Factor completely. 1. x2 – x – 12 ANSWER (x – 4)(x + 3) 2. 2x2 – 5x – 3 ANSWER (x – 3)(2x + 1)

Factor completely. 3. Use synthetic division to divide 2x3 – 3x2 – 18x – 8 by x – 4. ANSWER 2x2 + 5x + 2 4. The volume of a box is modeled by f (x) = x (x – 1)(x – 2), where x is the length in meters. What is the volume when the length is 3 meters. ANSWER 6m3

EXAMPLE 1 List possible rational zeros List the possible rational zeros of f using the rational zero theorem. a. f (x) = x3 + 2x2 – 11x + 12 Factors of the constant term: + 1, + 2, + 3, + 4, + 6, + 12 Factors of the leading coefficient: + 1 Possible rational zeros: + , + , + , + , + , + 1 2 3 4 6 12 Simplified list of possible zeros: + 1, + 2, + 3, + 4, + 6, + 12

EXAMPLE 1 List possible rational zeros b. f (x) = 4x4 – x3 – 3x2 + 9x – 10 Factors of the constant term: + 1, + 2, + 5, + 10 Factors of the leading coefficient: + 1, + 2, + 4 + , + , + , + , + , + , + , + , + , + , + Possible rational zeros: 1 2 5 10 4 + 10 Simplified list of possible zeros: + 1, + 2, + 5, + 10, + , + , + 5 2 1 4 +

GUIDED PRACTICE for Example 1 List the possible rational zeros of f using the rational zero theorem. 1. f (x) = x3 + 9x2 + 23x + 15 SOLUTION Factors of the constant term: + 1, + 3, + 5, + 15 Factors of the leading coefficient: + 1 Possible rational zeros: + , + , + , + 1 3 5 15 Simplified list of possible zeros: + 1, + 3, + 5 + 15

GUIDED PRACTICE for Example 1 2. f (x) =2x3 + 3x2 – 11x – 6 SOLUTION Factors of the constant term: + 1, + 2, + 3 Factors of the leading coefficient: + 1, + 2, + 4 Possible rational zeros: + , + , + , + , + , + 1 2 3 6 1 Simplified list of possible zeros: + 1, + 2, + 3, + 6 + 3 2 +

EXAMPLE 2 Find zeros when the leading coefficient is 1 Find all real zeros of f (x) = x3 – 8x2 +11x + 20. SOLUTION STEP 1 List the possible rational zeros. The leading coefficient is 1 and the constant term is 20. So, the possible rational zeros are: x = + , + , + , + , + , + 5 1 4 2 10 20

↑ ↑ EXAMPLE 2 Find zeros when the leading coefficient is 1 STEP 2 Test these zeros using synthetic division. 1 1 – 8 11 20 Test x =1: 1 – 7 4 1 – 7 4 24 1 is not a zero. ↑ Test x = –1: –1 1 –8 11 20 1 – 9 20 0 –1 9 20 –1 is a zero ↑

EXAMPLE 2 Find zeros when the leading coefficient is 1 Because –1 is a zero of f, you can write f (x) = (x + 1)(x2 – 9x + 20). STEP 3 Factor the trinomial in f (x) and use the factor theorem. f (x) = (x + 1) (x2 – 9x + 20) = (x + 1)(x – 4)(x – 5) The zeros of f are –1, 4, and 5. ANSWER

GUIDED PRACTICE for Example 2 Find all real zeros of the function. 3. f (x) = x3 – 4x2 – 15x + 18 SOLUTION Factors of the constant term: + 1, + 2, + 3, + 6, + 9 Factors of the leading coefficient: + 1 Possible rational zeros: + , + + 1 3 6 Simplified list of possible zeros: –3, 1 , 6

GUIDED PRACTICE for Example 2 4. f (x) + x3 – 8x2 + 5x+ 14 SOLUTION STEP 1 List the possible rational zeros. The leading coefficient is 1 and the constant term is 14. So, the possible rational zeros are: x = + , + , + 7 1 2

↑ ↑ GUIDED PRACTICE for Example 2 STEP 2 Test these zeros using synthetic division. 1 1 –8 5 14 Test x =1: 1 –7 –2 1 –7 –2 12 1 is not a zero. ↑ Test x = –1: –1 1 –8 5 14 1 –9 14 0 –1 9 14 –1 is a zero. ↑

GUIDED PRACTICE for Example 2 Because –1 is a zero of f, you can write f (x) = (x + 1)(x2 – 9x + 14). STEP 3 Factor the trinomial in f (x) and use the factor theorem. f (x) = (x + 1) (x2 – 9x + 14) = (x + 1)(x + 4)(x – 7) The zeros of f are –1, 2, and 7. ANSWER

EXAMPLE 3 Find zeros when the leading coefficient is not 1 Find all real zeros of f (x) =10x4 – 11x3 – 42x2 + 7x + 12. SOLUTION STEP 1 6 3 2 1 12 + , + , + , + , + , + , + , + , + 4 5 10 + , + , + , + , + , + , , + List the possible rational zeros of f :

EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 2 x = – , x = , x = , and x = 3 2 12 5 1 are reasonable based on the graph shown at the right. Choose reasonable values from the list above to check using the graph of the function. For f , the values

↑ EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 3 Check the values using synthetic division until a zero is found. 2 3 – 10 –11 –42 7 12 –15 39 – 9 69 4 10 – 26 – 3 – 23 21 10 – 11 – 42 7 12 – 5 8 17 –12 10 – 16 – 34 24 0 2 1 1 is a zero. 2 – ↑

Find zeros when the leading coefficient is not 1 EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 4 Factor out a binomial using the result of the synthetic division. f (x) = x + (10x3 – 16x2 – 34x + 24) 1 2 Write as a product of factors. = x + (2)(5x3 – 8x2 – 17x + 12) 1 2 Factor 2 out of the second factor. = (2x +1)(5x3 – 8x2 – 17x +12) Multiply the first factor by 2.

EXAMPLE 3 Find zeros when the leading coefficient is not 1 Repeat the steps above for g (x) = 5x3 – 8x2 – 17x + 12. Any zero of g will also be a zero of f. The possible rational zeros of g are: x = + 1, + 2, + 3, + 4, + 6, + 12, + , + , + , + , + , + 1 5 2 3 4 6 12 STEP 5 The graph of g shows that may be a zero. Synthetic division shows that is a zero and 3 5 g (x) = x – (5x2 – 5x – 20) 3 5 = (5x – 3)(x2 – x – 4). f (x) = (2x + 1) g (x) It follows that: = (2x + 1)(5x – 3)(x2 – x – 4)

Find zeros when the leading coefficient is not 1 EXAMPLE 3 Find zeros when the leading coefficient is not 1 STEP 6 Find the remaining zeros of f by solving x2 – x – 4 = 0. Substitute 1 for a, 21 for b, and 24 for c in the quadratic formula. x = – (– 1) + √ (– 1)2 – 4(1)(– 4) 2(1) x = 1 + √17 2 Simplify. ANSWER 1 3 The real zeros of f are , , 1 + √17, and 1 – √17. – 2 5 2 2

GUIDED PRACTICE for Example 3 Find all real zeros of the function. 5. f (x) = 48x3+ 4x2 – 20x + 3 1 2 3 4 6 , , ANSWER

GUIDED PRACTICE for Example 3 6. f (x) = 2x4 + 5x3 – 18x2 – 19x + 42 3 2 2, , 1, +2 √2 ANSWER

EXAMPLE 4 Solve a multi-step problem ICE SCULPTURES Some ice sculptures are made by filling a mold with water and then freezing it. You are making such an ice sculpture for a school dance. It is to be shaped like a pyramid with a height that is 1 foot greater than the length of each side of its square base. The volume of the ice sculpture is 4 cubic feet. What are the dimensions of the mold?

Solve a multi-step problem EXAMPLE 4 Solve a multi-step problem SOLUTION STEP 1 Write an equation for the volume of the ice sculpture. 4 = x2(x + 1) 1 3 Write equation. 12 = x3 + x2 Multiply each side by 3 and simplify. 0 = x3 + x2 – 12 Subtract 12 from each side.

EXAMPLE 4 Solve a multi-step problem STEP 2 + , + , + , + ,+ , + 1 2 3 4 6 12 List the possible rational solutions: STEP 3 Test possible solutions. Only positive x-values make sense. 1 1 1 0 – 12 1 2 2 1 2 2 – 10

↑ EXAMPLE 4 Solve a multi-step problem 2 1 1 0 – 12 2 6 12 1 3 6 0 2 1 1 0 – 12 2 6 12 1 3 6 0 ↑ 2 is a solution. STEP 4 x2 + 3x + 6 = 0, are x = and can be discarded because they are imaginary numbers. 2 Check for other solutions. The other two solutions, which satisfy – 3 + i √ 15

EXAMPLE 4 Solve a multi-step problem The only reasonable solution is x = 2. The base of the mold is 2 feet by 2 feet. The height of the mold is 2 + 1 = 3 feet. ANSWER

GUIDED PRACTICE for Example 4 WHAT IF? In Example 4, suppose the base of the ice sculpture has sides that are 1 foot longer than the height. The volume of the ice sculpture is 6 cubic feet. What are the dimensions of the mold? 7. Base side: 3 ft, height: 2 ft ANSWER

Daily Homework Quiz 1. List the possible rational zeros of f(x) = x3 + 8x2 – x + 4. ANSWER + 1, + 2, + 4 Find all real zeros of the functions 2. f(x) = x3 – 3x2 – 6x + 8. ANSWER – 2, 1, 4

Daily Homework Quiz 3. f(x) = 2x3 – 3x2 – 17x + 30. ANSWER – 3, 2, 5 2 4. The volume V of a storage shed with a triangular roof can be modeled by V = x3 + x2(6 – x). If the volume of the shed is 80 cubic feet, find x. 1 2 ANSWER 4