Transcript Polynomial Division

Polynomial and
Synthetic Division
Pre-Calculus
Mrs.Volynskaya
Polynomial Division

Polynomial Division is very similar to long
division.

Example:
3x 3  5 x 2  10x  3
3x  1
Polynomial Division
x
 2x
2
4

7
3x  1
3x  1 3x3  5x 2  10x  3
3x  x
3
2
Subtract!!
 6x  10 x
2
 6x  2x
12 x  3
12 x  4
2
7
Subtract!!
Subtract!!
Polynomial Division

Example:
2 x  9 x  15
2x  5
3

2
Notice that there is no x term. However, we
need to include it when we divide.
Polynomial Division
x
2
 2x  5

 10
2x  5
2 x  5 2 x3  9 x 2  0 x  15
2 x  5x
3
2
 4x  0 x
2
 4 x  10x
 10 x  15
 10 x  25
2
 10
Try This

Example:
x  5 x  10x  9 x  34
x2
4

3
2
Answer:
x 3  3x 2  4 x  17
What does it mean if a number
divides evenly into another??
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.


(SUSPENSE IS BUILDING)
Example
 Divide:
 Step 1:
x 2  5x  6
x 1
 Write the coefficients of the dividend in a
upside-down division symbol.
1
5
6
Example
 Step 2:
x 2  5x  6
x 1
 Take the zero of the divisor, and write it on
the left.
 The divisor is x – 1, so the zero is 1.
1 1
5
6
Example
 Step 3:
x 2  5x  6
x 1
 Carry down the first coefficient.
1 1
1
5
6
Example
 Step 4:
x 2  5x  6
x 1
 Multiply the zero by this number. Write the
product under the next coefficient.
1 1
1
5
1
6
Example
 Step 5:
x 2  5x  6
x 1
 Add.
1 1
5
1
1
6
6
Example
x 2  5x  6
x 1
 Step etc.:
 Repeat as necessary
1 1
1
5
1
6
6
6 12
Example
The numbers at the bottom represent the
coefficients of the answer. The new
polynomial will be one degree less than
the original.
x 2  5x  6

x 1
1 1 5 6
1 6
12
x6
x 1
1 6 12
Synthetic Division
The 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 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
cont’d

Then, use the synthetic division pattern by adding terms in columns
and multiplying the results by –3.

So, you have

.
Try These

Examples:
+ x3 – 11x2 – 5x + 30)  (x – 2)
 (x4 – 1)  (x + 1)
[Don’t forget to include the missing terms!]
 (x4

Answers:
+ 3x2 – 5x – 15
 x3 – x2 + x – 1
 x3
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) is
a factor of f (x). This means that
there exists a second-degree
polynomial q (x) such that
f (x) = (x – 2)  q(x).
To find q(x), you can use
long 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 – Solution
Think
Think
Think
Multiply: 6x2(x – 2).
Subtract.
Multiply: –7x(x – 2).
Subtract.
Multiply: 2(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 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 be
f (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
cont’d
Because the remainder is r = –9, you can conclude that
r = f(k)
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
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) = 0
and (x – 2) is a factor.
Example – Solution
cont’d
Take the result of this division and perform synthetic division again
using the factor (x + 3).
0 remainder, so f(–3) = 0
and (x + 3) is a factor.
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).
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.