# Factoring Word Problems

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## Factoring Word Problems

 Okay, I know I'm posting here a lot, however I have a test tomorrow and wanna be prepared haha. 1. Find Two consecutive odd integers whose product is 1 less than their sum. 3. The length of a rectangle is 3 feet more than twice the width. If the area of the rectangle is 20 square feet, what are the dimensions of the rectangle?
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## Re: Factoring Word Problems

 1. Find Two consecutive odd integers whose product is 1 less than their sum. x = 1st odd integer x + 2 = 2nd odd integer x(x + 2) = x + x + 2 - 1 {product is one less than sum} x² + 2x = 2x + 1 {used distributive property and combined like terms} x² = 1 {subtracted 2x from both sides} x = 1 or x = -1 {took square root of both sides} x + 2 = 3 or x + 2 = 1 {substituted 1 and -1 into x + 2} the two consecutive odd integers are 1 and 3.....also -1 and 13. The length of a rectangle is 3 feet more than twice the width. If the area of the rectangle is 20 square feet, what are the dimensions of the rectangle? x = width 2x + 3 = length {length is 3 feet more than twice width} Area of a rectangle is length x width x(2x + 3) = 20 {area is length x width} 2x² + 3x = 20 {used distributive property} 2x² + 3x - 20 = 0 {subtracted 20 from both sides} (2x - 5)(x + 4) = 0 {factored into two binomials} 2x - 5 = 0 or x + 4 = 0 {set each factor equal to 0} 2x = 5 or x = -4 {added 5 and subtracted 4, respectively} x = 5/2 or x = -4 {divided left equation by 2} width cannot be negativex = 5/2 2x + 3 = 8 {substituted 5/2, in for x, into 2x + 3} width = 5/2 ft length = 8 ft- Algebra House