Chapter 4 Factoring And Quadratic Equations

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Chapter 4Factoring and Quadratic EquationsLesson 1: Factoring by GCF, DOTS, and Case ILesson 2: Factoring by Grouping & Case IILesson 3: Factoring by Sum and Difference of Perfect CubesLesson 4: Solving Quadratic Equations by FactoringLesson 5: Solving Quadratic InequalitiesThis assignment is a teacher-modified version of Algebra 2 Common Core Copyright (c) 2016 eMathInstruction, LLC used by permission.

Chapter 4Lesson 1GCF, DOTS, and Case I FactoringFactoring:When we factor it is important to remember that ; we are simply rewriteit in an equivalent form.GCF or “Greatest Common Factor”:Factoring by GCF means that we “ ” what terms have in . Thiscan be a combination of numbers, variables, or both.Exercise #1: Factor out by GCF.(a) 3x2 6x(b) 20x – 5x2(c) 10x2y3 – 25xy4(d) 15x – 30x3(e) 2x2 8x 10(f) 8x3 – 12x2 20x

Exercise #2: Rewritten in factored form 20x2 – 36x is equivalent to(1) 2x(10x – 15)(3) 5x(4x 7)(2) 4x(5x – 9)(4) 9x(x – 4)Conjugate Multiplication Pattern“DOTS” Factoring:Another type of factoring stems from multiplication. This type of factoringis known as Difference of Two Squares or “DOTS” factoring. When we factor these types of expressions,we conjugate multiplication.Exercise #3: Write each of the following binomials as the product of a conjugate pair.(a) x2 – 9(b) 25 – 4x2(c)(d)

Exercise #4: Factor each expression completely.Factoring Completely: When we factor completely, it means that we factor until we. It is important to always look for a GCF first.(a) 28x2 – 7(b) 3x3 – 48x(c) 16x4 – 81(d) 2x4 – 162Trinomial Case I Factoring:Another type of factoring is trinomial factoring. This is when we have a trinomial. In Case I factoring, theleading coefficient is .To factor these, it is helpful to look at the term and the term.Exercise #5: Write each of the following trinomials in factored form.(a)x2 – 7x – 18(b) x2 14x 24

(c) x2 x – 12(d) x2 – 5x 6(e) x2 – 15x 44(f) x2 – 6x – 16Exercise #6: Factor each expression completely.(a) 4x2 8x 4(b) 5x2 - 25x – 30(c) 2x2 8x – 64(d) 3x2 18x 27

Chapter 4Lesson 1 HOMEWORKGCF, DOTS, and Case I FactoringFluency:1. Factor the following completely:(a) 30 x 2 35x(b) 20 x3 5x 2 15x(c) 14 x3 35x2 7 x(d) 27 x3 12 x(e) 8x 2 512(f) 25 x 2 (h)(i)(k) x 17 x 30219

2. If one factor of 56x4y3 – 42x2y6 is 14x2y3, what is the other factor?(1) 4x2 – 3y3(3) 4x2y – 3xy3(2) 4x2 – 3y2(4) 4x2y – 3xy23. When factored completely, x3 – 13x2 – 30x is(1) x(x 3)(x – 10)(3) x(x 2)(x – 15)(2) x(x – 3)(x – 10)(4) x(x – 2)(x 15)Applications4. The area of any rectangular shape is given by the product of its width and length. If the area of aparticular rectangular garden is given by A 15x 2 35x and its width is given by 5x , then find anexpression for the garden’s length. Justify your response.5. The volume of a particular rectangular box is given by the equation V 50 x 2 x3 . The height andlength of the box are shown on the diagram below. Find the width of the box in terms of x. Recall thatV L W H for a rectangular box.x 52x?

Chapter 4Lesson 2Grouping & Case II FactoringFactoring by Grouping:A new type of factoring is factoring by grouping. This type of factoring requires us to see structure inexpressions. We usually factor by grouping when we have a polynomial that has four or more terms.Examplex3 2x2 3x 6Steps1. terms together that havea factor.2. Factor each group3. Factor by4. Distribute to check (if needed)Exercise #1: Factor the expression 2x3 – 6x2 5x – 15. Justify each step with one of the three majorproperties of real numbers, i.e. the commutative, associative, or distributive.

Exercise #2: Use the method of factoring by grouping to completely factor the following expressions.(a) 3x3 2x2 - 27x - 18(b) 18x3 9x2 - 2x - 1(c) x5 4x3 2x2 8(d) 5x3 10x2 20x 40Exercise #3: Write the expression (x 3)(x - 4) 5(x 3) as the equivalent product of binomials.Exercise #4: Consider the expression x2 ab - ax - bx.(a) How can you rewrite the expression so that the first two terms share a common factor (other than1)?(b) Write this expression as an equivalent product of binomials.

Exercise #5: Louis factored the expression 2x3 10x2 7x 21 below. Is he correct? Explain.2x3 10x2 7x 21 2x2(x 5) 7(x 3) (2x2 7)(x 5 x 3) (2x2 7)(2x 8)Case II Trinomial Factoring:Case II Trinomial factoring is used when the leading coefficient is not 1, even after any GCF was takenout.Example2x2 – 7x 6Steps1. See if you can factor out a GCF.2. Multiply the coefficient of the first tem withthe last term.3. Split the middle term. You must determinethe signs and coefficients for the two terms.4.Factor by grouping.Exercise #6: Factor the following trinomials.(a) 3x2 19x - 40(b) 2x2 - 15x 18

(c) 15x2 13x 2(d) 10x2 13x – 30(e) 12x2 8x - 15(f) 36x2 - 35x 6Exercise #7: Factor the following trinomials completely.(a) 10x2 55x - 105(b) 12x2 57x - 15(c) 2x2 20x 50(d) 12x2 29x - 8

Chapter 4Lesson 2 HOMEWORKGrouping & Case II FactoringFluency:1. Rewrite each of the following as a product of binomials.(a) x 10 x 3 x 5 x 3 (b) 3x 7 x 5 x 5 2 x 4 (c) 10 x 6 x 35x 21(d) 12 x 3x 20 x 5(e) 2 x 29 x 15(f) 18x 25x 822222. Which of the following is the correct factorization of the trinomial 12 x 23x 10 ?2(1) 6 x 1 3x 10 (3) 4 x 5 3x 2 (2) 6 x 2 2 x 5 (4) 4 x 5 3x 2

3. Factor each expression completely.(a) 15x 110 x 120(b) 10 x 26 x 12 x232Reasoning:4. Consider the expression: x 5x 9 x 45 . Enter this expression on your calculator and find its32zeroes. Provide evidence. Then, factor it completely. Do you see the relationship between the factorsand the zeroes?

Chapter 4Lesson 3Factoring Sum & Difference of CubesTwo special factoring formulas are the sum and difference of perfect cubes. These formulas are:a3 b3 a3 – b3 Yes, these formulas have to be memorized!!! The quadratic portion usually cannot be factored.Examples:When you have a pair of cubes, carefully apply the appropriate rule. By “carefully,” I mean “usingparentheses to keep track of everything, especially the negative signs.” Here are some typical problems:Factor:1.) x3 – 8Hint: It helps to write the example in cubed form. For thisexample, we would write this as: (x)3 – (2)3. By doing this, wecan clearly see which is our “a” value and which is our “b” value.Now, finish factoring by using the rules.

Exercise #2: Factor each of the following.(a) a3 27(b) x3 – 64(c) 8b3 216c6(d) x9 – 125

Excerise #3: Factor each of the following completely.(a) 2x3 128(b) x9 – 512(c) x6 – 729(d) 2 – 686y3Exercise #4: Daniel incorrectly factored x6 – 125 as (x3 5)(x3 – 5). Where did Daniel go wrong? What isthe correct factorization?

Exercise #5: y6 – 64 can be first factored as a difference of cubes or as a difference of squares.(a) Factor y6 – 64 first as a difference of cubes, and then factor completely.(b) Factor y6 - 64 first as a difference of squares and then factor completely.(c) Based on your answers to part (a) and part (b), if a polynomial can be factored as a differenceof cubes or a difference of squares which one should you do first? Why?

Chapter 4Lesson 3 HomeworkFactoring Sum & Difference of CubesFluencyExercise #1: Each of the following expressions is either a difference of perfect cubes or a sum of perfectcubes. Factor appropriately.(a) 216z3 – w3(b) 27r3 1000s3(c) 250t3 16s3(d) 8k6 – 27q3(e) a6 – 8b3(f) 64y6 – 1

Exercise #2: Louis incorrectly factored x3 – 216 as (x2 6)(x – 36). Where did Louis go wrong? What is thecorrect factorization?Exercise #3: Factor 216s3 27t3.(1) (6s 3t)(36s2 – 18st 9t2)(3) (6s – 3t)(36s2 18st 9t2)(2) (9s 3t)(36s2 9t2)(4) (6s 3t)(36s2 18st 9t2)

Chapter 4Lesson 4Solving Quadratic EquationsSolving Quadratic EquationsThe Zero Product Law:If the product of multiple factors is equal to zero then at least one of the factors must be equal to zero.The Zero Product Law can be used to solve any quadratic equation that is (notprime). To utilize this technique, we must first set the equation to andthen factor the non-zero side.Exercise #1: Solve each of the following quadratic equations using the Zero Product Law.(a) x2 3x – 14 -2x 10(b) 3x2 12x – 7 x2 3x – 2

Exercise #2: Consider the system of equations shown below consisting of a parabola and a line.y 3x2 - 8x 5andy 4x 5(a) Find the intersection points of these curves algebraically.(b) Using your calculator, sketch the graph of this system on the axes on the axes below. Be sure to labelthe curves with equations, the intersection points, and the window.(c)Verify your answer to part (a) by using the Intersect command on your calculator.

Exercise #3: The parabola shown below has the equation y x2 – 2x – 3.(a) Write the coordinates of the two x-intercepts of thegraph.(b) Algebraically find the x-intercepts of the parabola.Exercise #4: Algebraically find the set of x-intercepts (zeros) for each parabola given below.(a) y 4x2 – 1(b) y 3x2 13x – 10

(c) y 4x2 – 10x(c) y x2 13x – 14Exercise #5: A quadratic function of the form y x2 bx c is shown below.(a) What are the x-intercepts of this parabola?(b) Based on your answer in part (a), write the equationof this quadratic function first in factored form and thenin trinomial form.

Chapter 4Lesson 4 HomeworkSolving Quadratic EquationsFluency:1. Solve each of the following equations for the value of x.(a) 12 x 8x 0(b) x 4 x 40 10 x 1522(c) 6 x 15x 2 2 x 10 x 422(d) 4 x 3x 11 3x 22

Applications:2. Consider the system of equations shown below consisting of one linear and one quadratic equation.y 4 x 5 and y 2 x 2 5x 10y(a) Find the intersection points of this system algebraically.x(b) Using your calculator, sketch a graph of this system to the right. Be sure to label the curves withequations, the intersection points, and the window.(c) Verify with the Intersect function on your calculator.

3. Algebraically, find the zeroes (x-intercepts) of each quadratic function given below.(a) y 12 x 2 18x(b) y 2 x 2 6 x 8Reasoning4. A quadratic function of the form y x 2 bx c is shown on the graph below.(a) What are the x-intercepts of this parabola?(b) Based on your answer to part (a), write the equation of this quadraticfunction first in factored form and then in trinomial form.yx

Chapter 4Lesson 5Solving Quadratic InequalitiesQuadratic Inequalities in One Variable:Quadratic inequalities are similar to quadratic equations. When solving these types of problems, it isimportant to first find the zeros of the quadratic equation. Then, we put these zeros on a number lineand test each interval to see where we must shade. Finally, we write our answer as an inequality. or : we need to put an open circle on our number line. or : we need to put a close circle on our number line.Exercise #1: Solve each of the following quadratic inequalities. Graph your solutions on a number lineand write your final answer in set-builder notation.(a) x2 – 5x – 36 0(b) x2 – x – 12 0(c) 2x2 – 13x – 7 0(d) 5x2 28x – 12 0

(e) 2x2 – 4x – 8 10x – 8(f) x2 14x – 6 14x 19Exercise #2: The number line graphis the solution to which of the followinginequalities?(1) x2 – 2x – 8 0(3) x2 – 2x – 8 0(2) x2 2x – 8 0(4) x2 2x – 8 0Exercise #3: Which of the following represents the solution set of the inequality -2x2 7x – 3 0?(1)(3)(2)(4)

Chapter 4Lesson 5 HomeworkSolving Quadratic InequalitiesFluency1. Which of the following values of x is in the solution set of the inequality x x 2 0 ?2(1) 1(3) 0(2) 2(4) 42. The solution set of the inequality x 25 is which of the following?2(1) 5, (3) , 5 5, (2) 5, 5 (4) , 5 3. The solution to the inequality x 9 0 can be expressed graphically as2(1)(2)(3)-505-505(4)-505-505

4. Find the solution set to each of the quadratic inequalities shown below. Represent your solution setusing any acceptable notation and graphically on a number line.(a) x 5x 6(b) x 10 x 24(c) 8x 50 x 5 10 x 5(d) 7 x 4 x 3 3x 4 x 422222

Grouping & Case II Factoring Factoring by Grouping: A new type of factoring is factoring by grouping. This type of factoring requires us to see structure in expressions. We usually factor by grouping when we have a polynomial that has four or more terms. Example Steps x3 2x2 3x 6 1. _ terms together that have

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