Collecting Like Terms

Transcription

1 MPM1D Unit 2: Algebra Lesson 5 Learning goal: how to simplify algebraic expressions by collecting like terms. Date: Collecting Like Terms WARM-UP Example 1: Simplify each expression using exponent laws. a) (3!!!)( 7!! ) b) (4!!!!! )(6!!! ) c)!!! (!!! )!!!!! Recall, like terms are two or more terms that have the same variable raised to the same exponent. Algebraic expressions that contain like terms can be simplified by combining each group of like terms into a single term. Examples: 3x + 4x 9x 2 6x 2 12x 3 y 2-5x 3 y 2 Why can t you simplify? 4x 2 + 4x x 2 7 6x 3 y + 5xy 3 Example 2: Simplify the following algebraic expressions. a) 7! + 5 3! b) 6!! + 11! + 8!! 15! c) 6! ! d) 12! 5 7! 11 e) 2!! 3! + 7 3!! + 4! 7 f) 11!!! 12!!!

2 1. Are the terms in each pair like or unlike? Assignment 2.5: Collecting Like Terms a) 5a and 2a b) 3x 2 and x 3 c) 2p 3 and p 3 2 d) 4ab and ab e) 3b 4 and 4b 3 f) 6a 2 b and 3a 2 b g) 9pq 3 and p 3 q h) 2x 2 y and 3x 2 y 2 2. Simplify. a) 4 + v + 5v 10 b) 7a 2b a 3b c) 8k k 5k k d) 2x 2 4x + 8x 2 + 5x e) 12 4m 2 8 m 2 + 2m 2 f) 6y + 4y y 6 y 3. Simplify. a) 2a + 6b 2 + b 4 + a b) 4x + 3xy + y + 5x 2xy 3y c) m 4 m m 2 + m 4 d) x 2 + 3xy + 2y 2 x 2 + 2xy y Find the perimeter of each figure below. a) b) n 2.5 Answers 1. a) like b) unlike c) like d) like e) unlike f) like g) unlike h) unlike 2. a) 6v 6 b) 6a 5b c) 7k + 5 d) 10x 2 + x e) 4 3m 2 f) 5y a) 3a + 7b 6 b) 9x + xy 2y c) 2m 4 3m d) 5xy + y 2 4. a) 8n + 4 b) 8y + 2

3 MPM1D Unit 2: Algebra Lesson 6 Learning goal: how to add and subtract polynomials. Date: Adding and Subtracting Polynomials WARM-UP Example 1: Simplify each algebraic expression. a) 6! ! b) 3! + 7 3!! + 4! 7 c) 11!!! 12!!! Example 2: Determine a simplified expression the perimeter of the following rectangle. ADDING POLYNOMIALS In order to distinguish one polynomial from another polynomial in an algebraic expression, the polynomials are often placed in separate pairs of brackets. ie. (3!! + 5! 1) + (4!! 2!) Example 3: Add the following polynomials. a) (4! 7) + ( 3! + 2) b)!! + 2! + 6!! + 10! + (5! + 1)

4 SUBTRACTING POLYNOMIALS Suppose you were asked to evaluate the following expression: 4 ( 2). Explain the procedure for subtracting integers. This same idea can be used to subtract polynomials. Example 4: Subtract the following polynomials. a) 3! 2 (4! + 9) b) 5!! 12! ( 4! 8) c) 12 4!! + 5! 1 d) 8!! + 7! 4!! 3 ( 5! + 9) Example 5: The measures of two sides of a triangle are given. The perimeter is P = 4x 2 + 5x + 5. Find the measure of the third side. x 2 + 3x 5 2x 2 + 3x + 6

5 Assignment 2.6: Adding and Subtracting Polynomials 1. Simplify the following expressions. a) (y 2 + 6y 5) + ( 7y 2 + 2y 2) b) ( 2n + 2n 2 + 2) + ( 1 7n 2 + n) c) (3m 2 + m) + ( 10m 2 m 2) d) ( 3d 2 + 2) + ( 2 7d 2 + d) e) (4 8w) (7w + 1) f) (mn 5m 7) ( 6n + 2m + 1) g) (xy x 5y + 4y 2 ) (6y 2 + 9y xy) h) (2a + 3b 3a 2 + b 2 ) ( a 2 + 8b 2 + 3a b) 2. For each shape below, write the perimeter as a sum of polynomials and in simplest form. i) ii) iii) iv) 3. A student subtracted (3y 2 + 5y + 2) (4y 2 + 3y + 2) like this: = 3y 2 5y 2 4y 2 3y 2 = 3y 2 4y 2 5y 3y 2 2 = y 2 8y 4 a) Explain why the student s solution is incorrect. b) What is the correct answer? Show your work. 4. The difference between two polynomials is (5x + 3). One of the two polynomials is (4x + 1 3x 2 ). What is the other polynomial? Explain how you found your answer. 5. The sum of the perimeters of two shapes is represented by 13x + 4y. The perimeter of one shape is represented by 4x 2y. Determine an expression for the perimeter of the other shape. Show your work. 6. A rectangular field has a perimeter of 10a 6 meters. The width is 2a meters. Determine an expression for the length of this field. 2.6 Answers 1. a) 6y 2 + 8y 7 b) n 5n c) 7m 2 2 d) 10d 2 + d e) 3 15w f) mn 7m 8 + 6n g) 2xy x 14y 2y 2 h) a + 4b 2a 2 7b 2 2. i) (2n + 2) + (n + 1) + (2n + 2) + (n + 1) = 6n + 6 ii) (3p + 4) + (3p + 4) + (3p + 4) = 9p + 12 iii) (4y + 1) + (4y + 1) + (4y + 1) + (4y + 1) = 16y + 4 iv) (a + 8) + (a + 3) + (12) = 2a a) The student is incorrect because he changed the signs in the first polynomial. b) (3y 2 + 5y + 2) (4y 2 + 3y + 2) = 3y 2 + 5y + 2 4y 2 3y 2 = 3y 2 4y 2 + 5y 3y = y 2 + 2y 4. (4x + 1 3x 2 ) (5x + 3) = 3x 2 x 2, or (5x + 3) + (4x + 1 3x 2 ) = 3x 2 + 9x x+6y 6. 3a-3

6 MPM1D Unit 2: Algebra Lesson 7 Learning goal: how to multiply a polynomial by a monomial to simplify expressions. Date: Multiplying a Polynomial by a Monomial WARM-UP Example 1: Simplify the following algebraic expressions. a) 6! 4 + (2! + 4) b) 2!! +! + 12 (3!! + 4! 6) c) ( 4!!! )(3!!!! ) d) (4!! )(2!! ) e)! 6 2 5! + (! + 4) f) (!!!!! )(!!!! ) (!!!)! THE DISTRIBUTIVE PROPERTY A rectangle has an unknown length and a width of 4 units. If the length is increased by 7 units to create a larger rectangle, write a simplified algebraic expression for the area of the new, larger rectangle. x 7 What is the width of the new rectangle? 4 original x 4 7 What is the length of the new rectangle? What is the area of the new rectangle? This property is known as the distributive property. This is also known as expanding. Distributive Property:! (! +!) =!" +!"

7 Example 2: Expand the following. a) 3(! + 4) b) 7(! + 3) c) (2! 1) d) 4(! 5) Now lets try expanding with a variable Simplify:!(2! + 5) Example 3: Expand the following. a)!(! + 1) b) 3!(! + 4) c)!( 5! + 2) d) 3!(2! 1) Now lets try expanding with a coefficient and a variable Simplify: 3!(9!! 4!) Example 4: Expand the following. a) 3!( 4!! + 2!! ) b) 5!! (3! 1) c) 2!! (!! 3! + 9) d) (! 1)(11!) Now try combining everything you learned about simplifying expressions Simplify: 2!! 4! !(! + 4)

8 Example 5: Expand the following. a) 3! 2 + 6(! + 1) b)!!"! 4! !(!!! +! + 4) c) 3[ 2 6! + 5!] d) 5!! + 5 2(3!! 4! 7)

9 Assignment 2.7: Multiplying a Polynomial by a Monomial 1. Determine each product. a) 4(3a + 2) b) (d 2 + 2d)( 3) c) 2(4c 2 2c + 3) d) 4(b 2 2b 3) e) 5c(c 2 6c 1) f) 3h(4 h 2 ) 2. Here is a student s solution for a multiplication question. ( 5k 2 k 3)( 2) = 2(5k 2 ) 2(k) 2(3) = 10k 2 2k 6 a) Explain why the student s solution is incorrect. b) What is the correct answer? Show your work. 3. Write a simplified expression for the area of the following rectangles a) b) 4. Expand a) 4x 2 (3x + 2) b) 2n(2n 3) c) pq(3p + 2q) d) 3d(2d 2 4d + 1) e) 2(x + 4) 4(2x + 3) f) 3a(2a + 4b 3) 2b(3a + 2ab) g) 2p(p 4) + 6(p 2 + 4p 3) 2.7 Answers 1. a)12a + 8 b) 3d 2 6d c) 8c 2 4c + 6 d) 4b 2 + 8b + 12 e) 5c 3 30c 2 5c f) 12h + 3h 3 2. a) The negative signs were omitted on the first polynomial when ( 2) was distributed; ( 5k 2 )( 2) + ( k)( 2) + ( 3)( 2) = 10k 2 + 2k a) 2d(3d + 4) = 6d 2 + 8d b) y(4y + 6) = 4y 2 + 6y 4. a) 12x 3 + 8x 2 b) 4n 2 6n c) 3p 2 q + 2pq 2 d) 6d 3-12d 2 +3d e) -6x 4 f) 6a 2 + 6ab 9a 4ab 2 g) 8p p 18

10 MPM1D Unit 2: Algebra Lesson 8 Learning goal: how to apply knowledge of simplifying expressions to geometric problems. Date: Simplifying Algebraic Expressions WARM-UP Example 1: Simplify and expand the following. a)!(! 5) b) 2!(3! + 1) c) ( 3!! )(3! + 1 2!! ) d) 2 3! (! 4) e) 4! 3!! 2! + 1 3!(2!! 5) f) 3! 4! 5! 2!(2! + 3!) g) 2! 3![5 2! 1 ] h)!!!! 3!!!! +!!!! i) 5!!! + 6 2[! !! ] Example 2: Find the perimeter.

11 Example 4: Find the missing side length given the perimeter below. Example 5: Find the area of the shaded region.

12 Assignment 2.8: Simplifying Algebraic Expressions 1. Simplify the following: a) 3p 4q + 2p q 21 g) 2x 2 + 3x 7x (-5x 2 ) b) -3b(5a 3b) + 4(-3ab 5b 2 ) h) 3x(x - 2y) 4(-3x 2-2xy) c) -3(x 2 + 3y) + 5(-6y x 2 ) i) -3(7xy - 11y 2 ) 2y(-2x + 3y) d) x y x + y j) s t s t e) 36 [ 2( x y) ] + k) 2x(x 2y) [3 2x(x y)] 37 [ ] f) { x 2 x (2x 1) } l) 2 3!!! +! 3[! + 4! 3! 2! ] 2. A triangle has sides of length 2a centimeters, 7b centimeters, and 5a + 3 centimeters. What is the perimeter of the triangle? 3. A square has a side of length 9x 2 inches. Each side is shortened by 3 inches. What is the perimeter of the new smaller square? 4. A triangle has sides of length 4a 5 feet, 3a + 8 feet, and 9a + 2 feet. Each side is doubled in length. What is the perimeter of the new enlarged triangle? 5. Find the area of the shaded region. a) b) 2.8 Answers 1. a) 5p+q-18 b) -27ab-11b 2 c) -8x 2-39y d)!!!"!" e) -6x-6y+18 f) -27x+6 g) 7x2-4x h) 15x 2 +2xy i) -17xy+27y 2 j)!!"!" 4x2-6xy-3 l) 3x-38y 2 +22yw 2. 7a+7b x a a) 116x 2 +53x b) 11x 2-44x

13 MPM1D Unit 2: Algebra Lesson 9 Learning goal: how to find the greatest common factor to common factor a polynomial. Date: Common Factoring WARM-UP Example 1: Simplify the following. a)!!"!!!! b)!!!!!! c)!!"!!!!!!!!! d)!"!!!!!"!!! DIVIDING A POLYNOMIAL BY A MONOMIAL Rule: when dividing by a monomial, each term must be divided by the monomial. Example 2: Simplify. a)!!!!!!! b)!!!!"!!!"! c)!"!!!!!"!!!!!"!"!!" FACTORING A factor is a number or term that divides evenly into each term or polynomial. In algebra, to factor means to express a polynomial as a product of factors, usually a monomial x polynomial. The monomial factor is the Greatest Common Factor (GCF) and the polynomial factor is the result of dividing each term in the original polynomial by the monomial factor. Recall the Greatest Common Factor (GCF) is the highest value that divides exactly into two or more values. Example 3: Given one factor of a polynomial, determine the other factor for each of the following. a) 5 is a factor of 15!. b) 3! is a factor of 15!". c) 8! is the GCF of 24!" 16!. d)!!! is the GCF of 2!!!! + 5!!! -4!!!.

14 Example 4: Determine the GCF of each of the following terms. a) 8!! and 16! b) 15!"#, 25!", 10!!!! c) 7!!!!!! and 2!!!!! Factoring is the opposite of expanding. EXPANDING Factored Form Expanded Form 3(10! + 3) 30! + 9 FACTORING Now lets combine everything to factor a binomial Consider : 20x 2 +15x First let s determine the GCF = Now, divide each term in the original expression by the GCF To complete this reverse distributive process, = ( ) Write the GCF in front of the brackets, and GCF what's left after What is left over after dividing in the brackets. Example 5: Common factor the following. a) 10! 20 b) 22! 99! c)!!! d) 10!! 25! e) 13! + 12! 4! f) 4!! + 8! 10!" g) 12!!!!! 8!"# h) 16!! + 8! + 4! i) 35!!!! 21!!! + 7!!!!

15 Example 6: The area of a tennis court is represented by 60x x. What are the dimensions of the tennis court? 2 Example 7: A triangle has an area of 8x + 12x and a height of 4x. What is the length of its base? bh The formula for the area of a triangle is A = 2

16 Assignment 2.9: Common Factoring 1. Using the greatest common factor, write the binomial in factored form. a) 4x + 20 b) 5x + 30x 2 c) 12x 2 48x d) 21x 2 49x e) 18x + 33 f) 20x 50x 2 g) 48x 2 63x h) 36x 3 72x 2 2. Common Factor a) 4x x + 8 b) 3x 2 + 6x 9 c) 5x x 2 120x d) 3x 4 36x x e) 18x 50y f) 100z + 50z 75z g) 36rs 108r s 7 10 h) ab a i) 2cd 3c + 4c j) 3 2 c + c c 3. The area of a chalkboard is represented by 21x 2 + 6x. What are the dimensions of chalkboard? 4. Challenge: Common Factor a) a 2 b 3 c ab 2 c 2 + a 2 b 2 c 2 b) 3x(x + y) + 2y(x + y) c) 5x(2x 3) (2x 3) 2.9 Answers 1. a) 4(x + 5) b) 5x(1 + 6x) c) 12x(x 4) d) 7x(3x 7) e) 3(6x + 11) f) 10x(2 5x) g) 3x(16x + 21) h) 36x 2 (x + 2) 2. a) 4(x 2 + 3x + 2) b) 3(x 2 + 2x 3) c) 5x(x 2 + 2x 24) d) 3x 2 (x 2 12x + 35) e) 2(9x 25 y ) f) 25 z (4z + 2z 3) g) 36 rs (1 3 rs) 7 3 h) a ( b a ) i) c (2c d 3c+ 4) 2 j) cc ( + c 1) 3. 3x by 7x a) ab 2 c (ab c + ac) b) (x+y)(3x + 2y) c) (2x 3)(5x 1)