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

http://www.youtube.com/watch?v=5dBAdzl2Mns James Sousa: Dividing Polynomials-Synthetic 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 \begin{align*}n\end{align*} must have exactly \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 \begin{align*}\left(x=\frac{p}{q}\right)\end{align*}, where \begin{align*}p\end{align*} is an integer factor of the constant term and \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 \begin{align*}f(x)\end{align*} divided by a linear divisor \begin{align*}(x-a)\end{align*} is equal to \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.

\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*}.

\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:

\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:

\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 \begin{align*}x\end{align*} go into \begin{align*}x^3\end{align*}?” which in this case is \begin{align*}x^2\end{align*}.

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

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

\begin{align*}& \overset{x^2}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\\ & \quad \ \underline{-(x^3-3x^2)}\end{align*}

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

\begin{align*}& \overset{x^2+5x+10}{x-3 \overline{ ) {x^3+2x^2-5x+7}}}\\ & \quad \ \underline{-(x^3-3x^2)}\\ & \qquad \qquad \ \ 5x^2-5x\\ & \qquad \quad \ \underline{-(5x^2-15x)}\\ & \qquad \qquad \qquad \quad \ 10x+7\\ & \qquad \qquad \qquad \underline{-(10x-30)}\\ & \qquad \qquad \qquad \qquad \qquad 37\end{align*}

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

\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 \begin{align*}x\end{align*}.

\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.

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

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

\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.

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

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

\begin{align*}& \overset{}{+3 \overline{ ) {1 \ \ 2 \ -5 \ \ 7}}}\\ & \quad \ \ \underline{\downarrow \ 3 \quad 15 \ \ 30}\\ & \quad \ \ 1 \ \ 5 \quad 10 \ \ 37 \end{align*}

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

\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.

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

2. Completely factor the following polynomial.

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

3. Divide the following polynomials.

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

**Answers:**

1. \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 \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.

\begin{align*}& x^4+6x^3+3x^2-26x-24\\ & =(x+1)(x^3+5x^2-2x-4)\\ & =(x+1)(x-2)(x^2+7x+12)\\ &= (x+1)(x-2)(x+3)(x+4)\end{align*}

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

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

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

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

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

Completely factor the following polynomials.

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

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

8. \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*}