\frac{x^3+2x^2-4x+8}{x-2}=x^2+4x+4+\frac{16}{x-2}\end{align*}

2. Notice that possible roots are ±1,2,3,4,6,8,24\begin{align*}\pm 1, 2, 3, 4, 6, 8, 24\end{align*}.  Of these 14 possibilities, four will yield a remainder of zero.  When you find one, repeat the process.

x4+6x3+3x226x24=(x+1)(x3+5x22x4)=(x+1)(x2)(x2+7x+12)=(x+1)(x2)(x+3)(x+4)

3. 3x52x2+10x5x1=3x4+3x3+3x2+x+11+6x1\begin{align*}\frac{3x^5-2x^2+10x-5}{x-1}=3x^4+3x^3+3x^2+x+11+\frac{6}{x-1}\end{align*}

#### Practice

Identify all possible rational solutions of the following polynomials using the Rational Root Theorem.

1. 15x1412x13+x12++2x25x+5=0\begin{align*}15x^{14}-12x^{13}+x^{12}+ \cdots+2x^2-5x+5=0\end{align*}

2. 18x11+42x10+x9++x23x+7=0\begin{align*}18x^{11}+42x^{10}+x^9+\cdots+x^2-3x+7=0\end{align*}

3. 12x16+11x15+3x14++6x22x+11=0\begin{align*}12x^{16}+11x^{15}+3x^{14}+\cdots+6x^2-2x+11=0\end{align*}

4. 14x77x6+x5++x2+6x+3=0\begin{align*}14x^7-7x^6+x^5+\cdots+x^2+6x+3=0\end{align*}

5. 9x910x8+3x7++4x22x+2=0\begin{align*}9x^9-10x^8+3x^7+\cdots+4x^2-2x+2=0\end{align*}

Completely factor the following polynomials.

6. 2x4x321x226x8\begin{align*}2x^4-x^3-21x^2-26x-8\end{align*}

7. x4+7x3+5x231x30\begin{align*}x^4+7x^3+5x^2-31x-30\end{align*}

8. x4+3x38x212x+16\begin{align*}x^4+3x^3-8x^2-12x+16\end{align*}

9. \begin{align*}4x^4+19x^3-48x^2-117x-54\end{align*}

10. \begin{align*}2x^4+17x^3-8x^2-173x+210\end{align*}

Divide the following polynomials.

11. \begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+4}\end{align*}

12. \begin{align*}\frac{x^4+7x^3+5x^2-31x-30}{x+2}\end{align*}

13. \begin{align*}\frac{x^4+3x^3-8x^2-12x+16}{x+3}\end{align*}

14. \begin{align*}\frac{2x^4-x^3-21x^2-26x-8}{x^3-x^2-10x-8}\end{align*}

15. \begin{align*}\frac{x^4+8x^3+3x^2-32x-28}{x^3+10x^2+23x+14}\end{align*}

### Vocabulary Language: English

Denominator

Denominator

The denominator of a fraction (rational number) is the number on the bottom and indicates the total number of equal parts in the whole or the group. $\frac{5}{8}$ has denominator $8$.
Dividend

Dividend

In a division problem, the dividend is the number or expression that is being divided.
divisor

divisor

In a division problem, the divisor is the number or expression that is being divided into the dividend. For example: In the expression $152 \div 6$, 6 is the divisor and 152 is the dividend.
Polynomial long division

Polynomial long division

Polynomial long division is the standard method of long division, applied to the division of polynomials.
Rational Expression

Rational Expression

A rational expression is a fraction with polynomials in the numerator and the denominator.
Rational Root Theorem

Rational Root Theorem

The rational root theorem states that for a polynomial, $f(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$, where $a_n, a_{n-1}, \cdots a_0$ are integers, the rational roots can be determined from the factors of $a_n$ and $a_0$. More specifically, if $p$ is a factor of $a_0$ and $q$ is a factor of $a_n$, then all the rational factors will have the form $\pm \frac{p}{q}$.
Remainder Theorem

Remainder Theorem

The remainder theorem states that if $f(k) = r$, then $r$ is the remainder when dividing $f(x)$ by $(x - k)$.
Synthetic Division

Synthetic Division

Synthetic division is a shorthand version of polynomial long division where only the coefficients of the polynomial are used.