2.6. Solve Factorable Polynomial Inequalities Algebraically. Solve Linear Inequalities. Solution

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1 .6 Solve Factorable Polynomial Inequalities Algebraically A rectangular in-ground swimming pool is to be installed. The engineer overseeing the construction project estimates that at least 148 m of earth and rocks needs to be excavated. What are the minimum dimensions of the excavation if the depth must be m more than one quarter of the width, and the length must be 1 m more than four times the width? The solution to this problem can be found by solving a cubic polynomial inequality algebraically. Example 1 Solve Linear Inequalities Solve each inequality. Show the solution on a number line. a) x 8 b) 4 x 1 Solution a) x 8 x 8 x Solving linear inequalities is similar to solving linear equations; however, when both sides of an inequality are multiplied or divided by a negative number, the inequality sign must be reversed. b) 4 x 1 x 1 4 x 16 x _ 16 x 8 Divide both sides by. Reverse the inequality MHR Advanced Functions Chapter

2 Example Solve Polynomial Inequalities Algebraically Solve each inequality. a) (x )(x ) b) x 6x 1x 16 Solution a) (x )(x ) Method 1: Consider All Cases (x )(x ) A product mn is positive when m and n are both positive or both negative Case 1 x x x x x _ x _ is included in the inequality x. So, the solution is x _. Case x x x x x _ x is included in the inequality x _. So, the solution is x. Combining the results of the two cases, the solution is x or x _. Method : Use Intervals (x )(x ) The roots of the equation (x )(x ) are x and x _. Use the roots to break the number line into three intervals. x x x 1 1 Test arbitrary values of x for each interval. For x, test x 4. (4 )[(4) ] 11 Since 11, x is a solution..6 Solve Factorable Polynomial Inequalities Algebraically MHR 1

3 For x _, test x. ( )[() ] 9 Since 9, x _ is not a solution. For x _, test x. ( )[() ] 5 Since 5, x _ is a solution. All of the information can be summarized in a table: Interval x x x _ x _ x _ Factor (x ) (x ) (x )(x ) The solution is x or x _. This can be shown on a number line. x x 1 1 b) x 6x 1x 16 Factor x 6x 1x 16 using the factor theorem. x 6x 1x 16 (x 4)(x )(x 1) So, the inequality becomes (x 4)(x )(x 1). You will obtain the same final result if you divide the inequality by and consider (x 4)(x )(x 1). Method 1: Consider All Cases (x 4)(x )(x 1) Since is a constant factor, it can be combined with (x 4) to form one factor. Thus, the three factors of (x 4)(x )(x 1) are (x 4), x, and x 1. A product abc is negative when all three factors, a, b, and c, are negative, or when two of the factors are positive and the third one is negative. There are four cases to consider. Case 1 (x 4) x x 1 x 4 x x 1 x 4 14 MHR Advanced Functions Chapter

4 The broken lines indicate that 4 x 1 is common to all three intervals The values of x that are common to all three inequalities are x 4 and x 1. So, 4 x 1 is a solution. Case (x 4) x x 1 x 4 x x 1 x There are no x-values that are common to all three inequalities. Case has no solution. Case (x 4) x x 1 x 4 x x 1 x There are no x-values that are common to all three inequalities. Case has no solution..6 Solve Factorable Polynomial Inequalities Algebraically MHR 15

5 Case 4 (x 4) x x 1 x 4 x x 1 x The broken line indicates that x is common to all three intervals. x is included the intervals x 4 and x 1. So, x is a solution. Combining the results of the four cases, the solution is 4 x 1 or x. Method : Use Intervals (x 4)(x )(x 1) The roots of (x 4)(x )(x 1) are x 4, x 1, and x. Use the roots to break the number line into four intervals. x 4 4 x 1 1 x x Test arbitrary values of x in each interval. For x 4, test x 5. (5 4)(5 )(5 1) 56 Since 56, x 4 is not a solution. For 4 x 1, test x. ( 4)( )( 1) Since, 4 x 1 is a solution. For 1 x, test x. ( 4)( )( 1) 16 Since 16, 1 x is not a solution. For x, test x. ( 4)( )( 1) 56 Since 56, x is a solution. 16 MHR Advanced Functions Chapter

6 Interval Factor x 4 x 4 4 x 1 x 1 1 x x x (x 4) (x ) (x 1)(x ) (x 4)(x ) (x 1) The solution is 4 x 1 or x. This can be shown on a number line. x 4 4 x 1 1 x x Example Solve a Problem Involving a Factorable Polynomial Inequality A rectangular in-ground pool is to be installed. The engineer overseeing the construction project estimates that at least 148 m of earth and rocks needs to be excavated. What are the minimum dimensions of the excavation if the depth must be m more than one quarter of the width, and the length must be 1 m more than four times the width? Representing Connecting Reasoning and Proving Problem Solving Communicating Selecting Tools Reflecting Solution From the given information, h _ 1 w and l 4w 1, with l, 4 w, and h. V lwh (4w 1)(w) ( 1 _ 4 w ) w 11w 4w Since the volume must be at least 148 m, V 148; that is, w 11w 4w 148. Solve w 11w 4w 148. Factor the corresponding polynomial function. w 11w 4w 148 (w 8)(w 19w 176) Then, solve (w 8)(w 19w 176). w 19w 176 cannot be factored further. Case 1 Both factors are non-negative. w 8 w 19w 176 w 8 w 19w 176 is true for all values of w. These include values for w 8. So, w 8 is a solution. The formula for the volume, V, of a rectangular prism is V lwh, where l is the length, w is the width, and h is the height. The discriminant can be used to test for factors. If b 4ac is a perfect square then the quadratic can be factored. Here b 4ac 4, so the expression has no real roots..6 Solve Factorable Polynomial Inequalities Algebraically MHR 17

7 Case Both factors are non-positive, and w is positive (because w represents the width). w 8 w 19w 176 w 19w 176 is not possible for any values of w. There is no solution. So, the possible solution is w 8. When w 8, h _ 1 (8) 4 and l 4() 1. 4 The dimensions of the excavation that give a volume of at least 148 cm are width 8 m, depth 4 m, and length m. << >> KEY CONCEPTS Factorable inequalities can be solved algebraically by considering all cases using intervals and then testing values in each interval Tables and number lines can help organize intervals to provide a visual clue to solutions. Communicate Your Understanding C1 C C Why is it necessary to reverse an inequality sign when each side is multiplied or divided by a negative value? Support your answer with examples. What are the similarities between solving a linear inequality and solving a polynomial inequality? Which method is more efficient for solving factorable inequalities algebraically, using cases or using intervals? Explain. A Practise For help with question 1, refer to Example Solve each inequality. Show each solution on a number line. a) x 5 b) x 1 4 c) 5 x 6 d) 7x 4 x e) 4x 5x f) (1 x) x 8 For help with questions to 4, refer to Example.. Solve by considering all cases. Show each solution on a number line. a) (x )(x 4) b) (x )(4 x). Solve using intervals. Show each solution on a number line. a) (x )(x ) b) (x 6)(x 9) c) (4x 1)( x) 4. Solve. a) (x )( x)(x 1) b) (x 1)(x 1)(x 7) c) (7x )(1 x)(x 5) d) (x 4)(x 1)(x ) 18 MHR Advanced Functions Chapter