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EXAMPLE 3 Standardized Test Practice SOLUTION 8x 3 y 2x y 2 7x4y37x4y3 4y4y 56x 7 y 4 8xy 3 = Multiply numerators and denominators. 8 7 x x 6 y 3 y 8 x y 3 = Factor and divide out common factors. 7x6y7x6y = Simplified form The correct answer is B. ANSWER

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EXAMPLE 4 Multiply rational expressions Multiply: x 2 + x – 20 3x3x 3x –3x 2 x2 + 4x – 5 x 2 + x – 20 3x3x 3x –3x 2 x2 + 4x – 5 3x(1– x) (x –1)(x +5) = (x + 5)(x – 4) 3x3x Factor numerators and denominators. 3x(1– x)(x + 5)(x – 4) = (x –1)(x + 5)(3x) Multiply numerators and denominators. 3x(–1)(x – 1)(x + 5)(x – 4) = (x – 1)(x + 5)(3x) Rewrite 1– x as (– 1)(x – 1). 3x(–1)(x – 1)(x + 5)(x – 4) = (x – 1)(x + 5)(3x) Divide out common factors. SOLUTION

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EXAMPLE 4 Multiply rational expressions = (–1)(x – 4) Simplify. = –x + 4 Multiply. ANSWER –x + 4

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EXAMPLE 5 Multiply a rational expression by a polynomial Multiply: x + 2 x 3 – 27 (x 2 + 3x + 9) x + 2 x 3 – 27 (x 2 + 3x + 9) Write polynomial as a rational expression. = x + 2 x 3 – 27 x 2 + 3x + 9 1 (x + 2)(x 2 + 3x + 9) (x – 3)(x 2 + 3x + 9) = Factor denominator. (x + 2)(x 2 + 3x + 9) (x – 3)(x 2 + 3x + 9) = Divide out common factors. = x + 2 x – 3 Simplified form SOLUTION ANSWER x + 2 x – 3

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GUIDED PRACTICE for Examples 3, 4 and 5 Multiply the expressions. Simplify the result. 3x 5 y 2 8xy 6xy 2 9x3y9x3y 8.8. 3x 5 y 2 2xy 6xy 2 9x3y9x3y 18x 6 y 4 72x 4 y 2 = Multiply numerators and denominators. Factor and divide out common factors. Simplified form 18 x 4 y 2 x 2 y 2 18 4 x 4 y 2 = = x2y2x2y2 4 SOLUTION

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GUIDED PRACTICE for Examples 3, 4 and 5 9. x + 3 2x22x2 2x 2 – 10x x 2 – 25 Factor numerators and denominators. 2x(x –5) (x + 3) = (x –5)(x + 5)2x (x) Multiply numerators and denominators. x + 3 = x(x + 5) Divide out common factors. x + 3 2x22x2 2x 2 – 10x x 2 – 25 2x(x –5) (x –5)(x +5) = x + 3 2x2x (x)(x) = 2x(x –5) (x + 3) (x –5)(x + 5)2x (x) Simplified form SOLUTION

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GUIDED PRACTICE for Examples 3, 4 and 5 10. Factor denominators. Multiply numerators and denominators. x + 5 = x – 1 Divide out common factors. x 2 +x + 1 x + 5 x 3 – 1 Simplified form x 2 +x + 1 x + 5 x 3 – 1 = x 2 +x + 1 x + 5 (x – 1) (x 2 +x + 1) 1 = (x + 5) (x 2 +x + 1) (x – 1) (x 2 +x + 1) (x + 5) (x 2 +x + 1) (x – 1) (x 2 +x + 1) = SOLUTION

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