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1 Topic Factoring Quadratics ax ² + bx + c Factoring Quadratics ax ² + bx + c

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2 Lesson California Standard: 11.0 Students apply basic factoring techniques to second- and simple third-degree polynomials. These techniques include finding a common factor for all terms in a polynomial, recognizing the difference of two squares, and recognizing perfect squares of binomials. What it means for you: Youll learn how to factor more complicated quadratic expressions. Factoring Quadratics ax ² + bx + c Topic Key words: quadratic polynomial factor trial and error

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3 Lesson Topic Factoring Quadratics ax ² + bx + c The method in Topic for factoring a quadratic expression only works if the x 2 -term has a coefficient of 1. Its a little more complicated when the x 2 -coefficient ( a ) isnt 1 but only a little.

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4 Lesson You Can Take Out a Common Factor from Each Term Topic Factoring Quadratics ax ² + bx + c If you see a common factor in each term, such as a number or a variable, take it out first. ax 2 + abx + ac a ( x 2 + bx + c )

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5 Example 1 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 3 x x x x + 18 = 3( x x + 6) The expression in parentheses can be factored using the method in Topic 6.6.1: Take out the common factor = 3( x + 2)( x + 3) But if the expression in parentheses still has a 1, then the expression will need to be factored using the second method, shown in Example 2.

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6 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each expression completely x x y 2 – 12 y – t 2 – 22 t r 2 – x x p p m m x x –2 k 2 – 20 k – –3 m 2 – 30 m – 72 3( x + 1)( x + 4) 4( y + 4)( y – 7) 2( t – 5)( t – 6) 3( r – 5)( r + 5) 4( x + 4)( x + 4) 7( p + 1)( p + 9) 5( m + 1)( m + 3) 6( x + 2)( x + 5) –2( k + 2)( k + 8) –3( m + 4)( m + 6)

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7 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each expression completely x – x –3 x 2 – 84 x – x 2 y – 33 xy – 126 y x x x – 25 x n 2 y ny – 5600 y a 2 – 8 x 2 a a – 80 – a x 2 m x 2 m – 102 x a 2 b 2 x a 2 bx + 63 a 2 –1( x + 1)( x – 2) –3( x + 3)( x + 25) 3 y ( x + 3)( x – 14) 10( x + 4)( x + 25) –25( x – 4)( x + 1) 100 y ( n – 7)( n + 8) 8 a 2 (2 – x )(2 + x ) –1( a – 5)( a – 16) 2 x 2 ( m + 17)( m – 3) 3 a 2 ( bx + 3)( bx + 7)

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8 Lesson You Can Factor ax 2 + bx + c by Trial and Error Topic Factoring Quadratics ax ² + bx + c If you cant see a common factor, then you need to get ax 2 + bx + c into the form ( a 1 x + c 1 )( a 2 x + c 2 ), where a 1 and a 2 are factors of a, and c 1 and c 2 are factors of c. ax 2 + bx + c = ( a 1 x + c 1 )( a 2 x + c 2 ) a 1 c 2 + a 2 c 1 = b a = a 1 a 2 c = c 1 c 2

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9 Lesson Topic Factoring Quadratics ax ² + bx + c Note that if a number is positive then its two factors will be either both positive or both negative. If a number is negative, then its two factors will have different signs one positive and one negative. These facts give important clues about the signs of c 1 and c 2.

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10 Example 2 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 3 x x + 6. Write down pairs of factors of a = 3: a 1 = 1 and a 2 = 3 Write down pairs of factors of c = 6: c 1 = 1 and c 2 = 6, c 1 = 2 and c 2 = 3 Now find the combination of factors that gives a 1 c 2 + a 2 c 1 = 11. Put the x -terms into parentheses first, with the pair of coefficients 3 and 1: (3 x )( x ). Now try all the pairs of c 1 and c 2 in the parentheses and find the possible values of a 1 c 2 + a 2 c 1 (and a 1 c 2 – a 2 c 1 )… Solution continues… 3 x x + 6

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11 Solution continues… Example 2 Topic Factoring Quadratics ax ² + bx + c (3 x 1)( x 6) multiplies to give 18 x and x, which add/subtract to give 19 x or 17 x. (3 x 6)( x 1) multiplies to give 3 x and 6 x, which add/subtract to give 9 x or 3 x. (3 x 2)( x 3) multiplies to give 9 x and 2 x, which add/subtract to give 11 x or 7 x. (3 x 3)( x 2) multiplies to give 6 x and 3 x, which add/subtract to give 9 x or 3 x. So (3 x 2)( x 3) is the combination that gives 11 x (so b = 11). Solution (continued) Factor 3 x x + 6.

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12 Example 2 Topic Solution (continued) Factoring Quadratics ax ² + bx + c Factor 3 x x + 6. Now fill in the + / – signs. Both c 1 and c 2 are positive (since c = c 1 c 2 and b = a 1 c 2 + a 2 c 1 are positive), so the final factors are (3 x + 2)( x + 3). Check by expanding the parentheses to make sure they give the original equation: (3 x + 2)( x + 3) = 3 x x + 2 x + 6 = 3 x x + 6 Thats what you started with, so (3 x + 2)( x + 3) is the correct factorization.

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13 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x a a y y + 3 Possible combinations: (2 x 1)( x 2) multiply to 4 x and x, which add/subtract to 5 x or 3 x (2 x 2)( x 1) multiply to 2 x and 2 x, which add/subtract to 4 x or 0 x So 2 x x + 2 = (2 x + 1)( x + 2) Possible combinations: (2 y 1)( y 3) multiply to 6 y and y, which add/subtract to 7 y or 5 y (2 y 3)( y 1) multiply to 2 y and 3 y, which add/subtract to 5 y or y So 2 y y + 3 = (2 y + 1)( y + 3) Possible combinations: (2 a 1)( a 11) multiply to 22 a and 1 a, which add/subtract to 23 a or 21 a (2 a 11)( a 1) multiply to 2 a and 11 a, which add/subtract to 13 a or 9 a So 2 a a + 11 = (2 a + 11)( a + 1)

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14 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x x x x x x x + 12 (2 x 7)(2 x 7) multiply to 14 x and 14 x, which add/subtract to 28 x or 0 x So 4 x x + 49 = (2 x + 7)(2 x + 7) (2 x 3)(2 x 3) multiply to 6 x and 6 x, which add/subtract to 12 x or 0 x So 4 x x + 9 = (2 x + 3)(2 x + 3) (2 x 7)(3 x 1) multiply to 2 x and 21 x, which add/subtract to 23 x or 19 x So 6 x x + 7 = (2 x + 7)(3 x + 1) (3 x 4)( x 3) multiply to 9 x and 4 x, which add/subtract to 13 x or 5 x So 3 x x + 12 = (3 x + 4)( x + 3)

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15 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x a a p p a a + 11 (2 x 1)( x 5) multiply to 10 x and x, which add/subtract to 11 x or 9 x So 2 x x + 5 = (2 x + 1)( x + 5) (2 a 1)(2 a 7) multiply to 14 a and 2 a, which add/subtract to 16 a or 12 a So 4 a a + 7 = (2 a + 1)(2 a + 7) Take out a common factor of 2: 2( p p + 6), then factor the rest. ( p 6)( p 1) multiply to p and 6 p, which add/subtract to 7 p or 5 p So 2 p p + 12 = 2( p + 6)( p + 1) (4 a 1)(2 a 11) multiply to 44 a and 2 a, which add/subtract to 46 a or 42 a So 8 a a + 11 = (4 a + 1)(2 a + 11)

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16 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial g g x x a a b b + 55 (3 x 2)(3 x 2) multiply to 6 x and 6 x, which add/subtract to 12 x or 0 x So 9 x x + 4 = (3 x + 2) 2 Take out a common factor of 3: 3( g g + 72), then factor the rest. ( g 8)( g 9) multiply to 9 g and 8 g, which add/subtract to 17 g or g So 3 g p = 3( g + 8)( g + 9) (2 a 9)(2 a 9) multiply to 18 a and 18 a, which add/subtract to 36 a or 0 a So 4 a a + 81 = (2 a + 9) 2 (2 b 5)(2 b 11) multiply to 22 b and 10 b, which add/subtract to 32 b or 12 b So 4 b b + 55 = (2 b + 5)(2 b + 11)

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17 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x a a t t + 20 (2 a 3)(5 a 4) multiply to 8 a and 15 a, which add/subtract to 23 a or 7 a So 10 a a + 12 = (2 a + 3)(5 a + 4) (3 x 4)( x 6) multiply to 18 x and 4 x, which add/subtract to 22 x or 14 x So 3 x x + 24 = (3 x + 4)( x + 6) (2 t 5)(3 t 4) multiply to 8 t and 15 t, which add/subtract to 23 t or 7 t So 6 t t + 20 = (2 t + 5)(3 t + 4)

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18 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x b b + 36 Take out a common factor of 2: 2(2 x x + 8), then factor the rest. (2 x 1)( x 8) multiply to 16 x and x, which add/subtract to 17 x or 15 x So 4 x x + 16 = 2(2 x + 1)( x + 8) Take out a common factor of 3: 3(5 b b + 12), then factor the rest. (5 b 2)( b 6) multiply to 30 b and 2 b, which add/subtract to 32 b or 28 b So 15 b b + 36 = 3(5 b + 2)( b + 6)

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19 Example 3 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 6 x x – 6. Write down pairs of factors of a = 6: a 1 = 6 and a 2 = 1, a 1 = 2 and a 2 = 3, Write down pairs of factors of c = –6 (ignoring the minus sign for now): c 1 = 1 and c 2 = 6, c 1 = 2 and c 2 = 3 Put the x -terms into parentheses first, with the first pair of possible values for a 1 and a 2, 6 and 1: (6 x )(1 x ). Now try all the pairs of c 1 and c 2 in the parentheses and find the possible values of a 1 c 2 + a 2 c 1 and a 1 c 2 – a 2 c 1 as before… Solution continues…

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20 Solution continues… Example 3 Topic Solution (continued) Factoring Quadratics ax ² + bx + c Factor 6 x x – 6. (6 x 1)( x 6) multiplies to give 36 x and x, which add/subtract to give 37 x or 35 x. (6 x 6)( x 1) multiplies to give 6 x and 6 x, which add/subtract to give 12 x or 0 x. (6 x 2)( x 3) multiplies to give 18 x and 2 x, which add/subtract to give 20 x or 16 x. (6 x 3)( x 2) multiplies to give 12 x and 3 x, which add/subtract to give 15 x or 9 x. None of these combinations works, so try again with (2 x )(3 x )…

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21 Solution continues… Example 3 Topic Solution (continued) Factoring Quadratics ax ² + bx + c Factor 6 x x – 6. (2 x 1)(3 x 6) multiplies to give 12 x and 3 x, which add/subtract to give 15 x or 9 x. (2 x 6)(3 x 1) multiplies to give 2 x and 18 x, which add/subtract to give 20 x or 16 x. (2 x 3)(3 x 2) multiplies to give 4 x and 9 x, which add/subtract to give 13 x or 5 x. 5 x is what you want, so you can stop there so (2 x 3)(3 x 2) is the right combination.

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22 Example 3 Topic Solution (continued) Factoring Quadratics ax ² + bx + c Factor 6 x x – 6. Now fill in the + / – signs to get b = +5. One of c 1 and c 2 must be negative, to give c = –6, so the final factors are either (2 x + 3)(3 x – 2) or (2 x – 3)(3 x + 2). The x -term of (2 x + 3)(3 x – 2) will be 9 x – 4 x = 5 x, whereas the x -term for (2 x – 3)(3 x + 2) will be 4 x – 9 x = –5 x. So the correct factorization is (2 x + 3)(3 x – 2).

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23 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x x – y 2 – y – k k – 6 (2 x 1)( x 2) multiply to 4 x and x, which add/subtract to 5 x or 3 x (2 x 2)( x 1) multiply to 2 x and 2 x, which add/subtract to 4 x or 0 x So 2 x x – 2 = (2 x – 1)( x + 2) (3 y 1)( y 2) multiply to 6 y and y, which add/subtract to 7 y or 5 y (3 y 2)( y 1) multiply to 3 y and 2 y, which add/subtract to 5 y or y So 3 y 2 – y – 2 = (3 y + 2)( y – 1) (5 k 2)( k 3) multiply to 15 k and 2 k, which add/subtract to 17 k or 13 k (5 k 3)( k 2) multiply to 10 k and 3 k, which add/subtract to 13 k or 7 k So 5 k k – 6 = (5 k – 2)( k + 3)

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24 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x 2 – x – b 2 – 23 b x 2 – 5 x + 2 (3 x 5)( x 2) multiply to 6 x and 5 x, which add/subtract to 11 x or x (3 x 2)( x 5) multiply to 15 x and 2 x, which add/subtract to 17 x or 13 x So 3 x 2 – x – 10 = (3 x + 5)( x – 2) (3 b 1)(2 b 7) multiply to 21 b and 2 b, which add/subtract to 23 b or 19 b (3 b 7)(2 b 1) multiply to 3 b and 14 b, which add/subtract to 17 b or 11 b So 6 b 2 – 23 b + 7 = (3 b – 1)(2 b – 7) (2 x 2)( x 1) multiply to 2 x and 2 x, which add/subtract to 4 x or 0 x (2 x 1)( x 2) multiply to 4 x and x, which add/subtract to 5 x or 3 x So 2 x 2 – 5 x + 2 = (2 x – 1)( x – 2)

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25 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial k 2 – 2 k – v 2 – 16 v x – 2 x 2 (3 k 1)( k 1) multiply to 3 k and k, which add/subtract to 4 k or 2 k So 3 k 2 – 2 k – 1 = (3 k + 1)( k – 1) (3 v 5)( v 1) multiply to 3 v and 5 v, which add/subtract to 8 v or 2 v (3 v 1)( v 5) multiply to 15 v and v, which add/subtract to 16 v or 10 v So 3 v 2 – 16 v + 5 = (3 v – 1)( v – 5) Take out a factor of –1: –(2 x 2 – 5 x – 18), then factor the rest. (2 x 9)( x 2) multiply to 4 x and 9 x, which add/subtract to 13 x or 5 x (2 x 2)( x 9) multiply to 18 x and 2 x, which add/subtract to 20 x or 16 x So x – 2 x 2 = –(2 x – 9)( x + 2)

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26 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x – 2 x x x a 2 – 26 a – 8 Take out a factor of –1: –(2 x 2 – x – 28), then factor the rest. (2 x 7)( x 4) multiply to 8 x and 7 x, which add/subtract to 15 x or x (2 x 4)( x 7) multiply to 14 x and 4 x, which add/subtract to 18 x or 10 x So 28 + x – 2 x 2 = –(2 x + 7)( x – 4) (3 x 2)(3 x 2) multiply to 6 x and 6 x, which add/subtract to 12 x or 0 x So 9 x x + 4 = (3 x + 2)(3 x + 2) (7 a 2)( a 4) multiply to 28 a and 2 a, which add/subtract to 30 a or 26 a (7 a 4)( a 2) multiply to 14 a and 4 a, which add/subtract to 18 a or 10 a So 7 a 2 – 26 a – 8 = (7 a + 2)( a – 4)

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27 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x 2 – 7 x – x x – x x – 20 (4 x 1)(3 x 2) multiply to 8 x and 3 x, which add/subtract to 11 x or 5 x (4 x 2)(3 x 1) multiply to 4 x and 6 x, which add/subtract to 10 x or 2 x So 12 x x – 2 = (4 x – 1)(3 x + 2) Take out a factor of 2: 2(3 x 2 + x – 10), then factor the rest. (3 x 2)( x 5) multiply to 15 x and 2 x, which add/subtract to 17 x or 13 x (3 x 5)( x 2) multiply to 6 x and 5 x, which add/subtract to 11 x or x So 6 x x – 20 = 2(3 x – 5)( x + 2) (3 x 3)( x 2) multiply to 6 x and 3 x, which add/subtract to 9 x or 3 x (3 x 2)( x 3) multiply to 9 x and 2 x, which add/subtract to 11 x or 7 x So 3 x 2 – 7 x – 6 = (3 x + 2)( x – 3)

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28 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each polynomial x 2 + x – y 2 + y – m 2 – 3 m – 20 (2 x 1)(9 x 4) multiply to 8 x and 9 x, which add/subtract to 17 x or x (2 x 4)(9 x 1) multiply to 2 x and 36 x, which add/subtract to 38 x or 34 x So 18 x 2 + x – 4 = (2 x + 1)(9 x – 4) (3 y 3)(2 y 4) multiply to 12 y and 6 y, which add/subtract to 18 y or 6 y (3 y 4)(2 y 3) multiply to 9 y and 8 y, which add/subtract to 17 y or y So 6 y 2 + y – 12 = (3 y – 4)(2 y + 3) (3 m 4)(3 m 5) multiply to 15 m and 12 m, which add/subtract to 27 m or 3 m So 9 m 2 – 3 m – 20 = (3 m + 4)(3 m – 5)

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29 Lesson Use the + and – Signs in the Quadratic to Work Faster Topic Factoring Quadratics ax ² + bx + c Looking carefully at the signs in the quadratic that you are factoring can help to narrow down the choices for a 1, a 2, c 1, and c 2. The next three Examples show you how.

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30 Example 4 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 3 x x + 6. Here c = 6, which is positive so its factors c 1 and c 2 are either both positive or both negative. But since b = 11 is positive, you can tell that c 1 and c 2 must be positive (so that a 1 c 2 + a 2 c 1 is positive).

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31 Example 5 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 3 x 2 – 11 x + 6. Here c is also positive, so c 1 and c 2 are either both positive or both negative. But since b = –11 is negative, you can tell that c 1 and c 2 must be negative (so that a 1 c 2 + a 2 c 1 is negative).

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32 Example 6 Topic Solution Solution follows… Factoring Quadratics ax ² + bx + c Factor 6 x x – 6. In this expression, c is negative, so one of c 1 and c 2 must be positive and the other must be negative. So instead of looking at both the sums and differences of all the different combinations a 1 c 2 and a 2 c 1, you only need to look at the differences.

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33 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each expression n 2 + n – x 2 – 5 x – a a + 1 c is negative, so one of c 1 and c 2 must be positive and the other negative (2 n 3)( n 1) multiply to 2 n and 3 n. To add to get n, 3 n must be positive and 2 n negative. So 2 n 2 + n – 3 = (2 n + 3)( n – 1) c is negative, so one of c 1 and c 2 must be positive and the other negative (2 x 1)( x 3) multiply to 6 x and x. To add to get –5 x, x must be positive and 6 x negative. So 2 x 2 – 5 x – 3 = (2 x + 1)( x – 3) b and c are positive, so c 1 and c 2 must both be positive. (4 a 1)( a 1) multiply to 4 a and a and add to 5 a (2 a 1)(2 a 1) multiply to 2 a and 2 a and add to 4 a So 4 a a + 1 = (2 a + 1)(2 a + 1)

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34 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each expression x 2 – 4 x y y t 2 + t – 3 b and c are positive, so c 1 and c 2 must both be positive. (3 y 1)(3 y 1) multiply to 3 y and 3 y and add to 6 y So 9 y y + 1 = (3 y + 1) 2 c is negative, so one of c 1 and c 2 must be positive and the other negative (4 t 3)( t 1) multiply to 4 t and 3 t. To add to get t, 4 t must be positive and 3 t negative. So 4 t 2 + t – 3 = (4 t – 3)( t + 1) b is negative and c is positive, so c 1 and c 2 must both be negative. (3 x 1)( x 1) multiply to 3 x and x and add to 4 x So 3 x 2 – 4 x + 1 = (3 x – 1)( x – 1)

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35 Lesson Guided Practice Topic Solution follows… Factoring Quadratics ax ² + bx + c Factor each expression x 2 – x – x 2 – 6 x t 2 + t – b b + 21 c is negative, so one of c 1 and c 2 must be positive and the other negative (5 x 9)( x 2) multiply to 10 x and 9 x. To add to get – x, 9 x must be positive and 10 x negative. So 5 x 2 – x – 18 = (5 x + 9)( x – 2) b is negative and c is positive, so c 1 and c 2 must both be negative. (3 x 1)(3 x 1) multiply to 3 x and 3 x and add to 6 x. So 9 x 2 – 6 x + 1 = (3 x – 1) 2 c is negative, so one of c 1 and c 2 must be positive and the other negative (3 t 1)(2 t 1) multiply to 3 t and 2 t. To add to get t, 3 t must be positive and 2 t negative. So 6 t 2 + t – 1 = (3 t – 1)(2 t + 1) b and c are positive, so c 1 and c 2 must both be positive. ( b 7)( b 3) multiply to 3 b and 7 b and add to 10 b. So 9 y y + 1 = (3 y + 1) 2

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36 Independent Practice Solution follows… Topic Factoring Quadratics ax ² + bx + c Factor each polynomial k 2 – 7 k k 2 – 15 k t 2 – 11 t – 7 k – 2 k h – 3 h x 2 – 14 x – x – 6 x x x 2 – 3 (5 k – 2)( k – 1) (4 k – 3)( k – 3) (4 t – 1)(3 t – 2) –(2 k + 9)( k – 1) –(3 h + 5)( h – 2) (5 x + 2)(3 x – 4) –(3 x + 2)(2 x – 3) (3 x 2 – 1)( x 2 + 3)

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37 Independent Practice Solution follows… Topic Factoring Quadratics ax ² + bx + c 9. If the area of a rectangle is (6 x x + 14) square units and the length is (3 x + 2) units, find the width w in terms of x. 10. The area of a parallelogram is (12 x x – 10) cm 2, where x is positive. If the area is given by the formula Area = base × height, find the base length and the height of the parallelogram, given that they are both linear factors of the area. (2 x + 7) units (4 x + 5) cm, (3 x – 2) cm

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38 Independent Practice Solution follows… Topic Factoring Quadratics ax ² + bx + c Factor each of these expressions completely. 11. x 2 ab – 3 abx – 18 ab 12. ( d + 2) x 2 – 7 x ( d + 2) – 18( d + 2) 13. ( x + 1) m 2 – 2 m ( x + 1) + ( x + 1) 14. –2 dk 2 – 14 dk + 36 d ab ( x + 3)( x – 6) ( d + 2)( x – 9)( x + 2) ( x + 1)( m – 1) 2 –2 d ( k + 9)( k – 2)

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39 Independent Practice Solution follows… Topic Factoring Quadratics ax ² + bx + c 15. Five identical rectangular floor tiles have a total area of (15 x x – 40) m 2. Find the dimensions of each floor tile, if the length of each side can be written in the form ax + b, where a and b are integers. (3 x – 4) m by ( x + 2) m.

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40 Topic Round Up Factoring Quadratics ax ² + bx + c Now you can factor lots of different types of polynomials. In the next Section youll learn about another type quadratic expressions containing two different variables.

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