EXAMPLE 1 Simplify a rational expression 2 2

EXAMPLE 1 Simplify a rational expression 2 – 2 x – 15 x Simplify : x 2 – 9 SOLUTION x 2 – 2 x – 15 (x +3)(x – 5) = (x +3)(x – 3) x 2 – 9 ANSWER Factor numerator and denominator. (x +3)(x – 5) = (x +3)(x – 3) Divide out common factor. x– 5 = x– 3 Simplified form x– 5 x– 3

EXAMPLE 2 Solve a multi-step problem Packaging A company makes a tin to hold flavored popcorn. The tin is a rectangular prism with a square base. The company is designing a new tin with the same base and twice the height of the old tin. • Find the surface area and volume of each tin. • Calculate the ratio of surface area to volume for each tin. • What do the ratios tell you about the efficiencies of the two tins?

EXAMPLE 2 Solve a multi-step problem SOLUTION Old tin STEP 1 S = 2 s 2 + 4 sh New tin S = 2 s 2 + 4 s(2 h) Find surface area, S. = 2 s 2 + 8 sh V = s 2(2 h) = 2 s 2 h STEP 2 S = 2 s 2 + 4 sh S = 2 s 2 + 8 sh V s 2 h V 2 s 2 h 2 s(s + 4 h) s(22 + 4 h) = = s(sh) 2 s(sh) = 2 s + 4 h sh = s + 4 h sh Find volume, V. Write ratio of S to V. Divide out common factor. Simplified form

EXAMPLE 2 Solve a multi-step problem STEP 3 2 s + 4 h > s + 4 h sh sh Because the left side of the inequality has a greater numerator than the right side and both have the same (positive) denominator. The ratio of surface area to volume is greater for the old tin than for the new tin. So, the old tin is less efficient than the new tin.

GUIDED PRACTICE for Examples 1 and 2 Simplify the expression, if possible. 1. 2(x + 1)(x + 3) SOLUTION 2(x + 1) 2(x +1) = (x +1)(x + 3) (x + 1)(x + 3) 2 = x+3 ANSWER 2 x+3 Divide out common factor. Simplified form

GUIDED PRACTICE for Examples 1 and 2 2. 40 x + 20 10 x + 30 SOLUTION 20(2 x +1) 40 x + 20 = 10(x + 3) 10 x + 30 20(2 x +1) = 10(x + 3) = ANSWER 2(2 x +1) x+3 Factor numerator and denominator. Divide out common factor. Simplified form

GUIDED PRACTICE 3. for Examples 1 and 2 4 x(x + 2) SOLUTION 4 x(x + 2) ANSWER 4 x(x + 2) Simplified form

GUIDED PRACTICE 4. for Examples 1 and 2 x+4 x 2 – 16 SOLUTION (x + 4) x+4 = (x + 4)(x – 4) x 2 – 16 = (x + 4)(x – 4) 1 = x– 4 ANSWER 1 x– 4 Factor numerator and denominator. Divide out common factor. Simplified form

GUIDED PRACTICE 5. for Examples 1 and 2 x 2 – 2 x – 3 x 2 – x – 6 SOLUTION x 2 – 2 x – 3 (x – 3)(x + 1) x 2 – x – 6 = (x – 3)(x + 2) ANSWER Factor numerator and denominator. (x – 3)(x + 1) = (x – 3)(x + 2) Divide out common factor. x+1 = x+2 Simplified form x+1 x+2

GUIDED PRACTICE 6. for Examples 1 and 2 2 x 2 + 10 x 3 x 2 + 16 x + 5 SOLUTION 2 x 2 + 10 x 2 x(x + 5) = 3 x 2 + 16 x + 5 (3 x + 1)(x + 5) = 2 x(x + 5) (3 x + 1)(x + 5) 2 x = 3 x + 1 ANSWER 2 x 3 x + 1 Factor numerator and denominator. Divide out common factor. Simplified form

GUIDED PRACTICE 7. for Examples 1 and 2 What If? In Example 2, suppose the new popcorn tin is the same height as the old tin but has a base with sides twice as long. What is the ratio of surface area to volume for this tin? SOLUTION New tin Old tin STEP 1 S = 2 s 2 + 4 sh S = 2 (2 s)2 + 4(2 s)h Find surface area, S. = 8 s 2 + 8 sh V = s 2 h V = (2 s)2 h = 4 s 2 h Find volume, V.

GUIDED PRACTICE for Examples 1 and 2 STEP 2 S = 2 s 2 + 4 sh S = 8 s 2 + 8 sh Write ratio of S to V. s 2 h V 2 s 2 h V 4 s(2 s + 2 h) Divide out common factor. s(2 s + 4 h) = = s(sh) 4 s(sh) 2 s + 4 h = =sh ANSWER 2 s + 4 h =sh 2 s + 4 h = =sh Simplified form