Welcome to Math 19500 Video Lessons Prof. Department of Mathematics The City College of New York Fall 2013 An important feature of the following Beamer slide presentations is that you, the reader, move step-by-step and at your own pace through these notes. To do so, use the arrow keys or the mouse to move from slide to slide, forwards or backwards. Also use the index dots at the top of this slide (or the index at the left, accessible from the Adobe Acrobat Toolbar) to access the different sections of this document. To prepare for the Chapter 1.3 Quiz (September 18th at the start of class), please Read all the following material carefully, especially the included Examples. Memorize and understand all included Definitions and Procedures. Work out the Exercises section, which explains how to check your answers. Do the Quiz Review and check your answers by referring back to the Examples.
Section 1.3 Algebraic Expressions To do arithmetic, we combine expressions built with numbers. To do algebra, we combine expressions built with variables. We need to set up a basic dictionary of algebra expressions. For example, we plan to study polynomials such as 3x 3 + 2x 2 + 7x + 9. Why would anyone be interested in this weird collection of symbols? The usual answer is that polynomials arise in physics, biology, and many other settings. For example, throw a rock upward from the ground with initial speed 96 feet per second. Then, after t seconds, the rock is 96t 16t 2 feet above the ground. One non-standard answer is that polynomials are closely related to the decimal (or any other base) number system. Specifically, the value of the expression 3x 3 + 2x 2 + 7x + 9 when we substitute 10 for x is 3 10 3 + 2 10 2 + 7 10 + 9. = 3(1000) + 2(100) + 7(10) + 9 = three thousand two hundred seventy nine = 3,279 in decimal notation.
The basic algebra expressions are monomials and polynomials, defined as follows. Definition A monomial is a product of numbers and variables. x 2 3 2 x 3 2 is a monomial in one variable. x 2 5 y 3 2x 2y 4 2 is a monomial in two variables. Definition A standard form monomial is a number times a product of powers of variables, none of which are repeated, and which appear in alphabetical order. Converting monomials to standard form: x 2 3 2 x 3 2 = 18x 5 x 2 5 y 3 2x 2y 4 2 = 5 2 2 2x 2 x y 3 y 4 = 40x 3 y 7 x 2y 3x 4y 5y 2 7x 2 = x 3x 7x 2 2y 4y 5y 2 = 21xxx 2 40yyy 2 = 840x 4 y 4
Reminder: The word subtract simply means add the negative of. 4 7 means 4 + ( 7) = 3. Avoid writing 4 + 7. 4x 3 7x 3 = 4x 3 + ( 7x 3 ) = (4 + 7)x 3 = 3x 3. However, it s easier to write 4x 3 7x 3 = (4 7)x 3 = 3x 3. Procedure To add (or subtract) monomials: Rewrite them in standard form. If their variable parts are identical, add (or subtract) their coefficients. If not, their sum does not simplify. Examples: 3x 3 + 4x 3 = (3 + 4)x 3 = 7x 3 3xyx + 4x 2 y = 3x 2 y + 4x 2 y = 7x 2 y 5x 3 7x 3 = (5 7)x 3 = 2x 3 4xyx + 5yxy = 4x 2 y + 5xy 2 does not simplify.
Definition A polynomial is a sum of monomials. The terms of a polynomial are the monomials in the sum. A polynomial is collected provided the monomials are in standard form and no two monomials have the same variable part. A polynomial in one variable is in (unique) standard form if the powers of the variable decrease from left to right. Here a few examples of polynomials: 0 : Every number is both a monomial and a polynomial. x 3 : Every monomial is a polynomial. 3 x x 3 y + 787x 3x 50 Procedure To convert a polynomial to standard form: Rewrite all monomials in standard form. Collect terms: add monomials with the same variable part. If the polynomial has just one variable, reorder the terms so that the powers decrease from left to right.
Example 1: Convert each polynomial to standard form: x 2 + 2x 3 + 6x 2 + 5 = 7x 2 + 2x 3 + 5 (Collected terms) = 2x 3 + 7x 2 + 5 (Standard form) 3x 2 + 5x 3 + 2x 2 + 7x 1 + 8x = 5x 3 + 5x 2 + 15x 1 3x 2 y 2 x+3x 2 y 3 +5xyxyx y 3 x 2 = 3x 3 y 2 +3x 2 y 3 +5x 3 y 2 x 2 y 3 = 8x 3 y 2 +2x 2 y 3 In this course, we will discuss standard forms for polynomials in several variables. Definition In a standard form polynomial in one variable: the leading term is the term with the highest power of the variable. the leading coefficient is the coefficient of the leading term. the degree of the polynomial is the exponent of the first term. Special cases: The degree of a number is zero since, for example, 4 can be written as 4x 0. The degree of the zero polynomial is undefined. To find the leading term, leading coefficient, and degree of any polynomial, first rewrite it in standard form.
Example 2: Let P be the polynomial 3x 2 4x 5 7x 3 + 2x 5 + 777 + 2x 5. The terms with variable part x 5 add to zero. Reorder the other terms to write P = 7x 3 + 3x 2 + 777 in standard form. Then P has leading term 7x 3, leading coefficient 7, and degree 3. Rough definition An algebraic expression combines polynomials by using parentheses, addition, subtraction, multiplication, and powers with positive integer exponents. To clarify the definition, we give examples of the combinations mentioned. In the following list, parentheses are crucial. Definition If P and Q are polynomials, then their sum is P + Q and their their difference is P (Q); their product is (P )(Q) and their quotient is (P )/(Q) = P Q ; the 3rd power of P is (P ) 3.
Procedure To simplify a sum of monomials, rewrite them in standard form and then add monomials with the same variable part by adding their coefficients. Example 3: 3x 3 + 4x 3 = 7x 3. 3x 2 + 4x 3 = 4x 3 + 3x 2 4xyx + 7x 2 y = 4x 2 y + 7x 2 y = 11x 2 y 4xyx + 7yxy + 12x 2 y = 4x 2 y + 7xy 2 + 12x 2 y = 16x 2 y + 7xy 2 Procedure To multiply monomials in standard form, multiply coefficients and multiply powers. Example 4: (3x 3 )(4x 4 )(5x 5 ) = 3 4 5 x 3 x 4 x 5 = 60x 3+4+5 = 60x 12. 3x 3 4y 4 5x 5 6y 5 = 3x 3 5x 5 4y 4 6y 5 = 15x 3+5 24y 5+4 = 360x 8 y 9
Procedure To add polynomials, remove parentheses and combine terms. To subtract polynomials, distribute the minus sign and then combine terms. To multiply a constant by a polynomial, multiply the constant by each term of the polynomial. Example 5: (x + y) + (2x 3y) = x + y + 2x 3y = x + 2x + y 3y = 3x 2y Example 6: The true meaning of subtraction is add the negative, as follows: (x+y) (2x 3y) = x+y+ 1(2x 3y) = x+y+ 2x+3y = x 2x+y+3y = x+4y But we always do this more quickly by distributing the minus sign (x + y) (2x 3y) = x + y 2x + 3y = x 2x + y + 3y = x + 4y In the same way, we speed up the following problem by distributing the -5 : Example 7: 3(x 3 + 3x + 1) 5(2x 3 7x 2 + 2) = 3x 3 + 9x + 3 10x 3 + 35x 2 10 = 7x 3 + 35x 2 + 9x 7
Procedure To multiply polynomials P and Q, multiply each term of P by each term of Q and add the results. Then the degree of P Q is the degree of P + degree of Q. Example 8: Rewrite the product (2x + 3)(x 2 5x + 4) as a standard form polynomial. Method 1: Start with the chart at the right: x 2 5x +4 the terms of the two factors are written in the 2x 2x 3 10x 2 +8x top row and the left column. +3 3x 2 15x +12 Each blue term is the product of the terms at the top of its column and the left of its row. Add and collect these six terms to get the answer: 2x 3 7x 2 7x + 12 Method 2: You could also find the product without a chart: Multiply 2x by (x 2 5x + 4) : 2x 3 10x 2 + 8x Multiply +3 by (x 2 5x + 4) : 3x 2 15x + 12 Add the results: = 2x 3 10x 2 + 8x + 3x 2 15x + 12 Rearrange: = 2x 3 10x 2 + 3x 2 + 8x 15x + 12 Collect terms: = 2x 3 7x 2 7x + 12
Rewriting an expression as a simplified sum An important part of working with polynomial expressions is becoming comfortable with rewriting those expressions as either a simplified sum or a simplified product. Procedure To rewrite an expression as a simplified sum, multiply out and collect terms. To rewrite an expression as a simplifed product, factor the expression completely. In practice, to rewrite an expression as a simplified sum, multiply polynomials to get rid of parentheses and then rewrite the polynomial in standard form. One important tool for doing that is to use special product formulas such as (A + B) 2 = A 2 + 2AB + B 2 and A 2 B 2 = (A + B)(A B) Example 9: Use (A + B) 2 = A 2 + 2AB + B 2 to find and simplify (3x 4y) 2. Solution: substitute 3x for A and 4y for B to get: (3x 4y) 2 = ((3x) + ( 4y) 2 = (3x) 2 + 2(3x)( 4y) + ( 4y) 2 = 9x 2 24xy + 16x 2
Example 10: Rewrite 4(a + b) 2 3(a + c) 2 as a simplified sum. Solution: We have already done this sort of thing in Chapter 0. The original problem: 4(a + b) 2 3(a + c) 2 Use (A + B) 2 = A 2 + 2AB + B 2 twice: = 4(a 2 + 2ab + b 2 ) 3(a 2 + 2ac + c 2 ) Use the distributive law: = 4a 2 + 8ab + 4b 2 3a 2 6ac 3c 2 Collect terms as best you can: = a 2 + 4b 2 3c 2 + 8ab 6ac Example 11: Use A 2 B 2 = (A + B)(A B) to find and simplify (x 2 + 1) 2 (x 2 1) 2. Solution: Substitute x 2 + 1 for A and x 2 1 for B in A 2 B 2 = (A + B)(A B) to get (x 2 + 1) 2 (x 2 1) 2 = ((x 2 + 1) + (x 2 1))((x 2 + 1) (x 2 1)) = (x 2 + 1 + x 2 1)(x 2 + 1 x 2 + 1) = (2x 2 )(2) = 4x 2
Basic factoring: Finding the GCF Here are some basic examples of rewriting an expression as a simplified product, which simply means: keep on factoring until you can t go any further! Example 12: Factoring numbers and simple sums 20 = 4 5 = 2 2 5 1000 = 10 3 = (2 5) 3 = 2 3 5 3 4x + 6y = 2 (2x + 3y) 60x + 108y = 2(30x + 54y) = 2(2)(15x + 27y) = 2(2)(3)(5x + 9y) = 12(5x + 9y) In the last two examples, we followed a simple strategy: to factor a sum of terms, pull out common factors until you are forced to stop. But this isn t very systematic. What if you had to factor something more difficult such as 48x 5 y 2 + 20x 2 y 3 z + 50x 7 z 3 For this sort of problem, we need to discuss factoring in a systematic way. The basic idea is: first pull out the biggest possible common factor of the coefficients 48, 20, and 50. In other words, find the greatest common factor (GCF) of 48, 20, and 50. Then work on factoring out the GCF of the variable parts x 2 y 3 z 2, x 2 y 3 z, and x 2 y 3.
Procedure To find the Greatest Common Factor (GCF) of two or more whole numbers Factor each number completely into powers of primes. For each prime that appears in all the factorizations, choose the lowest power that appears. The GCF is the product of those lowest powers. If no prime appears in all factorizations, the GCF is 1. Example 13: Find the GCF of 200, 216, and 180. Step 1: 200 = 2 3 5 2 216 = 2 3 3 3 180 = 2 2 3 2 5 1 Step 2: The prime 2 appears in all factorizations. The lowest power is 2 2 (in 200). The prime 3 does not appear in all factorizations: leave it out of the GCF The prime 5 does not appear in all factorizations: leave it out of the GCF Answer: The GCF is the product 2 2 = 4 Check: 4 is a common factor since 200 = 4 50, 216 = 4 54, and 180 = 4 45. Further check: The remaining factors 50 = 2 5 2, 54 = 2 3 3, and 45 = 3 2 5 have no common factor.
Example 14: Find the GCF of 81 and 64. Here 81 = 3 4 and 64 = 2 6 There isn t any prime that appears in all factorizations and so the GCF is 1. Example 15: Find the GCF of 360, 450, and 1500. Step 1: 360 = 2 3 3 2 5 1 450 = 2 3 3 5 2 1500 = 2 2 3 5 3 Step 2: The prime 2 appears in all factorizations. The lowest power is 2 1 = 2 (in 450). The prime 3 appears in all factorizations: The lowest power is 3 1 = 3 (in 1500). The prime 5 does not appear in all factorizations: The lowest power is 5 1 = 5 (in 360). Answer: The GCF is the product 2 3 5 = 30. Check: 360 is a common factor since 360 = 30 12, 450 = 30 15, and 1500 = 30 50. Further check: The remaining factors 12 = 2 2 3, 15 = 3 5, and 502 5 2 have no common factor.
The same idea applies to monomials: Example 16: Find the GCF of the 3 monomials x 3 y 4, x 2 y 6 z 2, and xy 8 z 4 Solution: The primes that appear are x, y, z x appears in all monomials, with lowest power x 1 = x y appears in all monomials, with lowest power y 4 z doesn t appear in all monomials, leave it out of the GCF Answer: The GCF is the product of lowest powers: xy 4. Sometimes we will phrase the solution method slightly differently: Solution (version 2): The 3 monomials can be rewritten x 3 y 4 z 0, x 2 y 6 z 2, and x 1 y 8 z 4. The GCF is the product of lowest powers: x 1 y 4 z 0 = xy 4.
Factoring polynomials Procedure To factor a polynomial P : Rewrite P in standard form as a sum of monomials; Find the GCF of the coefficients and factor it out of the sum; Then find the GCF of the variable parts and factor it out. Example 17: Factor 10x 5 y 2 + 20x 2 y 3 z + 30x 7 y 5 z 3 The GCF of the coefficients 10, 20, 30 is 10. Factor it out to get 10(x 5 y 2 + 2x 2 y 3 z + 3x 7 y 5 z 3 ) The GCF of the variable parts x 5 y 2, x 2 y 3 z, and x 7 y 5 z 3 is x 2 y 2. Factor it out: 10x 2 y 2 (x 3 + 2yz + 3x 5 y 3 z 3 ) Additional Factoring Examples: 2x 5 + 3x 2 = x 2 (2x 3 + 3) We factored out the lowest power of x 2x 5 + 3x 2 + 7 can t be factored further, since x is missing from the last term. x 3 y 5 + x 2 y 8 = x 2 y 5 (x + y 3 ) We factored out lowest powers of x and y.
Example 18: Factor 60x 2 y 5 + 108x 3 y 4 + 120x 8 y completely. We need to find the GCF of the three monomials. To do this, first factor and find the GCF of their coefficients: 60 = 2 2 3 5 108 = 2 2 3 3 5 0 120 = 2 3 3 5 The GCF of the coefficients is the product of lowest powers: 2 2 3 1 = 12. Now find the GCF of the monomials variable parts x 2 y 5, x 3 y 4, and x 8 y: The GCF of the variable parts is the product of lowest powers x 2 y. Factor out each of these GCF s in turn: Solution: 60x 2 y 5 + 108x 3 y 4 + 120x 8 y = 12(5x 2 y 5 + 9x 3 y 4 + 10x 8 y = 12x 2 y(5y 4 + 9xy 3 + 10x 6 )
Factoring polynomials You have probably seen examples of factoring trinomials (another name for degree 2 polynomials.) Factoring trinomials with leading coefficient 1 uses the identity x 2 + (r + s)x + rs = (x + r)(x + s). Procedure To factor x 2 + bx + c, try to find r and s with rs = c and r + s = b If you can do that, then x 2 + bx + c = (x + r)(x + s). Example 19: To factor x 2 + 7x + 12, find r and s with rs = 12 and r + s = 7. Fool around a bit to get r = 3 and s = 4. Therefore x 2 + 7x + 12 = (x + 3)(x + 4). Example 20: Factor x 2 5x 50. You want to list all pairs r and s with rs = 50, then choose the pair with r + s = 5. The factors of 50 are 1, 2, 5, 10... If r = 1 and s = 50, then r + s = 49 If r = 2 and s = 25, then r + s = 25 If r = 5 and s = 10, then r + s = 5. Good! x 2 5x 50 = (x + 5)(x 10).
Example 21: Factor 4x 3 20x 2 200x. Solution: 4x 3 20x 2 200x = 4(x 3 5x 2 50x) = 4x(x 2 5x 50) = 4x(x 10)(x + 5) Factoring quadratic polynomials such as 6x 2 + 13x + 6 with leading coefficients other than 1 is done in high school with a messy trial and error method. We will explain a better method in Section 1.5.
Procedure To factor any polynomial : Find and factor out the GCF of the coefficients of the monomials. Find and factor out the GCF of the variable parts of the monomials. Factor the remaining polynomial. Example 22: Factor 60x 2 y 3 z 2 + 120x 2 y 3 z + 60x 2 y 3 Solution: The expression to be factored is a sum of 3 terms. Preliminary work: Find the two GCF s The GCF of 60 = 2 2 3 5, 120 = 2 3 3 5 and 60 = 2 2 3 5 is 2 2 3 5 = 60 The GCF of the variable parts of the 3 terms is clearly x 2 y 3 (product of lowest powers that appear in all terms). Now put everything together: Original problem: Factor 60x 2 y 3 z 2 + 120x 2 y 3 z + 60x 2 y 3 1. GCF of coefficients is 60; factor it out: = 60(x 2 y 3 z 2 + 2x 2 y 3 z + x 2 y 3 ) 2. GCF of variable parts is x 2 y 3 ; factor it out: = 60x 2 y 3 (z 2 + 2z + 1) 3. Factor the remaining polynomial z 2 + 2z + 1 = 60x 2 y 3 (z + 1) 2
Nothing changes when you factor expressions with fractional exponents: just be careful to figure out correctly which (fractional) exponent is the lowest! Example 23: Factor 12x 1/2 36x 1/2 + 24x 3/2 Solution: Here is the original problem 12x 1/2 36x 1/2 + 24x 3/2 Factor out GCF of 12, 36, 24 : = 12(x 1/2 3x 1/2 + 2x 3/2 ) Factor out lowest power of x: = 12x 3 2 (x 1/2 ( 3/2) 3x 1/2 ( 3/2) + 2x 3/2 ( 3/2) ) = 12x 3/2 (x 2 3x 1 + 2x 0) Factor the remaining part: = 12x 3/2 (x 2 3x + 2) Here is the answer: = 12x 3/2 (x 1)(x 2) Example 15 in SRW 1.3 is an artificial method of factoring, and so we omit it.
Example 24: Factor 40(x 1) 3 (2x + 1) 4 24(x 1) 4 (2x + 1) 3 (*) Step 1. The GCF of the coefficients 40 = 2 3 5 and 24 = 2 3 3 is the product of lowest prime powers 2 3 3 0 5 0 = 8. Step 2: The GCF of the variable parts (x 1) 3 (2x + 1) 4 and (x 1) 4 (2x + 1) 3 is the product of lowest powers (x 1) 3 (2x + 1) 3 Step 3: First factor out 8 from (*); then factor out (x 1) 3 (2x + 1) 3 Solution: Expression to factor is 40(x 1) 3 (2x + 1) 4 24(x 1) 4 (2x + 1) 3 Factor out GCF of coefficients: = 8(5(x 1) 3 (2x + 1) 4 3(x 1) 4 (2x + 1) 3 ) Factor out GCF of variable parts: = 8(x 1) 3 (2x + 1) 3 (5(2x + 1)) (3(x 1)) Expand the cofactor = 8(x 1) 3 (2x + 1) 3 (10x + 5 3x + 3) Simplify the cofactor = 8(x 1) 3 (2x + 1) 3 (7x + 8)
Exercises for Chaper 1.3 Click on Wolfram Calculator to find an answer checker. Click on Wolfram Algebra Examples to see how to check various types of algebra problems. 1. Do the WebAssign exercises for Chapter 1.3. Rewrite each product as a simplified sum: 2. ( x + 7)( x 7) 3. 2(2x + 1)(x 1) 2 4. (x + 2)(x 2 + x + 1) 5. Factor each polynomial completely: a) x 2 17x + 60 b) x 2 10x 75 c) 4x 4 24x 3 + 32x 2 d) y 4 (y + 2) 4 y 3 (y + 2) 5 e) 5(x 2 + 4) 4 (2x)(x 2) 4 + (x 2 + 4) 5 (4)(x 2) 3 f) (x 2 + 3) 1/3 2 3 x2 (x 2 + 3) 4/3 g) 100x 2 yz + 125xyz 2 + 200xy 2 z h) 40x 3 (x 1) 7 (2x + 1) 4 24(x 1) 4 x 3 (2x + 1) 8.
Quiz Review Example 1: Convert each polynomial to standard form: x 2 + 2x 3 + 6x 2 + 5 3x 2 + 5x 3 + 2x 2 + 7x 1 + 8x 3x 2 y 2 x + 3x 2 y 3 + 5xyxyx y 3 x 2 x 3 y 2 + 2x 2 y 3 Example 2: Rewrite 3x 2 4x 5 7x 3 + 2x 5 + 777 + 2x 5 in standard form. Example 3: Rewrite each polynomial in standard form 3x 3 + 4x 3 3x 2 + 4x 3 4xyx + 7x 2 y. 4xyx + 7yxy + 12x 2 y Example 5: Rewrite as a simplified sum: (x + y) + (2x 3y). Example 6: Rewrite as a simplified sum: (x + y) (2x 3y). Example 7: Rewrite as a simplified sum: 3(x 3 + 3x + 1) 5(2x 3 7x 2 + 2). Example 8: Rewrite the product (2x + 3)(x 2 5x + 4) as a standard form polynomial. Example 9: Use (A + B) 2 = A 2 + 2AB + B 2 to find and simplify (3x 4y) 2. Example 10: Rewrite 4(a + b) 2 3(a + c) 2 as a simplified sum. Example 11: Use A 2 B 2 = (A + B)(A B) to simplify (x 2 + 1) 2 (x 2 1) 2. Example 12: Factor each of the following completely: 20 1000 4x + 6y 60x + 108y.
Example 13: Find the GCF of 200, 216, and 180. Example 14: Find the GCF of 81 and 64. Example 15: Find the GCF of 360, 450, and 1500. Example 16: Find the GCF of the 3 monomials x 3 y 4, x 2 y 6 z 2, and xy 8 z 4. Example 17: Factor 10x 5 y 2 + 20x 2 y 3 z + 30x 7 y 5 z 3. Example 18: Factor 60x 2 y 5 + 108x 3 y 4 + 120x 8 y completely. Example 19: Factor x 2 + 7x + 12. Example 20: Factor x 2 5x 50. Example 21: Factor 4x 3 20x 2 200x. Example 22: Factor 60x 2 y 3 z 2 + 120x 2 y 3 z + 60x 2 y 3. Example 23: Factor 12x 1/2 36x 1/2 + 24x 3/2. Example 24: Factor 40(x 1) 3 (2x + 1) 4 24(x 1) 4 (2x + 1) 3.