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• • • • • • • P.159: 5-16 (Some) P. 159: 19-36 (Some) P. 160: 46 P. 160: 49-56 (Some) P. 160: 57-64 (Some) P. 161: 79 Homework Supplement Use long division to divide 5 into 3462. 1. Write your answer without decimals. 2. What would it mean if the remainder was zero? 692 5 3462 30 46 45 12 10 2 Use long division to divide 5 into 3462. Divisor 692 5 3462 30 46 45 12 10 2 Quotient Dividend Remainder Use long division to divide 5 into 3462. Dividend Divisor 3462 2 692 5 5 Quotient Remainder Divisor If you are lucky enough to get a remainder of zero when dividing, then the divisor divides evenly into the dividend. This means that the divisor is a factor of the dividend. For example, when dividing 3 into 192, the remainder is 0. Therefore, 3 is a factor of 192. Objectives: 1. To divide polynomials using long and synthetic division 2. To apply the Factor and Remainder Theorems to find real zeros of polynomial functions • • • • • • • Assignment: P.159: 5-16 (Some) P. 159: 19-36 (Some) P. 160: 46 P. 160: 49-56 (Some) P. 160: 57-64 (Some) P. 161: 79 Homework Supplement As a class, use your vast mathematical knowledge to define each of these words without the aid of your textbook. Quotient Remainder Dividend Divisor Divides Evenly Factor Dividing polynomials works just like long division. In fact, it is called long division! Before you start dividing: 1. Make sure the divisor and dividend are in standard form (highest to lowest powers). 2. If your polynomial is missing a term, add it in with a coefficient of 0 as a place holder. 2x x 3 3 2x 0x x 3 3 2 Divide x + 1 into x2 + 3x + 5 x 2 x 1 x22 3x 5 −x − x 2x 5 −2 x − 2 3 How many times does x go into x2? Multiply x by x + 1 Multiply 2 by x + 1 Line up the first term of the quotient with the term of the dividend with the same degree. Divide x + 1 into x2 + 3x + 5 Quotient x 2 x 1 x22 3x 5 −x − x 2x 5 −2 x − 2 Divisor 3 Dividend Remainder Divide x + 1 into x2 + 3x + 5 Dividend x 3x 5 3 x2 x 1 x 1 2 Quotient Divisor Remainder Divisor Divide 6x3 – 16x2 + 17x – 6 by 3x – 2 Use long division to divide x4 – 10x2 + 2x + 3 by x–3 When your divisor is of the form x k, where k is a constant, then you can perform the division quicker and easier using just the coefficients of the dividend. This is called fake division. I mean, synthetic division. Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. = Add terms k a b c d ka a = Multiply by k Remainder Coefficients of Quotient (in decreasing order) Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. = Add terms k a b c d ka = Multiply by k a Important Note: You are always adding columns using synthetic division, whereas you subtracted columns in long division. Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. = Add terms k a b c d ka = Multiply by k a Important Note: k can be positive or negative. If you divide by x + 2, then k = -2 because x + 2 = x – (-2). Synthetic Division (of a Cubic Polynomial) To divide ax3 + bx2 + cx + d by x – k, use the following pattern. = Add terms k a b c d ka = Multiply by k a Important Note: Add a coefficient of zero for any missing terms! Use synthetic division to divide x4 – 10x2 + 2x + 3 by x+3 Evaluate f (−3) for f (x) = x4 – 10x2 + 2x + 3. If a polynomial f (x) is divided by x – k, the remainder is r = f (k). This means that you could use synthetic division to evaluate f (5) or f (−2). Your answer will be the remainder. Divide 2x3 + 9x2 + 4x + 5 by x + 3 using synthetic division. Use synthetic division to divide f(x) = 2x3 – 11x2 + 3x + 36 by x – 3. Since the remainder is zero when dividing f(x) by x – 3, we can write: f ( x) 2 x 2 5 x 12, x3 so f ( x) ( x 3)(2 x 2 5x 12) This means that x – 3 is a factor of f(x). A polynomial f(x) has a factor x – k if and only if f(k) = 0. This theorem can be used to help factor/solve a polynomial function if you already know one of the factors. Factor f(x) = 2x3 – 11x2 + 3x + 36 given that x – 3 is one factor of f(x). Then find the zeros of f(x). Given that x – 4 is a factor of x3 – 6x2 + 5x + 12, rewrite x3 – 6x2 + 5x + 12 as a product of two polynomials. Find the other zeros of f(x) = 10x3 – 81x2 + 71x + 42 given that f(7) = 0. Objectives: 1. To divide polynomials using long and synthetic division 2. To apply the Factor and Remainder Theorems to find real zeros of polynomial functions • • • • • • • Assignment: P.159: 5-16 (Some) P. 159: 19-36 (Some) P. 160: 46 P. 160: 49-56 (Some) P. 160: 57-64 (Some) P. 161: 79 Homework Supplement