Objectives:1. To divide polynomials using long and synthetic

division2. To apply the Factor and Remainder Theorems to

find real zeros of polynomial functions

As a class, use your vast mathematical knowledge to define each of these words without the aid of your textbook.

Quotient Remainder

Dividend Divisor

Divides Evenly

Factor

Use long division to divide 5 into 3462.

5 34626

30-

46

9

45-

12

2

10-

2

Use long division to divide 5 into 3462.

5 34626

30-

46

9

45-

12

2

10-

2

Divisor Dividend

Quotient

Remainder

Use long division to divide 5 into 3462.

3462 2692

5 5

Dividend

Divisor

Quotient

Remainder

Divisor

If you are lucky enough to get a remainder of zero when dividing, then the divisor divides evenlydivides evenly into the dividend.

This means that the divisor is a factorfactor of the dividend.

For example, when dividing 3 into 192, the remainder is 0. Therefore, 3 is a factor of 192.

Dividing polynomials works just like long division. In fact, it is called long divisionlong division!

Before you start dividing:

1. Make sure the divisor and dividend are in standard form (highest to lowest powers).

2. If your polynomial is missing a term, add it in with a coefficient of 0 as a place holder.

32 3x x 3 22 0 3x x x

Divide x + 1 into x2 + 3x + 5

Line up the first term of the quotient with the term of the dividend with the same degree.

21 3 5x x x

How many times does x go into x2?x

Multiply x by x + 1

2 x x-

2x-

5

2

Multiply 2 by x + 1

2 2x - -

3

Divide x + 1 into x2 + 3x + 5

21 3 5x x x x

2 x x-

2x-

5

2

2 2x - -

3Divisor

Dividend

Quotient

Remainder

Divide x + 1 into x2 + 3x + 5

2 3 5 32

1 1

x xx

x x

Divisor

Dividend

Quotient

Remainder

Divisor

Divide 6x3 – 16x2 + 17x – 6 by 3x – 2

Use long division to divide x4 – 10x2 + 2x + 3 by x – 3

When your divisor is of the form x - k, where k is a constant, then you can perform the division quicker and easier using just the coefficients of the dividend.

This is called fake division. I mean, synthetic divisionsynthetic division.

Synthetic Division Synthetic Division (of a Cubic Polynomial)To divide ax3 + bx2 + cx + d by x – k, use the

following pattern.k a b c d

a

ka

= Add terms

= Multiply by k

Coefficients of Quotient (in decreasing order)

Remainder

Synthetic DivisionSynthetic Division (of a Cubic Polynomial)To divide ax3 + bx2 + cx + d by x – k, use the

following pattern.

Important Note: Important Note: You are always adding columns using synthetic division, whereas you subtracted columns in long division.

k a b c d

a

ka

= Add terms

= Multiply by k

Synthetic Division Synthetic Division (of a Cubic Polynomial)To divide ax3 + bx2 + cx + d by x – k, use the

following pattern.

Important Note: Important Note: k can be positive or negative. If you divide by x + 2, then k = -2 because x + 2 = x – (-2).

k a b c d

a

ka

= Add terms

= Multiply by k

Synthetic Division Synthetic Division (of a Cubic Polynomial)To divide ax3 + bx2 + cx + d by x – k, use the

following pattern.

Important Note: Important Note: Add a coefficient of zero for any missing terms!

k a b c d

a

ka

= Add terms

= Multiply by k

Use synthetic division to divide x4 – 10x2 + 2x + 3 by x – 3

Evaluate f (3) for f (x) = x4 – 10x2 + 2x + 3.

If a polynomial f (x) is divided by x – k, the remainder is r = f (k).

This means that you could use synthetic division to evaluate f (5) or f (-2). Your answer will be the remainder.

Divide 2x3 + 9x2 + 4x + 5 by x + 3 using synthetic division.

Use synthetic division to divide f(x) = 2x3 – 11x2 + 3x + 36 by x – 3.

Since the remainder is zero when dividing f(x) by x – 3, we can write:

This means that x – 3 is a factorfactor of f(x).

2( )2 5 12,

3

f xx x

x

2 so ( ) ( 3)(2 5 12)f x x x x

A polynomial f(x) has a factor x – k if and only if f(k) = 0.

This theorem can be used to help factor/solve a polynomial function if you already know one of the factors.

Factor f(x) = 2x3 – 11x2 + 3x + 36 given that x – 3 is one factor of f(x). Then find the zeros of f(x).

Given that x – 4 is a factor of x3 – 6x2 + 5x + 12, rewrite x3 – 6x2 + 5x + 12 as a product of two polynomials.

Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42 given that f(7) = 0.

Rational Zero Test: we use this to find the rational zeros for a polynomial f(x). It says that if f(x) is a polynomial of the form:

1 2 11 2 1 0( ) n n

n nf x a x a x a x a x a

Then the rational zeros of f(x) will be of the form:p

qRational zero =

Possible rational zeros = factors of the constant term___factors of the leading coefficient

Where p = factor of the constant &

q = factor of leading coefficient

•Keep in mind that a polynomial can have rational zeros, irrational zeros and complex zeros.

p

q

Ex 1: Find all of the possible rational zeros of f(x)

4 3 2( ) 3 6f x x x x x

Ex 2: Find the rational zeros of: 4 3 2( ) 3 6f x x x x x

Let’s start by listing all of the possible rational zeros, then we will use synthetic division to test out the zeros:

1. Start with a list of factors of -6 (the constant term): p =

2. Next create a list of factors of 1 (leading coefficient): q =

3. Now list your possible rational zeros: p/q =

Testing all of those possibilities could take a while so let’s use the graph of f(x) to locate good possibilities for zeros.

Use your trace button!

4 3 2( ) 3 6f x x x x x

Ex 2 continued: Find all of the rational zeros of the function

Ex 3: Find all the real zeros of :

p = Factors of 3:

q = Factors of 2:

Candidates for rational zeros: p/q =

Let’s look at the graph: Which looks worth trying?

Now use synthetic division to test them out.

3 22 3 8 3 0x x x

Homework

Dividing Polynomials Worksheet

Page 127-128

36,38, 49-59 odd