MATH 21
Lesson Four

CHAPTER #4 - POLYNOMIALS (pp. 160 to 217)

LESSON #4.1 - The General Objective of this lesson is to define polynomials and discuss the addition and subtraction of polynomials.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to; Identify polynomials, and their degree. Add and subtract polynomial expressions.

WORD PROBLEMS:

1. Two airplanes left airports that are 600 miles apart and flew toward each other. One plane flew 20 mph faster than the other one. If they both left at 1:00 PM and passed each other at 2:12 PM, what were their speeds?
   
2. If a freight train traveling at 30 mph is 300 miles ahead of a passenger train traveling at 55 mph, how long will it take the passenger train to catch up with the freight train?

4.1 - Polynomials: Sums and Differences - In Section 1.4 we learned that an algebraic expression is a combination of literal numbers and/or numerals joined together by a finite number of arithmetic operations. This definition means that just about anything we see in this course is an algebraic expression. A polynomial is an algebraic expression using only the operations of addition, subtraction and multiplication. Some examples of polynomials are;

x + y

x² - 2x + 1

xy -2z + yz

Definitions -

1.  A polynomial in one variable can be written as -

a_nxⁿ  +   a_n-1x^n-1 +  a_n-2x^n-2  + … +  a₂x²  +  a₁x  + a₀

where the a's are in R and n is in N.

2.  The coefficients are the a's which are real numbers multiplying the variables.

3.  A term is a quantity completely set off from other parts of the polynomial by either a plus or minus sign.

4.  A factor is a multiplying part of a term or of a polynomial. For instance -

3x  has the factors  3  and  x

x² - 4  has the factors  (x+2)  and  (x-2)

5.  A monomial is a polynomial containing only one term.

6.  A binomial is a polynomial containing exactly two terms.

7.  A trinomial is a polynomial containing exactly three terms.

(All polynomials containing four or more terms are simply called polynomials.)

8.  The degree of a polynomial is the largest sum of exponents of the literal factors in any one of the terms of the polynomial.

Combining Similar Terms - Similar or like terms are terms that have the same literal factors. Since polynomials are real numbers they are added and subtracted using the same rules as for real numbers. If we let P , Q, and R represent three polynomials, the rules for addition are;

P + Q = Q + P

P + (Q + R) = (P + Q) + R

P(Q + R) = PQ + PR

LESSON #4.2 - The General Objective of this lesson is to introduce the properties of exponents, and show how they are used in the multiplication and division of monomials.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to multiply and divide monomials using the properties of exponents.

4.2 - Products and Quotients of Monomials - As discussed in Section 3.1, a monomial is a polynomial containing only one term. The multiplication and division of monomials is facilitated by using the properties of exponents.

Properties of Natural Number Exponents - By definition, the product of n x's can be written as xⁿ, where x is called the base, and n is the power or exponent the number x is raised to. As a consequence of this definition we have the following rules.

If x and y are real numbers, and m and n are natural numbers. then;

Board and Class Problem - Page 172/66, 90.

LESSON #4.3- The General Objective of this lesson is to learn how to multiply polynomials.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to:

• Multiply one polynomial by another.
• Identify four special products.
• Use Pascal's triangle to find powers of binomials.

4.3 - Multiplying Polynomials - Since polynomials are real numbers they are multiplied using the same rules as for real numbers. If we let P , Q, and R represent three polynomials, the rules for multiplication are:

PQ = QP

P(QR) = (PQ)R

P(Q + R) = PQ + PR

Board and Class Problem - Find the product of (2x2 +2xy + y2)(2x - 3y).

Special Products - There are some multiplication products that arise so often in algebra that they are worth memorizing. They are -

(x + y)² = x² + 2xy + y²

(x - y)² = x² - 2xy + y²

(x + y)(x - y) = x² - y²

(x + y)³ = x³ + 3x²y + 3xy² + y³

Pascal's Triangle - It becomes increasingly tedious to rise binomials to higher and higher powers. There is a technique which simplifies this task. Consider products of the form (x + y)ⁿ.

An examination of these products reveals the following features -

1. The degree of each term is the same.
2. Looking at the terms from left to right, we see that as the degree of the x factor drops, the degree of the y factor increases the same amount.
3. The first and last coefficients both equal one.
4. The increasing products can be displayed in the form of a pyramid, similar to that of a group of bowling pins.
5. The number of terms is one more than the degree of the polynomial.

Assuming that this pattern holds, we can write -

(x + y)⁴ = x⁴ + C₁x³y + C₂x²y² + C₃xy³ + y⁴

where our problem is reduced to finding the values of the coefficients. We could do this by carrying out the indicated multiplication, but there is a much easier way using an array of numbers named after the French mathematician, Blaise Pascal (1623 - 1662 A.D.)

Pascal's Triangle

Explain how the numbers on one line are obtained from those on the preceding line. Using the numbers on the fifth line we can now write that -

(x + y)⁴ = x⁴ + 4x³y + 6x²y² + 4xy³ + y⁴

Board and Class Problem - Expand (2x -y)⁶ .

LESSON #4.4- The General Objective of this lesson is to introduce the concept of factoring.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to -

• Explain what is meant by prime factorization.
• Factor using common factors.
• Factor by grouping.
• Solve equations using factoring.

4.4 - Factoring: Use of the Distributive Property - Multiplying polynomials is a fairly straightforward procedure, once you know the rules. Reversing the process, or factoring however, is not as direct. It is akin to the division algorithm used for numbers, where you guess and check. As in division, experience in multiplication is necessary to improve the quality of your guesses.

Prime Factors - A polynomial is in completely factored form when it is written as a product of prime factors. A polynomial with integral coefficients is in completely factored form if:

• It is expressed as a product of polynomials with integral coefficients.
• No polynomial, other than a monomial, within the factored form can be further factored into polynomials with integral coefficients.

Common Factors - The simplest type of factoring occurs when there are common factors. The distributive property,

P(Q + R) = PQ + PR

provides a bridge between factors and products. Rewriting it in reverse order yields -

PQ + PR = P(Q + R)

where P is the common factor.

Board and Class Problem - Page 187/ 40.

Factoring by Grouping - This is a variation of common factors, where the factors are polynomials.

Board and Class Problem - Page 187/ 64.

Equations and Problem Solving - One reason that factoring is important is in its application to solving certain types of equations, when combined with the following property of real numbers.

Property of a Zero Product - Let a and b be two real numbers. If

ab = 0

then a = 0 or b = 0.

Board and Class Problem - Page 187/ 76.

LESSON #4.5 - The General Objective of this lesson is to extend our list of special products.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to;

• Write down the rules for the difference of two squares, and for the sum and difference of two cubes.
• Factor polynomials using these rules.
• Apply these rules to the solution of equations.

4.5 -Factoring: The Difference of Two Squares and Sum or Difference of Two Cubes - In Section 4.3 we examined some special multiplication patterns. They were -

(x + y)² = x² + 2xy + y²

(x - y)² = x² - 2xy + y²

(x + y)³ = x3 + 3x²y + 3xy² + y²

Board and Class Problems - Carry out the following operations.

(x + y) (x - y) =

(x + y)(x² - xy + y²) =

(x - y)(x² + xy + y²) =

With these results we now have a list of  seven special products. They are -

Of these seven special products, the first three should be committed to memory, and the pattern of the other four should be studied.

Board and Class Problems - Page 195/76.

LESSON #4.6 - The General Objective of this lesson is to study trinomial factoring.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to factor trinomials.

4.6 - Factoring Trinomials - Expressing a trinomial as the product of two binomials is one of the most common types of factoring in algebra. The general form for such a trinomial is -

ax² + bx + c

Trinomials Whose Leading Coefficient Is One - Let us first consider trinomials where the leading coefficient is equal to one; i.e.

x² + bx + c

This expression can always be written as a product of two binomials -

x² + bx + c = (x+d)(x+e)

x² + bx + c = x² +ex + dx +de

x² + bx + c = x² +x(e+d)+de

Thus we see that the coefficients b and c must equal (d+e) and de respectively. The numbers d and e exist in the set of real numbers, however, when factoring we usually only consider the set of integers. This means that we can only factor a small set of trinomials of this form. Consider the following example.

Factor -

4x + x² - 12

Step 1 - List the trinomial in descending powers.

x² + 4x - 12

Step 2 - List all pairs of factors of the last term of the trinomial.

-12 = 1(-12) or (-1)12

-12 = 2(-6) or (-2)6

-12 = 3(-4) or (-3)4

Step 3 - Choose the product pair whose sum equals the middle coefficient.

6 + (-2) = 4

Step 4 - Write the trinomial in factored form.

x² + 4x - 12 = (x + 6)(x - 2)

Trinomials of the form ax² + bx + c - When the leading coefficient is not equal to one, our task is made more complicated, since then the trinomial can be expressed as -

ax² + bx + c = (dx + e)(fx + g)

We must now determine four numbers d, e, f and g, whereas when a equals one we only had to find two numbers. We will now investigate two different ways of factoring trinomials of this form.

Factor -

23x + 6x² + 20

Standard Technique -

Step 1 - List the trinomial in descending powers.

6x² + 23x + 20

Step 2 - Write the trinomial in factored form with blanks for possible coefficients.

6x² + 23x + 20 = ( __ x + __ )( __ x + __ )

Step 3 - Choose values for the blanks that yield the first and last coefficients.

Table of Possible Choices for Coefficients

6 = 1*6	20 = 1*20,    20 = 2*10,     20 = 4*5
6 = 6*1	20 = 1*20,    20 = 2*10,     20 = 4*5
6 = 2*3	20 = 1*20,    20 = 2*10,     20 = 4*5
6 = 3*2	20 = 1*20,    20 = 2*10,     20 = 4*5

Step 4 - Check each of these possible combinations in the blanks in Step 2, until the right combination yields the coefficient of the middle term. If no combination yields the middle coefficient, the trinomial is not factorable. For this problem -

6x² + 23x + 20 = (3x + 4)(2x + 5)

As you can see, this is a trial and error approach. If the first and last coefficients are large, or have many product combinations it can be a tedious technique. Another approach, which is more direct, is illustrated next.

Master Product Technique -

Step 1 - List the trinomial in descending powers.

6x² + 23x + 20

Step 2 - Find the Master Product (MP) which is defined to be the product of the first and last coefficients.

MP = 6 * 20 = 120

Step 3 - Write down all possible pairs of products of the

MP. = 1(120) = 2(60) = 3(40) = 5(24) = 6(20) = 8(15) = 10(12)

Step 4 - Choose the pair of factors whose sum is the coefficient of the middle term.

23 = 8 + 15

Note: If you can't find such a pair, the trinomial is not factorable.

Step 5 - Rewrite the given trinomial, replacing the middle term by the sum of two terms whose coefficients are the pair of factors found in Step 4.

6x² + 23x + 20 = 6x² + 8x + 15x + 20

Step 6 - Factor this expression by grouping.

6x² + 23x + 20 = 2x(3x + 4) + 5(3x + 4)

6x² + 23x + 20 = (3x + 4)(2x + 5)

Step 7 - Check your factoring by multiplying the binomial factors to ensure that their product is the given trinomial.

Board and Class Problems - Page 203/ 70.

Summary of Factoring Techniques - Quite often more than one method of factoring has to be employed to completely factor a polynomial. When using more than one technique it is usually best to proceed as follows:

1. Factor out any common factors that exist.
2. Use special products if appropriate.
3. Use trinomial factoring if appropriate.
4. Use grouping.

Occasionally, you might also try some unconventional techniques if you are convinced that the polynomial is factorable. Consider the following problem.

Board and Class Problem - Factor  -  x⁴ + x² + 1.

x⁴ + x² + 1 = x⁴ + 2x² + 1 - x²

x⁴ + x² + 1 = (x² +1)² - x²

x⁴ + x² + 1 = [(x² +1) + x][ (x² +1) - x]

x⁴ + x² + 1 = (x² + x +1) (x² - x +1)

LESSON #4.7 - The General Objective of this lesson is to apply factoring to problem solving.

PERFORMANCE OBJECTIVES - At the end of this lesson the student will be able to use factoring techniques to solve some second degree equations.

4.7 - Equations and Problem Solving - Factoring allows us to solve some second degree equations. Let us consider the following problem.

Board and Class Problem - Page 209/ 59.

HOMEWORK - Lesson Four Assignment.

COMMENTS

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