Polynomial and Synthetic Division

Pre-CalculusLesson 2-3

• 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 FunctionLet’s say we want to analyze this function and graph it:

f(x) = x7 - 8x5 - 2x4 - 21x3 + 10x2 + 108x + 72We know the left and right behaviorWe know the y- interceptTo 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 divisionLearn about a couple of theorems to help in factoring and solving higher level polynomials

Division of PolynomialsIn 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:

1331053 23

+−+−

xxxx

Polynomial Division

3105313 23 −+−+ xxxx

2x

233 xx +26x− x10+

x2−

xx 26 2 −−x12 3−

4+

x12 4+

7−

137+

−+

x

Subtract!!

Subtract!!

Subtract!!

Polynomial Division

Example:

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

521592 23

−+−

xxx

Polynomial Division

159252 23 ++−− xxx

2x

23 52 xx −24x− x0+

x2−

xx 104 2 +−x10− 15+

5−

x10− 25+

10−

5210−

−+

x

x0

Try This

Example:

Answer:

2349105 234

+−−++

xxxxx

1743 23 −++ xxx

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)

ExampleExample

Divide:Divide:Step 1:Step 1:

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

1 5 6

1652

−++

xxx

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 The divisor is x –– 1, so the zero is 1.1, so the zero is 1.

1 5 61

1652

−++

xxx

ExampleExample

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

1 5 61

1

1652

−++

xxx

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

11

1652

−++

xxx

ExampleExample

Step 5:Step 5:Add.Add.

1 5 61

116

1652

−++

xxx

ExampleExample

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

1 5 61

116

612

1652

−++

xxx

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

116

612 1

126−

++x

x

=−++

1652

xxx

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 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 – SolutionThen, 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 off (x) = 6x3 – 19x2 + 16x – 4.

Long Division of PolynomialsNotice 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 – 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 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 – SolutionBecause the remainder is r = –9, you can conclude thatf (–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 off (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 – SolutionTake the result of this division and perform synthetic division again

using the factor (x + 3).

Because the resulting quadratic expression factors as2x2 + 5x + 3 = (2x + 3)(x + 1)

the complete factorization of f (x) isf (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…