Algebra 2: Module 4 Lesson 1 In this lesson, you will review how to factor quadratics and solve quadratic equations by factoring. Solving Quadratic Equations by Factoring

What are the dimensions of the garden? Q: Suppose you are trying to create a garden. The length of the garden needs to be six feet longer than the width. You will be given 40 square feet of space. A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals length times width, then (x)(x+6) = 40. We want to find the values of that which make this equation true. We can use the method of factoring to solve this problem. Let’s learn how to solve by factoring so we can solve this problem without guess-and-check. Image courtesy of Tom Curtis @ FreeDigitalPhotos.net What are the dimensions of the garden? When you think of spring, you may think of the beautiful flowers or maybe the rain. Mother’s Day and Valentine’s Day conjure up visions of roses and chocolate. Have you ever eaten fresh vegetables from the garden? Ever think of planting a garden? Let’s look at this situation. Suppose you are trying to create a garden. The length of the garden needs to be six feet longer than the width. You will be given 40 square feet of space. A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals length times width, then (x)(x+6) = 40. We want to find the values of that make this equation true. We can use the method of factoring to solve this problem. Let’s learn how to solve by factoring so we can solve this problem without guess-and-check.

Distributive Property Distributive Property a(b + c) = ab + ac Using the Distributive Property EX 1: 3ac(2a – 7c) = 6a²c – 21ac² EX 2: (4x – 5y)(4x + 5y) = 16x² - 25y² EX 3: (2x – 5)(x – 3) = 2x² - 11x + 15 Work for Example 2: Work for Example 3: Let’s recall the distributive property. We use this in math to remove parentheses by multiplying the term outside by each term inside. Here are some examples. Distributive Property 𝑎 𝑏+𝑐 =𝑎𝑏+𝑎𝑐 Using the Distributive Property EX 1: 3𝑎𝑐 2𝑎−7𝑐 =6 𝑎 2 𝑐−21𝑎 𝑐 2 EX 2: 4𝑥−5𝑦 4𝑥+5𝑦 =16 𝑥 2 −25 𝑦 2 EX 3: 2𝑥−5 𝑥−3 =2 𝑥 2 −11𝑥+15 FOIL 4𝑥 −5𝑦 16 𝑥 2 −20𝑥𝑦 +5𝑦 20𝑥𝑦 −25 𝑦 2 FOIL 2𝑥 −5 𝑥 2 𝑥 2 −5𝑥 −3 −6𝑥 15

Methods for Factoring Quadratics Greatest Common Factor- remove the greatest common factor from each term using the distributive property. GCF (leftover) Difference of Two Perfect Squares- factor into two binomials with opposite signs and the square root of each term. x² - y² = (x + y)(x – y) Trinomial- factor into two binomials x² + bx + c = (x + p)(x + q) Factoring is a way of writing an expression as a product, in the form of a multiplication problem. There are 5 different types of factoring, 3 of which we will review and learn about in this lesson. The other methods you will learn later in the course. Greatest Common Factor- remove the greatest common factor from each term using the distributive property. GCF (leftover) Difference of Two Perfect Squares- factor into two binomials with opposite signs and the square root of each term. 𝑥 2 − 𝑦 2 =(𝑥+𝑦)(𝑥−𝑦) Trinomial- factor into two binomials 𝑥 2 +𝑏𝑥+𝑐=(𝑥+𝑝)(𝑥+𝑞)

Greatest Common Factor Factor 2x² + 4x = 2x(x + 2) You should check each term to see what they ALL have in common and then factor it out of each term. You can check by the distributive property. Factor 90an²+18a²n – 9an = 9an(10n + 2a – 1) GCF (leftover factor for each term) Greatest common factor is usually the first type of factoring you should look for in an expression. If there is a common factor and you do not take it out first, you may end up with an expression that is not factored completely or it may be more difficult to factor. It is also the one that students miss most often. Factor 2 𝑥 2 +4𝑥 = 2𝑥(𝑥+2) You should check each term to see what they ALL have in common and then factor it out of each term. You can check by the distributive property. Factor 90 𝑎𝑛 2 +18 𝑎 2 𝑛−9𝑎𝑛 = 9𝑎𝑛(10𝑛+2𝑎−1) GCF (leftover factor for each term)

Difference of Two Perfect Squares EX 1: Factor 81x² - 16 = (9x) ² - 4² = (9x + 4)(9x – 4) EX 2: Factor n² - 49 = (n) ² - 7² = (n + 7)(n – 7) *Notice that EX 2 could be written as n² + 0n - 49 and when you multiply the binomials, the middle terms are eliminated. *Notice that b = 0 and c = -49 *Let’s look at the X factor. Do you see the pattern? This method does NOT apply to the SUM a² + b². If you multiply two binomials that are the same except for the sign, the middle terms will be eliminated and you will end up with 2 perfect squares with a minus sign between. We call this the difference of two perfect squares, or just the difference of squares. In this lesson, you will learn how to take it apart and write it in factored form. EX 1: Factor 81 𝑥 2 −16= 9𝑥 2 − 4 2 =(9𝑥+4)(9𝑥−4) EX 2: Factor 𝑛 2 −49= 𝑛 2 − 7 2 =(𝑛+7)(𝑛−7) *Notice that EX 2 could be written as 𝑛 2 +0𝑛−49 and when you multiply the binomials, the middle terms are eliminated. *Notice that b = 0 and c = -49 *Let’s look at the X factor. Do you see the pattern? This method does NOT apply to the SUM a² + b².

How to Factor a Trinomial, a = 1 Set up your two binomial factors (x + p)(x + q) for the quadratic x² + bx + c, if it is factorable. Square root the first squared variable and put that in the front of each term Set up the X factor with c at the top and b at the bottom Find the values of p and q using the sum b and product c What 2 numbers multiply to = c AND add to = b? These are p and q! When you multiply two binomials, you most often end up with a trinomial. We are now going to look at how to take it apart and write a trinomial as two binomials multiplied. Set up your two binomial factors (x + p)(x + q) for the quadratic 𝑥 2 +𝑏𝑥+𝑐, if it is factorable. Square root the first squared variable and put that in the front of each term Set up the X factor with c at the top and b at the bottom Find the values of p and q using the sum b and product c What 2 numbers multiply to = c AND add to = b? These are p and q! Find this number p Find this number q

Factor Trinomial Examples, a = 1 EX 1: Factor x2 + 8x + 7 = ( )( ) -Write x in the front of each binomial -Find 2 numbers that multiply = 7 AND add = 8 *pq = 7 and p+q = 8 -Put those numbers into the binomials EX 2: Factor m2 – 2m – 80 = ( )( ) -Write m in the front of each binomial -Find 2 numbers that multiply = -80 AND add = -2 *pq = -80 and p+q = -2 EX 3: Factor c2 + 4c - 12 = ( )( ) -Write c in the front of each binomial -Find 2 numbers that multiply = -12 AND add = 4 *pq = -12 and p+q = 4 On this page, we will factor trinomials that have a leading coefficient of 1. If you recall from the review, there is an X factor pattern for the values of b (which is the x coefficient) and c (which is the constant). EX 1: Factor x2 + 8x + 7 = ( )( ) -Write x in the front of each binomial -Find 2 numbers that multiply = 7 AND add = 8 *pq = 7 and p+q = 8 -Put those numbers into the binomials EX 2: Factor m2 – 2m – 80 = ( )( ) -Write m in the front of each binomial -Find 2 numbers that multiply = -80 AND add = -2 *pq = -80 and p+q = -2 EX 3: Factor c2 + 4c - 12 = ( )( ) -Write c in the front of each binomial -Find 2 numbers that multiply = -12 AND add = 4 *pq = -12 and p+q = 4 -Put those numbers into the binomials

Solutions for Factoring Trinomial, a = 1 EX 1: Factor x2 + 8x + 7 = (x + 7)(x + 1) -Write x in the front of each binomial -Find 2 numbers that multiply = 7 AND add = 8 *pq = 7 and p+q = 8 7 x 1 = 7 AND 7 + 1 = 8 Check them first! -Put those numbers into the binomials EX 2: Factor m2 – 2m – 80 = (m - 10)(m + 8) -Write m in the front of each binomial -Find 2 numbers that multiply = -80 AND add = -2 *pq = -80 and p+q = -2 -10 x 8 = -80 AND -10 + 8 = -2 Check them first! EX 3: Factor c2 + 4c - 12 = (c + 6)(c - 2) -Write c in the front of each binomial -Find 2 numbers that multiply = -12 AND add = 4 *pq = -12 and p+q = 4 6 x -2 = 4 AND 6 + -2 = 4 Check them first! **NOTE: The binomial factors may be reversed because multiplication is commutative! 7 1 -10 8 Here you will find the solutions for the trinomial factoring examples on the previous slide. EX 1: Factor x2 + 8x + 7 = (x + 7)(x + 1) -Write x in the front of each binomial -Find 2 numbers that multiply = 7 AND add = 8 *pq = 7 and p+q = 8 7 x 1 = 7 AND 7 + 1 = 8 Check them first! -Put those numbers into the binomials EX 2: Factor m2 – 2m – 80 = (m - 10)(m + 8) -Write m in the front of each binomial -Find 2 numbers that multiply = -80 AND add = -2 *pq = -80 and p+q = -2 -10 x 8 = -80 AND -10 + 8 = -2 Check them first! EX 3: Factor c2 + 4c - 12 = (c + 6)(c - 2) -Write c in the front of each binomial -Find 2 numbers that multiply = -12 AND add = 4 *pq = -12 and p+q = 4 6 x -2 = 4 AND 6 + -2 = 4 Check them first! -Put those numbers into the binomials **NOTE: The binomial factors may be reversed because multiplication is commutative! 6 -2

How to Factor a Trinomial, a ≠1 Check to see if there is a GCF and factor it out You will have two binomial factors ( )( ) for the quadratic ax² + bx + c, if it is factorable. Set up the X factor with the product ac at the top and b at the bottom Find the values of p and q using the sum b and product ac *m x n = ac AND m + n = b Place m and n in the box with a and c terms Find the GCF of each row & column Use the signs from the m and n terms The GCFs are the binomial factors! Find this number m Find this number n What about trinomials that have a leading coefficient that is not equal to 1. Not only do we use the X factor method, but if you put the answers in the box you can use the Greatest Common Factor method to find the binomial factors. Check to see if there is a GCF and factor it out You will have two binomial factors ( )( ) for the quadratic 𝑎𝑥 2 +𝑏𝑥+𝑐, if it is factorable. Set up the X factor with the product ac at the top and b at the bottom Find the values of p and q using the sum b and product ac *m x n = ac AND m + n = b Place m and n in the box with a and c terms Find the GCF of each row & column Use the signs from the m and n terms The GCFs are the binomial factors! GCF ( ) ax² mx nx c Binomial Find GCF (Binomial) Find GCF

Factor the Trinomial, a ≠1 EX 1: Factor 2x2 + 13x - 7 =( )( ) -Find ac and b, put them in the X factor -Find m and n such that mxn= -14 AND m+n = 13 14 x -1 = -14 AND 14 + -1 = 13 -Put them in box and find the GCF for each row and column. 2x2 + 13x - 7 =(x + 7)(2x - 1) EX 2: Factor 6x2 - 19x + 3 =( )( ) mxn= 18 AND m+n = -19 -18 x -1 = 18 AND -18 + -1 = -19 -Put them in the box and find the GCF for each row and column. 6x2 - 19x + 3 =(6x - 1)(x - 3) GCF x + 7 2x 2x2 14x - 1 -1x - 7 14 -1 Here we can see the use of both the X factors and the Box methods to help us factor the trinomials. EX 1: Factor 2x2 + 13x - 7 =( )( ) -Find ac and b, put them in the X factor -Find m and n such that mxn= -14 AND m+n = 13 14 x -1 = -14 AND 14 + -1 = 13 -Put them in box and find the GCF for each row and column. 2x2 + 13x - 7 =(x + 7)(2x - 1) EX 2: Factor 6x2 - 19x + 3 =( )( ) mxn= 18 AND m+n = -19 -18 x -1 = 18 AND -18 + -1 = -19 6x2 - 19x + 3 =(6x - 1)(x - 3) -18 -1 GCF x - 3 6x 6x2 - 18x - 1 - 1x 3

Forms of a Quadratic Equation Standard Form of a Quadratic Equation ax² + bx + c = 0, a ≠ 0 Factored Form of a Quadratic Equation a(x + p)(x + q) = 0, a ≠ 0 Factoring means to write the terms in multiplication form (as a product). Zero Product Property If ab = 0 then either a = 0 or b = 0 (or both). The expression must be set equal to zero to use this property. Zero Product Example: Quadratic in Factored Form (x – 6) (x + 8) = 0 x – 6 = 0 or x + 8 = 0 x = 6 or x = 8 A quadratic equation can be written in 3 different forms, two of which we will study in this module and use to solve quadratic equations. Standard Form of a Quadratic Equation 𝑎 𝑥 2 +𝑏𝑥+𝑐=0, 𝑎≠0 Factored Form of a Quadratic Equation a(x + p)(x + q) = 0 Factoring means to write the terms in multiplication form (as a product). Zero Product Property If ab = 0 then either a = 0 or b = 0 (or both). The expression must be set equal to zero to use this property. Zero Product Example: Quadratic in Factored Form (x – 6) (x + 8) = 0 x – 6 = 0 or x + 8 = 0 x = 6 or x = 8

Solve a Quadratic in Factored Form Solve (x – 11)(x – 1) = 0. This equation is already in factored form. All we have to do is use the Zero Product Property and set each factor equal to zero. x – 11 = 0   or   x – 1 = 0 x = 11  or  x = 1 You may write your solutions as x = {1, 11}. Do not enclose the values of x in parentheses. They may be misconstrued as an ordered pair! But brackets may be used to signify a set of numbers. What if a quadratic is already factored? You just need to use the zero product property to find all possible values of the variable. Solve (x – 11)(x – 1) = 0. This equation is already in factored form. All we have to do use the Zero Product Property and set each factor equal to zero. x – 11 = 0   or   x – 1 = 0 x = 11  or  x = 1 You may write your solutions as x = {1, 11}. Do not enclose the values of x in parentheses. They may be misconstrued as an ordered pair! But brackets may be used to signify a set of numbers.

Solve by Factoring Write the quadratic equation in standard form: move all terms to one side of the equation, usually the left, using addition or subtraction such that one side of the equation is equal to zero. Simplify if needed. *To make the process easier, lets always try to make our quadratic term positive! [ax² + bx + c = 0] Write the quadratic equation in factored form: factor the equation completely. [a(x + p)(x + q) = 0] Set each factor equal to zero, and solve each equation. [If ab = 0 then either a = 0 or b = 0] Most quadratics are not written in factored form. So, we need to have a process by which we can untangle the quadratics to get them ready to solve. Write the quadratic equation in standard form: move all terms to one side of the equation, usually the left, using addition or subtraction such that one side of the equation is equal to zero. Simplify if needed. *To make the process easier, lets always try to make our quadratic term positive! [𝑎 𝑥 2 +𝑏𝑥+𝑐=0] Write the quadratic equation in factored form: factor the equation completely. [a(x + p)(x + q) = 0] Set each factor equal to zero, and solve each equation. [If ab = 0 then either a = 0 or b = 0]

Solve Quadratics by Factoring- Example 1 Solve x2 - 4x + 3 = 0. The quadratic equation is in standard form (set equal to zero). This needs to be factored first. Factor: x2 - 4x + 3 = (x - 1)(x - 3) *You may want to use the X factors for this. (x - 1)(x - 3) = 0 Set each factor equal to zero: x - 1 = 0 or x - 3 = 0 Solve: x = 1  or  x = 3, can also write {1, 3} This is one example of a quadratic trinomial to solve by factoring. This has a leading coefficient of 1 and we can use the X factor method. Solve x2 - 4x + 3 = 0. The quadratic equation is in standard form (set equal to zero). This needs to be factored first. Factor: x2 - 4x + 3 = (x - 1)(x - 3) *You may want to use the X factors for this. (x - 1)(x - 3) = 0 Set each factor equal to zero: x - 1 = 0 or x - 3 = 0 Solve: x = 1  or  x = 3, can also write {1, 3}

Solve Quadratics by Factoring- Example 2 Solve x2 – 8 = 2x. Write the equation in standard form (set the equation equal to zero). x2 – 2x – 8 = 0 Factor: x2 – 2x – 8 = (x – 4)(x + 2) (x – 4)(x + 2) = 0 Set each factor equal to zero: x – 4 = 0  or  x + 2 = 0 Solve: x = 4  or  x = -2 or {4, -2} In this next example, you will notice that the quadratic is not in standard form. We need to move all terms to one side and make the other side = to 0 before we factor. Solve x2 – 8 = 2x. Write the equation in standard form (set the equation equal to zero). x2 – 2x – 8 = 0 Factor: x2 – 2x – 8 = (x – 4)(x + 2) (x – 4)(x + 2) = 0 Set each factor equal to zero: x – 4 = 0  or  x + 2 = 0 Solve: x = 4  or  x = -2 or {4, -2}

Solve Quadratics by Factoring- Example 3 Solve x² + 3x = 0 Since it is written in standard form, factor it using GCF: x(x + 3) = 0. Set all the factors equal to zero, and solve: x(x + 3) = 0 x = 0  or  x + 3 = 0 x = 0  or  x = –3 OR {0, -3} Common mistakes on this type of problem: Do not use X factors- that is for Trinomials. Do not try to "solve" the equation for "x(x + 3) = 0" by dividing by x. Due to the fact that the zero product property says either factor could be zero or both dividing the x assumes that x is not zero. We don’t know that, and you lose one of your solutions. Therefore, we can't divide by zero…it against math laws! Even though you are used to factors having variables and numbers (like the factor, x + 3), a factor might contain only a variable, such as x. Be sure to set this equal to zero also. If it said 5(x-2)(x+4) = 0, 5 cannot = 0! This equation is a quadratic, but it is not a trinomial. We must determine another method to factor it. Solve x² + 3x = 0 Since it is written in standard form, factor it using GCF: x(x + 3) = 0. Set all the factors equal to zero, and solve: x(x + 3) = 0 x = 0  or  x + 3 = 0 x = 0  or  x = –3 OR {0, -3} Common mistakes on this type of problem: Do not use X factors- that is for Trinomials. Do not try to "solve" the equation for "x(x + 3) = 0" by dividing by x. Due to the fact that the zero product property says either factor could be zero or both, dividing the x assumes that x is not zero. We don’t know that, and you lose one of your solutions. Therefore, we can't divide by zero…it against math laws! Even though you are used to factors having variables and numbers (like the factor, x + 3), a factor might contain only a variable, such as x. Be sure to set this equal to zero also. If it said 5(x-2)(x+4) = 0, 5 cannot = 0!

Solve Quadratics by Factoring- Example 4 Solve (x + 2)(x + 3) = 20. It is very common to see this type of problem, and say: It's already factored!  So I'll set the factors equal to 0 and solve. Remember, the expression must be set equal to zero to use this property. Simplify by the distributive property (x + 2)(x + 3) = 20 (x + 2)(x + 3) -20 = 0 x2 + 5x + 6 - 20 = 0 x2 + 5x – 14 = 0 It is in standard form, ready to factor (x + 7)(x – 2) = 0 Factored, can use X factors x + 7 = 0  or   x – 2 = 0 Set each factor = to 0 and solve x = –7  or   x = 2 This example is not in standard form, but it also has parentheses. You will first need to simplify this quadratic by taking off the parentheses and moving it all to one side before you factor. Solve (x + 2)(x + 3) = 20. It is very common to see this type of problem, and say: It's already factored!  So I'll set the factors equal to 0 and solve. Remember, the expression must be set equal to zero to use this property. Simplify by the distributive property (x + 2)(x + 3) = 20 (x + 2)(x + 3) -20 = 0 x2 + 5x + 6 - 20 = 0 x2 + 5x – 14 = 0 It is in standard form, ready to factor (x + 7)(x – 2) = 0 Factored, can use X factors x + 7 = 0  or   x – 2 = 0 Set each factor = to 0 and solve x = –7  or   x = 2

Solve Quadratics by Factoring- Example 5 Solve 8x2 = 26x - 15. Write the equation in standard form (set the equation equal to zero). 8x2 - 26x + 15 = 0 Factor: (4x – 3)(2x - 5) = 0 Set each factor equal to zero: 4x – 3 = 0  or  2x - 5 = 0 Solve: 4x – 3 = 0 or  2x - 5 = 0 4x = 3 or 2x = 5 x = ¾ or x = 5/2 -20 -6 2x - 5 4x 8x2 -20x - 3 -6x 15 The final example is a trinomial with a leading coefficient that is not 1. First we need to write it in standard form. To factor and solve it, you can use both the X factors and the Box to find the greatest common factor. Solve 8x2 = 26x - 15. Write the equation in standard form (set the equation equal to zero). 8x2 - 26x + 15 = 0 Factor: (4x – 3)(2x - 5) = 0 Set each factor equal to zero: 4x – 3 = 0  or  2x - 5 = 0 Solve: 4x – 3 = 0 or  2x - 5 = 0 4x = 3 or 2x = 5 x = ¾ or x = 5/2

Back to the Garden Q: Suppose you are trying to create a garden. The length of the garden needs to be six feet longer than the width. You will be given 40 square feet of space. A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals length times width, then (x)(x+6) = 40. We want to find the values of that make this equation true. We can use the method of factoring to solve this problem. Simplify using the distributive property: x² + 6x = 40 Write it in standard form: x² + 6x – 40 = 0 Factor: (x + 10) (x – 4) = 0 Set each factor equal to zero: x + 10 = 0 or x – 4 = 0 Solve: x = -10 or x = 4 Since length cannot be negative, -10 cannot be a solution and the only solution is width = 4. Use 4 to find the length 4 + 6 = 10. The dimensions will be 4 feet x 10 feet. Image courtesy of Tom Curtis @ FreeDigitalPhotos.net Now we will revisit the garden question posed on the first slide. Whenever you solve a quadratic, you need to check your answers because they do not always work. We call answers that do not work extraneous solutions. Q: Suppose you are trying to create a garden. The length of the garden needs to be six feet longer than the width. You will be given 40 square feet of space. A: We know that the length is (x + 6) and the width is (x). The area is 40 sq. ft. If area equals length times width, then (x)(x+6) = 40. We want to find the values of that make this equation true. We can use the method of factoring to solve this problem. Simplify using the distributive property: x² + 6x = 40 Write it in standard form: x² + 6x – 40 = 0 Factor: (x + 10) (x – 4) = 0 Set each factor equal to zero: x + 10 = 0 or x – 4 = 0 Solve: x = -10 or x = 4 Since length cannot be negative, -10 cannot be a solution and the only solution is width = 4. Use 4 to find the length 4 + 6 = 10. The dimensions will be 4 feet x 10 feet.