Polynomials Functions Review (2)

Division Algorithm and Long Division of Polynomials

Example Set No.1 Find the quotient and remainder for each division problem.

4 2 2A. 8 6 3 1 2 2x x x x x

Be sure that the dividend and the divisor are written in descending order.

Account for each term in the dividend – physically write any missing term with a coefficient of 0.

3

4 2 2

2 4 2

8 6 3 1 2 2

2 2 8 6 30 1

x x x x x

x xx x x x

Rewrite the division problem in “long division” form

2 4 3 22 2 8 0 6 3 1x x x x x x

For the first step, divide the leading term of the divisor into the leading term of the dividend.

42

2

84

2

xx

x

The result is the first term of the quotient.

Multiply this result by the divisor and subtract from the dividend.

2 2 4 3 24 2 2 8 4 8x x x x x x

Remember that you are subtracting the entire product from the dividend.

Now divide the leading term of the divisor into the leading term of the current remainder.

3

2

42

2

xx

x This is the next term of

the quotient.

Multiply this result by the divisor and subtract from the current remainder in the dividend.

2 3 22 2 2 4 2 4x x x x x x

Since the degree of the divisor is greater than the degree of the current remainder, the division is complete.

4 2 2 28 6 3 1 2 2 4 2 7 1x x x x x x x x

4 2 2 22

7 18 6 3 1 2 2 4 2

2 2

xx x x x x x x

x x

You can also write the result of the division in the form:

dividend remainderquotient

divisor divisor

Check your work

Check your work (continued)

3B. 1 1x x

Noticing that the dividend is a difference of cubes, we know that we should obtain:

2 1x x

23 01 10xx x x 3

2xx

x 2 3 21x x x x

2xx

x 21x x x x

1x

x 1 1 1x x

3 21 1 1x x x x

Since the remainder of the division is 0, we can conclude that

1x is a factor of 3 1x

4 2 2C. 3 1 2 3x x x x

Take a few minutes to complete this division before the solution is shown.

4 2 2 23 1 2 3 2 4 2 11x x x x x x x

4 22

2 2

3 1 2 112 4

2 3 2 3

x x xx x

x x x x

Remember to check your answers either by hand or by using the graphing calculator.

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Synthetic Substitution and Synthetic Division

Direct Substitution

4 21 4 1 15 1 4P

4 1 15 1 4

4 15 4 15

Synthetic Substitution

Value of x being substituted into the function.

Coefficients of polynomial function in descending order, filling in 0 for any “missing” term.

First term drops straight down.

1 4 4 1 4 4

0 4 4 15 4 11

1 11 11

0 11 11

1 11 11

4 11 15

1 15P

Direct Substitution

3 21 2 1 3 1 4 1 4P

2 1 3 1 4 1 4

2 3 4 4 5

Synthetic Substitution

1 5P

Direct Substitution

3 22 3 2 5 2 10 2 1P

3 8 5 4 20 1

24 20 20 1 17

Synthetic Substitution

Why Synthetic Substitution Works

Consider evaluating the following: 3 2 for P a P x bx cx dx e

Direct Substitution

3 2P a b a c a d a e

Synthetic Substitution

3 2a b a c ad e

Synthetic DivisionThe Factor Theorem

The Remainder Theorem

The Factor Theorem

The Factor Theorem is a formal statement of the Fundamental Connections that we made between zeros of a polynomial function and factors of the polynomial.

1 0f

1 or 1 is a factor of x x f x

1 0f

1 is a factor of x f x

2 3 0f

2 is not a factor of x f x

The Remainder Theorem

f aa a ra q

When a polynomial is divided by x – a, then the remainder must either be 0 or a constant.

Consider the division algorithm for dividing the polynomial function f (x) by x – a.

f x x a q x r

Evaluating f (a), we have f a r

The Remainder Theorem tells us that synthetic substitution can be used to not only evaluate a function, but also to perform polynomial division when a polynomial is divided by a divisor in the form x – a .

When we interpret the process of synthetic substitution as the result of polynomial division, we typically refer to the process as synthetic division.

The divisor in synthetic division MUST be of the form x – a. If a polynomial is divided by x + a, you must be thinking of the divisor as x – (– a).

Coefficients of the quotient, which is one degree less than the dividend. These are listed in descending order.

Remainder.

4 3 22 7 4 27 18 2x x x x x 3 2quotient 2 11 18 9 remainder 0x x x

Check the result with your calculator.

Coefficients of the quotient, which is one degree less than the dividend. These are listed in descending order.

Remainder.

4 3 22 7 4 27 18 1x x x x x

3 2quotient 2 5 9 18 remainder 0x x x

Check the result with your calculator.

In the previous Example set, we found two zeros of this polynomial function, x = 2 and x = ‒1.

Since we need to write a complete factorization of this polynomial function, we will examine a procedure that utilizes repeated steps of synthetic division to find this factorization.

Solution

From Example Set No.2, we have the following synthetic division:

Now perform synthetic division again, using ‒1 on the quotient from above.

4 3 2

3 2

2

2 7 4 27 18

2 2 11 18 9

2 1 2 9 9

f x x x x x

x x x x

x x x x

We now know that the polynomial will factor into the product of two linear factors and a quadratic factor.

Now find the zeros of the quadratic factor by a method of your choice.

22 9 9 0

2 3 3 0

3 / 2 or 3

x x

x x

x x

3Zeros of the polynomial function are 3, , 1 and 2.

2x

Factorization: 2 1 2 3 3f x x x x x

Verify the zeros with your calculator.

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