Chapter 2- Factor Theorem and Inequalities

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Chapter 2- Factor Theorem and Inequalities Chapter 2- Factor Theorem and Inequalities Lesson Package MHF4U Chapter 2 Outline Unit Goal: By the end of this unit, you will be able to factor and solve polynomials up to degree 4 using the factor theorem, long division, and synthetic division. You will also learn how to solve factorable polynomial inequalities. Curriculum Section Subject Learning Goals Expectations - divide polynomial expressions using long division L1 Long Division C3.1 - understand the remainder theorem L2 Synthetic Division - divide polynomial expressions using synthetic division 3.1 - be able to determine factors of polynomial expressions by testing L3 Factor Theorem values C3.2 - solve polynomial equations up to degree 4 by factoring Solving Polynomial C3.2, C3.3, L4 - make connections between solutions and x-intercepts of the graph Equations C3.4, C3.7 - determine the equation of a family of polynomial functions given Families of Polynomial L5 the x-intercepts C1.8 Functions Solving Polynomial - Solve factorable polynomial inequalities L5 C4.1, C4.2, C4.3 Inequalities Assessments F/A/O Ministry Code P/O/C KTAC Note Completion A P Practice Worksheet F/A P Completion Quiz – Factor Theorem F P PreTest Review F/A P Test – Factor Theorem and C1.8 K(21%), T(34%), A(10%), Inequalities O C3.1, 3.2, 3.3, 3.3, 3.7 P C(34%) C4.1, 4.2, 4.3 L1 – 2.1 – Long Division of Polynomials and The Remainder Theorem Lesson MHF4U Jensen In this section you will apply the method of long division to divide a polynomial by a binomial. You will also learn to use the remainder theorem to determine the remainder of a division without dividing. Part 1: Do You Remember Long Division (divide, multiply, subtract, repeat)? 107 ÷ 4 can be completed using long division as follows: 4 does not go in to 1, so we start by determining how many times 4 goes in to 10. It goes in 2 times. So we put a 2 in the quotient above the 10. This is the division step. Then, multiply the 2 by 4 (the divisor) and put the product below the 10 in the dividend. This is the multiply step. Now, subtract 8 from the 10 in the dividend. Then bring down the next digit in the dividend and put it beside the difference you calculated. This is the subtract step. You then repeat these steps until there are no more digits in the dividend to bring down. Every division statement that involves numbers can be rewritten using multiplication and addition. We can express the results of our example in two different ways: +,- / 107 = 4 26 + 3 OR = 26 + . =26.75 =26 + 0.75 =26.75 Example 1: Use long division to calculate 753 ÷ 22 Part 2: Using Long Division to Divide a Polynomial by a Binomial The quotient of 3�/ − 5�3 − 7� − 1 ÷ � − 3 can be found using long division as well… Focus only on the first terms of the dividend and the divisor. Find the quotient of these terms. Since 3�/ ÷ � = 3�3, this becomes the first term of the quotient. Place 3�3 above the term of the dividend with the same degree. Multiply 3�3 by the divisor, and write the answer below the dividend. Make sure to line up ‘like terms’. 3�3(� − 3) = 3�/ − 9�3. Subtract this product from the dividend and then bring down the next term in the dividend. Now, once again, find the quotient of the first terms of the divisor and the new expression you have in the dividend. Since 4�3 ÷ � = 4�, this becomes the next term in the quotient. Multiply 4� by the divisor, and write the answer below the last line in the dividend. Make sure to line up ‘like terms’. 4�(� − 3) = 4�3 − 12�. Subtract this product from the dividend and then bring down the next term. Now, find the quotient of the first terms of the divisor and the new expression in the dividend. Since 5� ÷ � = 5, this becomes the next term in the quotient. Multiply 5(� − 3) = 5� − 15. Subtract this product from the dividend. The process is stopped once the degree of the remainder is less than the degree of the divisor. The divisor is degree 1 and the remainder is now degree 0, so we stop. The result in quotient form is: The result of the division of �(�) by a binomial of the form � − � is: 3�/ − 5�3 − 7� − 1 14 = 3�3 + 4� + 5 + � − 3 � − 3 �(�) � = �(�) + � − � � − � Where � is the remainder. The statement The expression that can be used to check the division is: that can be used to check the division is: 3�/ − 5�3 − 7� − 1 = � − 3 3�3 + 4� + 5 + 14 �(�) = (� − �)�(�) + � Note: you could check this answer by FOILing the product and collecting like terms. Example 2: Find the following quotients using long division. Express the result in quotient form. Also, write the statement that can be used to check the division (then check it!). a) �3 + 5� + 7 divided by � + 2 The result in quotient form is: �3 + 5� + 7 1 = � + 3 + � + 2 � + 2 The expression that can be used to check the division is: �3 + 5� + 7 = (� + 2)(� + 3) + 1 b) 2�/ − 3�3 + 8� − 12 divided by � − 1 The result in quotient form is: 2�/ − 3�3 + 8� − 12 −5 = 2�3 − � + 7 + � − 1 � − 1 The expression that can be used to check the division is: 2�/ − 3�3 + 8� − 12 = (� − 1)(2�3 − � + 7) − 5 c) 4�/ + 9� − 12 divided by 2� + 1 The result in quotient form is: 4�/ + 9� − 12 −17 = 2�3 − � + 5 + 2� + 1 2� + 1 The expression that can be used to check the division is: 4�/ + 9� − 12 = (2� + 1)(2�3 − � + 5) − 17 Example 3: The volume, in cubic cm, of a rectangular box is given by � � = �/ + 7�3 + 14� + 8. Determine expressions for possible dimensions of the box if the height is given by � + 2. Dividing the volume by the height will give an expression for the area of the base of the box. Factor the area of the base to get possible dimensions for the length and width of the box. Expressions for the possible dimensions of the box are � + 1, � + 2, and � + 4. Part 3: Remainder Theorem When a polynomial function �(�) is divided by � − �, the remainder is �(�); and when it is divided by @ �� − �, the remainder is � , where � and � are integers, and � ≠ 0. A Example 3: Apply the remainder theorem a) Use the remainder theorem to determine the remainder when � � = 2�/ + �3 − 3� − 6 is divided by � + 1 Since � + 1 is � − (−1), the remainder is �(−1). � −1 = 2(−1)/ + (−1)3 − 3 −1 − 6 = −2 + 1 + 3 − 6 = −4 Therefore, the remainder is −4 b) Verify your answer using long division Example 4: Use the remainder theorem to determine the remainder when � � = 2�/ + �3 − 3� − 6 is divided by 2� − 3 / The remainder is � 3 3 3 / 3 3 3 � = 2 + − 3 − 6 2 2 2 2 3- D / = 2 + − 3 − 6 C . 3 E. D D = + − − 6 C . 3 3- D +C 3. = + − − . Therefore, the remainder is − / / 3 = − 3 Example 5: Determine the value of � such that when 3�. + ��/ − 7� − 10 is divided by � − 2, the remainder is 8. The remainder is �(2). Solve for � when �(2) is set to equal 8. � 2 = 3 2 . + � 2 / − 7 2 − 10 8 = 3 2 . + � 2 / − 7 2 − 10 8 = 3 16 + 8� − 14 − 10 8 = 48 + 8� − 24 8 = 24 + 8� −16 = 8� −2 = � Therefore, the value of � is −2. L2 – 2.1 – Synthetic Division Lesson MHF4U Jensen In this section you will learn how to use synthetic division as an alternate method to dividing a polynomial by a binomial. Synthetic division is an efficient way to divide a polynomial by a binomial of the form � − �. IMPORTANT: When using Polynomial OR Synthetic division… • Terms must be arranged in descending order of degree, in both the divisor and the dividend. • Zero must be used as the coefficient of any missing powers of the variable in both the divisor and the dividend. Part 1: Synthetic division when the binomial is of the form � − � Divide 3�' − 5�) − 7� − 1 by � − 3. In this question, � = 3. List the coefficients of the dividend in the first row. To the left, write the � value (the zero of the divisor). Place a + sign above the horizontal line to represent addition and a × sign below the horizontal line to indicate multiplication of the divisor and the terms of the quotient. Bring the first term down, this is the coefficient of the first term of the quotient. Multiply it by the � value and write this product below the second term of the dividend. Now add the terms together. Multiply this sum by the � value and write the product below the third term of the dividend. Repeat this process until you have a completed the chart. The last number below the chart is the remainder. The first numbers are the coefficients of the quotient, starting with degree that is one less than the dividend. Don’t forget that the answer can be written in two ways… 3�' − 5�) − 7� − 1 �� = ��� + �� + � + � − 3 � − � OR 3�' − 5�) − 7� − 1 = � − � ��� + �� + � + �� Example 1: Use synthetic division to divide. Then write the multiplication statement that could be used to check the division. a) (�7 − 2�' + 13� − 6) ÷ (� + 2) = �7 − 2�' + 0�) + 13� − 6 Note: since the remainder is zero, both the quotient and divisor are factors of the dividend. b) (2�' − 5�) + 8� + 4) ÷ (� − 3) Part 2: Synthetic division when the binomial is of the form �� − � Divide 6�' + 5�) − 16� − 15 by 2� + 3 To use synthetic division, the divisor must be in the form � − �.
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