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Chapter 13 § 13.1

Factoring The Greatest Common Polynomials Factor

Chapter Sections Factors

13.1 – The Greatest Common Factor Factors (either numbers or polynomials) 13.2 – Factoring Trinomials of the Form x2 + bx + c When an integer is written as a product of 13.3 – Factoring Trinomials of the Form ax 2 + bx + c integers, each of the integers in the product is a factor of the original number. 13.4 – Factoring Trinomials of the Form x2 + bx + c When a polynomial is written as a product of by Grouping polynomials, each of the polynomials in the 13.5 – Factoring Perfect Square Trinomials and product is a factor of the original polynomial. Difference of Two Squares Factoring – writing a polynomial as a product of 13.6 – Solving Quadratic Equations by Factoring polynomials. 13.7 – Quadratic Equations and Problem Solving

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Greatest Common Factor Greatest Common Factor

Greatest common factor – largest quantity that is a Example factor of all the integers or polynomials involved. Find the GCF of each list of numbers. 1) 6, 8 and 46 6 = 2 · 3 Finding the GCF of a List of Integers or Terms 8 = 2 · 2 · 2 1) Prime factor the numbers. 46 = 2 · 23 2) Identify common prime factors. So the GCF is 2. 3) Take the product of all common prime factors. 2) 144, 256 and 300 144 = 2 · 2 · 2 · 3 · 3 • If there are no common prime factors, GCF is 1. 256 = 2 · 2 · 2 · 2 · 2 · 2 · 2 · 2 300 = 2 · 2 · 3 · 5 · 5 So the GCF is 2 · 2 = 4.

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Greatest Common Factor Greatest Common Factor

Example Example Find the GCF of each list of numbers. Find the GCF of each list of terms. 1) 12 and 8 1) x3 and x7 12 = 2 · 2 · 3 x3 = x · x · x 8 = 2 · 2 · 2 x7 = x · x · x · x · x · x · x So the GCF is 2 · 2 = 4. So the GCF is x · x · x = x3 2) 7 and 20 2) 6x5 and 4 x3 7 = 1 · 7 6x5 = 2 · 3 · x · x · x 20 = 2 · 2 · 5 3 There are no common prime factors so the 4x = 2 · 2 · x · x · x GCF is 1. So the GCF is 2 · x · x · x = 2 x3

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Greatest Common Factor Factoring out the GCF

Example Example Find the GCF of the following list of terms. Factor out the GCF in each of the following polynomials. a3b2, a2b5 and a4b7 a3b2 = a · a · a · b · b 1) 6 x3 – 9x2 + 12 x = a2b5 = a · a · b · b · b · b · b 3 · x · 2 · x2 – 3 · x · 3 · x + 3 · x · 4 = a4b7 = a · a · a · a · b · b · b · b · b · b · b 3x(2 x2 – 3x + 4) 2 2 So the GCF is a · a · b · b = a b 2) 14 x3y + 7 x2y – 7xy = Notice that the GCF of terms containing variables will use the 7 · x · y · 2 · x2 + 7 · x · y · x – 7 · x · y · 1 = smallest exponent found amongst the individual terms for each 7xy (2 x2 + x – 1) variable.

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Factoring Polynomials Factoring out the GCF

The first step in factoring a polynomial is to Example find the GCF of all its terms. Factor out the GCF in each of the following Then we write the polynomial as a product by polynomials. factoring out the GCF from all the terms. 1) 6( x + 2) – y(x + 2) = The remaining factors in each term will form a 6 · (x + 2) – y · (x + 2) = polynomial. (x + 2) (6 – y) 2) xy (y + 1) – (y + 1) = xy · (y + 1) – 1 · (y + 1) = (y + 1) (xy – 1)

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Factoring Factoring Trinomials

Remember that factoring out the GCF from the terms of Recall by using the FOIL method that a polynomial should always be the first step in factoring F O I L a polynomial. (x + 2)( x + 4) = x2 + 4x + 2x + 8 This will usually be followed by additional steps in the = x2 + 6x + 8 process. To factor x2 + bx + c into ( x + one #)( x + another #), Example note that b is the sum of the two numbers and c is the Factor 90 + 15 y2 – 18 x – 3xy 2. product of the two numbers. 90 + 15 y2 – 18 x – 3xy 2 = 3(30 + 5 y2 – 6x – xy 2) = So we’ll be looking for 2 numbers whose product is 3( 5 · 6 + 5 · y2 – 6 · x – x · y2) = c and whose sum is b. 3( 5(6 + y2) – x (6 + y2)) = Note: there are fewer choices for the product, so 3(6 + y2)(5 – x) that’s why we start there first. Martin-Gay, Developmental Mathematics 13 Martin-Gay, Developmental Mathematics 15

Factoring Polynomials

Example § 13.2 Factor the polynomial x2 + 13 x + 30. Since our two numbers must have a product of 30 and a sum of 13, the two numbers must both be positive. Positive factors of 30 Sum of Factors 1, 30 31 2, 15 17 Factoring Trinomials of the 3, 10 13 2 Note, there are other factors, but once we find a pair Form x + bx + c that works, we do not have to continue searching. So x2 + 13 x + 30 = ( x + 3)( x + 10).

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Factoring Polynomials Prime Polynomials

Example Example Factor the polynomial x2 – 11 x + 24. Factor the polynomial x2 – 6x + 10. Since our two numbers must have a product of 24 and a Since our two numbers must have a product of 10 and a sum of -11, the two numbers must both be negative. sum of – 6, the two numbers will have to both be negative. Negative factors of 24 Sum of Factors Negative factors of 10 Sum of Factors – 1, – 24 – 25 – 1, – 10 – 11 – 2, – 12 – 14 – 2, – 5 – 7 – 3, – 8 – 11 Since there is not a factor pair whose sum is – 6, 2 2 x – 6x +10 is not factorable and we call it a prime So x – 11 x + 24 = ( x – 3)( x – 8). polynomial .

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Factoring Polynomials Check Your Result!

Example You should always check your factoring Factor the polynomial x2 – 2x – 35. results by multiplying the factored polynomial Since our two numbers must have a product of – 35 and a to verify that it is equal to the original sum of – 2, the two numbers will have to have different signs. polynomial. Factors of – 35 Sum of Factors Many times you can detect computational – 1, 35 34 errors or errors in the signs of your numbers 1, – 35 – 34 by checking your results. – 5, 7 2 5, – 7 – 2 So x2 – 2x – 35 = ( x + 5)( x – 7).

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Factoring Polynomials

Example § 13.3 Factor the polynomial 25 x2 + 20 x + 4. Possible factors of 25 x2 are { x, 25 x} or {5 x, 5 x}. Possible factors of 4 are {1, 4} or {2, 2}. We need to methodically try each pair of factors until we find a combination that works, or exhaust all of our possible pairs of factors. Factoring Trinomials of Keep in mind that, because some of our pairs are not identical 2 factors, we may have to exchange some pairs of factors and the Form ax + bx + c make 2 attempts before we can definitely decide a particular pair of factors will not work. Continued.

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Factoring Trinomials Factoring Polynomials

Returning to the FOIL method, Example Continued F O I L We will be looking for a combination that gives the sum of the (3x + 2)(x + 4) = 3x2 + 12 x + 2x + 8 products of the outside terms and the inside terms equal to 20 x. = 3x2 + 14 x + 8 Factors Factors Resulting Product of Product of Sum of 2 of 25 x2 of 4 Binomials Outside Terms Inside Terms Products To factor ax + bx + c into (# 1·x + # 2)(# 3·x + # 4), note that a is the product of the two first coefficients, c is {x, 25 x} {1, 4} ( x + 1)( 25 x + 4) 4x 25 x 29 x the product of the two last coefficients and b is the sum of the products of the outside coefficients and (x + 4)( 25 x + 1) x 100 x 101 x inside coefficients. {x, 25 x} {2, 2} ( x + 2)( 25 x + 2) 2x 50 x 52 x Note that b is the sum of 2 products, not just 2 {5 x, 5 x} {2, 2} ( 5x + 2)( 5x + 2) 10 x 10 x 20 x numbers, as in the last section. Continued.

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Factoring Polynomials Factoring Polynomials

Example Continued Example Continued We will be looking for a combination that gives the sum of Check the resulting factorization using the FOIL method. the products of the outside terms and the inside terms equal F O I L to −41 x. (5 x + 2)(5 x + 2) = 5x(5 x) + 5 x(2) + 2(5 x) + 2(2) Factors Factors Resulting Product of Product of Sum of of 21 x2 of 10 Binomials Outside Terms Inside Terms Products = 25 x2 + 10 x + 10 x + 4 {x, 21 x}{1, 10}( x – 1)( 21 x – 10 ) –10 x −21 x – 31 x = 25 x2 + 20 x + 4 (x – 10 )( 21 x – 1) –x −210 x – 211 x So our final answer when asked to factor 25 x2 + 20 x + 4 {x, 21 x} {2, 5} ( x – 2)( 21 x – 5) –5x −42 x – 47 x will be (5 x + 2)(5 x + 2) or (5 x + 2) 2. (x – 5)( 21 x – 2) –2x −105 x – 107 x Continued. Martin-Gay, Developmental Mathematics 25 Martin-Gay, Developmental Mathematics 27

Factoring Polynomials Factoring Polynomials

Example Example Continued Factor the polynomial 21 x2 – 41 x + 10. Factors Factors Resulting Product of Product of Sum of of 21 x2 of 10 Binomials Outside Terms Inside Terms Products Possible factors of 21 x2 are { x, 21 x} or {3 x, 7 x}. {3 x, 7 x}{1, 10}( 3x – 1)( 7x – 10 ) −30 x −7x −37 x Since the middle term is negative, possible factors of 10 (3x – 10 )( 7x – 1) −3x −70 x −73 x must both be negative: {-1, -10} or {-2, -5}. {3 x, 7 x} {2, 5} ( 3x – 2)( 7x – 5) −15 x −14 x −29 x We need to methodically try each pair of factors until (3x – 5)( 7x – 2) −6x −35 x −41 x we find a combination that works, or exhaust all of our possible pairs of factors. Continued. Continued.

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Factoring Polynomials Factoring Polynomials

Example Continued Example Continued Check the resulting factorization using the FOIL method. We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to −7x. F O I L (3 x – 5)(7 x – 2) = 3x(7 x) + 3 x(-2) - 5(7 x) - 5(-2) Factors Resulting Product of Product of Sum of of 6 Binomials Outside Terms Inside Terms Products 2 = 21 x – 6x – 35 x + 10 {−1, −6} ( 3x – 1)( x – 6) −18 x −x −19 x 2 = 21 x – 41 x + 10 (3 x – 6)( x – 1) Common factor so no need to test. {−2, −3} ( 3x – 2)( x – 3) −9x −2x −11 x So our final answer when asked to factor 21 x2 – 41 x + 10 (3 x – 3)( x – 2) Common factor so no need to test. will be (3 x – 5)(7 x – 2). Continued. Martin-Gay, Developmental Mathematics 29 Martin-Gay, Developmental Mathematics 31

Factoring Polynomials Factoring Polynomials

Example Example Continued Factor the polynomial 3 x2 – 7x + 6. Now we have a problem, because we have The only possible factors for 3 are 1 and 3, so we know that, if exhausted all possible choices for the factors, factorable, the polynomial will have to look like (3 x )( x ) but have not found a pair where the sum of the in factored form, so that the product of the first two terms in the binomials will be 3 x2. products of the outside terms and the inside terms is –7. Since the middle term is negative, possible factors of 6 must both 2 be negative: { −1, − 6} or { − 2, − 3}. So 3 x – 7x + 6 is a prime polynomial and will We need to methodically try each pair of factors until we find a not factor. combination that works, or exhaust all of our possible pairs of factors. Continued.

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Factoring Polynomials Factoring Polynomials

Example Example Continued Factors Resulting Product of Product of Sum of Factor the polynomial 6 x2y2 – 2xy 2 – 60 y2. of -30 Binomials Outside Terms Inside Terms Products {-1, 30} ( 3x – 1)( x + 30 ) 90 x -x 89 x Remember that the larger the coefficient, the greater the (3 x + 30)( x – 1) Common factor so no need to test. probability of having multiple pairs of factors to check. {1, -30} ( 3x + 1)( x – 30 ) -90 x x -89 x So it is important that you attempt to factor out any (3 x – 30)( x + 1) Common factor so no need to test. common factors first. {-2, 15} ( 3x – 2)( x + 15 ) 45 x -2x 43 x 6x2y2 – 2xy 2 – 60 y2 = 2 y2(3 x2 – x – 30) (3 x + 15)( x – 2) Common factor so no need to test. The only possible factors for 3 are 1 and 3, so we know {2, -15} ( 3x + 2)( x – 15 ) -45 x 2x -43 x that, if we can factor the polynomial further, it will have to (3 x – 15)( x + 2) Common factor so no need to test. look like 2 y2(3 x )( x ) in factored form. Continued. Continued. Martin-Gay, Developmental Mathematics 33 Martin-Gay, Developmental Mathematics 35

Factoring Polynomials Factoring Polynomials

Example Continued Example Continued

Since the product of the last two terms of the binomials Factors Resulting Product of Product of Sum of will have to be –30, we know that they must be of –30 Binomials Outside Terms Inside Terms Products different signs. {–3, 10} (3 x – 3)( x + 10) Common factor so no need to test. Possible factors of –30 are { –1, 30}, {1, –30}, { –2, 15}, (3x + 10 )( x – 3) –9x 10 x x {2, –15}, { –3, 10}, {3, –10}, { –5, 6} or {5, –6}. {3, –10} (3 x + 3)( x – 10) Common factor so no need to test. (3 x – 10)( x + 3) 9x –10 x –x We will be looking for a combination that gives the sum of the products of the outside terms and the inside terms equal to –x. Continued. Continued.

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Factoring Polynomials Factoring by Grouping

Example Continued Factoring polynomials often involves additional techniques after initially factoring out the GCF. Check the resulting factorization using the FOIL method. One technique is factoring by grouping . F O I L Example (3 x – 10)( x + 3) = 3x(x) + 3 x(3) – 10( x) – 10(3) Factor xy + y + 2 x + 2 by grouping. = 3x2 + 9 x – 10 x – 30 Notice that, although 1 is the GCF for all four = 3x2 – x – 30 terms of the polynomial, the first 2 terms have a GCF of y and the last 2 terms have a GCF of 2. So our final answer when asked to factor the polynomial 6x2y2 – 2xy 2 – 60 y2 will be 2 y2(3 x – 10)( x + 3). xy + y + 2 x + 2 = x · y + 1 · y + 2 · x + 2 · 1 = y(x + 1) + 2(x + 1) = (x + 1) (y + 2)

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Factoring by Grouping

Factoring a Four-Term Polynomial by Grouping § 13.4 1) Arrange the terms so that the first two terms have a common factor and the last two terms have a common factor. 2) For each pair of terms, use the distributive property to factor out the pair’s greatest common factor. Factoring Trinomials of 3) If there is now a common binomial factor, factor it out. 2 4) If there is no common binomial factor in step 3, begin the Form x + bx + c again, rearranging the terms differently. • If no rearrangement leads to a common binomial by Grouping factor, the polynomial cannot be factored.

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Factoring by Grouping

Example Factor each of the following polynomials by grouping. § 13.5 1) x3 + 4 x + x2 + 4 = x · x2 + x · 4 + 1 · x2 + 1 · 4 = x(x2 + 4) + 1(x2 + 4) = (x2 + 4) (x + 1) 2) 2 x3 – x2 – 10 x + 5 = x2 · 2x – x2 · 1 – 5 · 2x – 5 · (– 1) = Factoring Perfect Square x2(2 x – 1) – 5(2 x – 1) = Trinomials and the (2 x – 1) (x2 – 5) Difference of Two Squares

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Factoring by Grouping Perfect Square Trinomials

Example Recall that in our very first example in Section Factor 2 x – 9y + 18 – xy by grouping. 4.3 we attempted to factor the polynomial 2 Neither pair has a common factor (other than 1). 25 x + 20 x + 4. So, rearrange the order of the factors. The result was (5 x + 2) 2, an example of a 2x + 18 – 9y – xy = 2 · x + 2 · 9 – 9 · y – x · y = binomial squared. 2(x + 9) – y(9 + x) = 2(x + 9) – y(x + 9) = (make sure the factors are identical) Any trinomial that factors into a single (x + 9) (2 – y) binomial squared is called a perfect square trinomial .

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Perfect Square Trinomials Difference of Two Squares In the last chapter we learned a shortcut for squaring a binomial Another shortcut for factoring a trinomial is when we (a + b)2 = a2 + 2 ab + b2 want to factor the difference of two squares. 2 2 (a – b)2 = a2 – 2ab + b2 a – b = ( a + b )( a – b) So if the first and last terms of our polynomial to be A binomial is the difference of two square if factored are can be written as expressions squared, and 1.both terms are squares and the middle term of our polynomial is twice the product of those two expressions, then we can use these two 2.the signs of the terms are different. previous equations to easily factor the polynomial. 9x2 – 25 y2 a2 + 2 ab + b2 = (a + b)2 a2 – 2ab + b2 = ( a – b)2 – c4 + d4

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Perfect Square Trinomials Difference of Two Squares

Example Example Factor the polynomial 16 x2 – 8xy + y2. Factor the polynomial x2 – 9. Since the first term, 16 x2, can be written as (4 x)2, and 2 the last term, y is obviously a square, we check the The first term is a square and the last term, 9, can be middle term. written as 3 2. The signs of each term are different, so 8xy = 2(4 x)( y) (twice the product of the expressions we have the difference of two squares that are squared to get the first and last terms of the Therefore x2 – 9 = ( x – 3)( x + 3) . polynomial) Note: You can use FOIL method to verify that the 2 2 2 Therefore 16 x – 8xy + y = (4 x – y) . factorization for the polynomial is accurate. Note: You can use FOIL method to verify that the factorization for the polynomial is accurate.

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Solving Quadratic Equations

Steps for Solving a Quadratic Equation by § 13.6 Factoring 1) Write the equation in standard form. 2) Factor the quadratic completely. 3) Set each factor containing a variable equal to 0. Solving Quadratic 4) Solve the resulting equations. 5) Check each solution in the original equation. Equations by Factoring

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Zero Factor Theorem Solving Quadratic Equations

Quadratic Equations Example • Can be written in the form ax 2 + bx + c = 0. Solve x2 – 5x = 24. • a, b and c are real numbers and a ≠ 0. • First write the quadratic equation in standard form. • This is referred to as standard form . x2 – 5x – 24 = 0 Zero Factor Theorem • Now we factor the quadratic using techniques from the previous sections. • If a and b are real numbers and ab = 0, then a = 0 2 or b = 0. x – 5x – 24 = ( x – 8)( x + 3) = 0 • This theorem is very useful in solving quadratic • We set each factor equal to 0. equations. x – 8 = 0 or x + 3 = 0, which will simplify to x = 8 or x = – 3 Continued.

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Solving Quadratic Equations Solving Quadratic Equations

Example Continued Example Continued • Check both possible answers in the original equation. • Check both possible answers in the original equation. 1 1 1 1 1 4( )( 8( )+= 9) 4( )() 19 += 4( ) (10) = (10)5 = 82 – 5( 8) = 64 – 40 = 24 true 8 8 8 8 2 true (–3)2 – 5( –3) = 9 – (–15) = 24 true 5 5 5 5 4(− )( 8(− )+=− 9) 4( )() −+=− 109 4( ) (1)(5)(1)5 −=−− = • So our solutions for x are 8 or –3. 4 4 4 4 true

1 5 • So our solutions for x are or − . 8 4

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Solving Quadratic Equations Finding x-intercepts

Example Recall that in Chapter 3, we found the x-intercept of Solve 4 x(8 x + 9) = 5 linear equations by letting y = 0 and solving for x. • First write the quadratic equation in standard form. The same method works for x-intercepts in quadratic 32 x2 + 36 x = 5 equations. 32 x2 + 36 x – 5 = 0 Note: When the quadratic equation is written in standard • Now we factor the quadratic using techniques from the previous sections. form, the graph is a parabola opening up (when a > 0) or 2 32 x2 + 36 x – 5 = (8 x – 1)(4 x + 5) = 0 down (when a < 0), where a is the coefficient of the x • We set each factor equal to 0. term. 8x – 1 = 0 or 4 x + 5 = 0 The intercepts will be where the parabola crosses the 1 5 8x = 1 or 4 x = – 5, which simplifies to x = or − . 8 4 x-axis. Continued. Martin-Gay, Developmental Mathematics 54 Martin-Gay, Developmental Mathematics 56

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Finding x-intercepts Strategy for Problem Solving

Example General Strategy for Problem Solving Find the x-intercepts of the graph of y = 4 x2 + 11 x + 6. 1) Understand the problem • Read and reread the problem The equation is already written in standard form, so • Choose a variable to represent the unknown we let y = 0, then factor the quadratic in x. • Construct a drawing, whenever possible 0 = 4 x2 + 11 x + 6 = (4 x + 3)( x + 2) • Propose a solution and check We set each factor equal to 0 and solve for x. 2) Translate the problem into an equation 4x + 3 = 0 or x + 2 = 0 3) Solve the equation 4) Interpret the result 4x = –3 or x = –2 • Check proposed solution in problem x = –¾ or x = –2 • State your conclusion So the x-intercepts are the points ( –¾, 0) and ( –2, 0).

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Finding an Unknown Number

Example § 13.7 The product of two consecutive positive integers is 132. Find the two integers. 1.) Understand Read and reread the problem. If we let Quadratic Equations x = one of the unknown positive integers, then and Problem Solving x + 1 = the next consecutive positive integer.

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Finding an Unknown Number Finding an Unknown Number

Example continued Example continued

2.) Translate 4.) Interpret Check: Remember that x is suppose to represent a positive The product of integer. So, although x = -12 satisfies our equation, it cannot be a solution for the problem we were presented. is 132 two consecutive positive integers If we let x = 11, then x + 1 = 12. The product of the two numbers is 11 · 12 = 132, our desired result. State: The two positive integers are 11 and 12. x • (x + 1) = 132

Continued

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Finding an Unknown Number The Pythagorean Theorem

Example continued Pythagorean Theorem 3.) Solve In a right triangle, the sum of the squares of the lengths of the two legs is equal to the square of the x(x + 1) = 132 length of the hypotenuse. x2 + x = 132 (Distributive property) (leg a)2 + (leg b)2 = (hypotenuse) 2 x2 + x – 132 = 0 (Write quadratic in standard form) hypotenuse (x + 12)( x – 11) = 0 (Factor quadratic polynomial) leg a x + 12 = 0 or x – 11 = 0 (Set factors equal to 0) x = –12 or x = 11 (Solve each factor for x) leg b Continued

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The Pythagorean Theorem The Pythagorean Theorem

Example Example continued Find the length of the shorter leg of a right triangle if the longer leg 4.) Interpret is 10 miles more than the shorter leg and the hypotenuse is 10 miles Check: Remember that x is suppose to represent the length of less than twice the shorter leg. the shorter side. So, although x = 0 satisfies our equation, it 1.) Understand cannot be a solution for the problem we were presented. Read and reread the problem. If we let 2 x - 10 If we let x = 30, then x + 10 = 40 and 2 x – 10 = 50. Since 302 + x = the length of the shorter leg, then x 402 = 900 + 1600 = 2500 = 502, the Pythagorean Theorem checks out. x + 10 = the length of the longer leg and x + 10 State: The length of the shorter leg is 30 miles. (Remember that 2x – 10 = the length of the hypotenuse. is all we were asked for in this problem.)

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The Pythagorean Theorem

Example continued 2.) Translate By the Pythagorean Theorem, (leg a)2 + (leg b)2 = (hypotenuse) 2 x2 + ( x + 10) 2 = (2 x – 10) 2 3.) Solve x2 + ( x + 10) 2 = (2 x – 10) 2 x2 + x2 + 20 x + 100 = 4 x2 – 40 x + 100 (multiply the binomials) 2x2 + 20 x + 100 = 4 x2 – 40 x + 100 (simplify left side) 0 = 2 x2 – 60 x (subtract 2 x2 + 20 x + 100 from both sides) 0 = 2 x(x – 30) (factor right side) x = 0 or x = 30 (set each factor = 0 and solve) Continued

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