Polynomial Long Division and Synthetic Division

• Use long division to divide polynomials by other polynomials.

• Use synthetic division to divide polynomials by binomials of the form (x – k).

• Use the Remainder Theorem and the Factor Theorem.

What You Should Learn

When our factoring techniques do not easily work… Analyzing and Graphing a Function

Let’s say we want to analyze this function and graph it:f(x) = x7 - 8x5 - 2x4 - 21x3 + 10x2 + 108x + 72 We know the left and right behavior We know the y- intercept To get a good approximation of the graph, we need to know the

x-intercepts or the “zeros”. To find all the real zeros of the function we must factor it completely.

Determining if one polynomial is a factor of another polynomial

Factoring a polynomial Polynomial division will help with this.

Today

We are going to learn about the process of division

Learn about a couple of theorems to help in factoring and solving higher level polynomials

Division of Polynomials In this section, we will study two procedures for

dividing polynomials.

These procedures are especially valuable in factoring and finding the zeros of polynomial functions.

Polynomial Division

Polynomial Division is very similar to long division.

Example:

13

31053 23

x

xxx

Polynomial Division

3105313 23 xxxx

2x

233 xx 26x x10

x2

xx 26 2 x12 3

4

x12 4

7

13

7

x

Subtract!!

Subtract!!

Subtract!!

Polynomial Division

Example:

Notice that there is no x term. However, we need to include it when we divide.

52

1592 23

x

xx

Polynomial Division

159252 23 xxx

2x

23 52 xx 24x x0

x2

xx 104 2 x10 15

5

x10 25

10

52

10

x

x0

Try This

Example:

Answer:

2

349105 234

x

xxxx

1743 23 xxx

Now let’s look at another method to divide… Why??? Sometimes it is easier…

Synthetic Division

Synthetic Division is a ‘shortcut’ for polynomial division that only works when dividing by a linear factor (x + b).

It involves the coefficients of the dividend, and the zero of the divisor.

ExampleExample

Divide:Divide: Step 1:Step 1:

Write the Write the coefficientscoefficients of the dividend in a of the dividend in a upside-down division symbol.upside-down division symbol.

1 5 6

1

652

x

xx

ExampleExample

Step 2:Step 2: Take the Take the zerozero of the divisor, and write it on of the divisor, and write it on

the left.the left. The divisor is x – 1, so the zero is 1.The divisor is x – 1, so the zero is 1.

1 5 61

1

652

x

xx

ExampleExample

Step 3:Step 3: Carry down the first coefficient.Carry down the first coefficient.

1 5 61

1

1

652

x

xx

ExampleExample

Step 4:Step 4: Multiply the zero by this number. Write the Multiply the zero by this number. Write the

product under the next coefficient.product under the next coefficient.

1 5 61

1

1

1

652

x

xx

ExampleExample

Step 5:Step 5: Add.Add.

1 5 61

1

1

6

1

652

x

xx

ExampleExample

Step etc.:Step etc.: Repeat as necessaryRepeat as necessary

1 5 61

1

1

6

6

12

1

652

x

xx

ExampleExample

The numbers at the bottom represent the The numbers at the bottom represent the coefficients of the answer. The new coefficients of the answer. The new polynomial will be one degree less than polynomial will be one degree less than the original.the original.

1 5 61

1

1

6

6

12 1

126

x

x

1

652

x

xx

Synthetic DivisionThe pattern for synthetic division of a cubic polynomial is summarized

as follows. (The pattern for higher-degree polynomials is similar.)

Synthetic Division

This algorithm for synthetic division works only for divisors of the form x – k.

Remember that x + k = x – (–k).

Using Synthetic DivisionUse synthetic division to divide x4 – 10x2 – 2x + 4 by x + 3.

Solution:

You should set up the array as follows. Note that a zero is included for the missing x3-term in the dividend.

Example – Solution Then, use the synthetic division pattern by adding terms in columns and

multiplying the results by –3.

So, you have

.

cont’d

Try These

Examples:(x4 + x3 – 11x2 – 5x + 30) (x – 2)(x4 – 1) (x + 1)

[Don’t forget to include the missing terms!]

Answers:x3 + 3x2 – 5x – 15x3 – x2 + x – 1

Application of Long Division

To begin, suppose you are given the graph of

f (x) = 6x3 – 19x2 + 16x – 4.

Long Division of Polynomials

Notice that a zero of f occurs at x = 2.

Because x = 2 is a zero of f,you know that (x – 2) isa factor of f (x). This means thatthere exists a second-degree polynomial q (x) such that

f (x) = (x – 2) q(x).

To find q(x), you can uselong division.

Example - Long Division of Polynomials

Divide 6x3 – 19x2 + 16x – 4 by x – 2, and use

the result to factor the polynomial

completely.

Example 1 – SolutionThink

Think

Think

Multiply: 6x2(x – 2).

Subtract.

Multiply: 2(x – 2).

Subtract.

Multiply: –7x(x – 2).

Subtract.

Example – Solution

From this division, you can conclude that

6x3 – 19x2 + 16x – 4 = (x – 2)(6x2 – 7x + 2)

and by factoring the quadratic 6x2 – 7x + 2,

you have

6x3 – 19x2 + 16x – 4 =

(x – 2)(2x – 1)(3x – 2).

cont’d

Long Division of Polynomials

Example – Factoring a Polynomial: Repeated Division

Show that (x – 2) and (x + 3) are factors of

f (x) = 2x4 + 7x3 – 4x2 – 27x – 18.

Then find the remaining factors of f (x).

Solution:Using synthetic division with the factor (x – 2), you obtain the following.

0 remainder, so f (2) = 0and (x – 2) is a factor.

Example – Solution

Take the result of this division and perform synthetic division again using the factor (x + 3).

Because the resulting quadratic expression factors as

2x2 + 5x + 3 = (2x + 3)(x + 1)

the complete factorization of f (x) is

f (x) = (x – 2)(x + 3)(2x + 3)(x + 1).

0 remainder, so f (–3) = 0and (x + 3) is a factor.

cont’d