Polynomial and Synthetic Division

Pre-Calculus

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

What does it mean if a number divides evenly into another??

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

Using Synthetic Division

Use 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

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

The Remainder and Factor Theorems

The Remainder and Factor Theorems

The remainder obtained in the synthetic division process has an important interpretation, as described in the Remainder Theorem.

The Remainder Theorem tells you that synthetic division can be used to evaluate a polynomial function. That is, to evaluate a polynomial function f (x) when x = k, divide f (x) by x – k. The remainder will bef (k).

Example Using the Remainder Theorem

Use the Remainder Theorem to evaluate the following function at x = –2.

f (x) = 3x3 + 8x2 + 5x – 7

Solution: Using synthetic division, you obtain the following.

Example – Solution

Because the remainder is r = –9, you can conclude that

f (–2) = –9.

This means that (–2, –9) is a point on the graph of f. You can check this by substituting x = –2 in the original function.

Check:

f (–2) = 3(–2)3 + 8(–2)2 + 5(–2) – 7

= 3(–8) + 8(4) – 10 – 7

= –9

r = f (k)

cont’d

The Remainder and Factor Theorems

Another important theorem is the Factor Theorem, stated below.

This theorem states that you can test to see whether a polynomial has (x – k) as a factor by evaluating the polynomial at x = k.

If the result is 0, (x – k) is a factor.

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

The Remainder and Factor Theorems

For instance, if you find that x – k divides evenly into f (x) (with no remainder), try sketching the graph of f.

You should find that (k, 0) is an x-intercept of the graph.

• 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.

Can you…

Questions?

OK, what if you are given a polynomial to factor without any clues as to possible roots?

Whatcha gonna do?

Rational Root Theorem …

Homework

The Usual for section 2-3