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Dividing Polynomials. 1. 2. 3. 4. 5. 6. – 4. Answer: x + 2 +. quotient. ( x + 2). ( x + 1). + (– 4). = x 2 + 3 x – 2. divisor. remainder. dividend. x. + 2. x 2 + x. 2 x. – 2. 2 x + 2. – 4. Check:.
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Dividing Polynomials 1. 2. 3. 4. 5. 6. –4 Answer: x + 2 + quotient (x + 2) (x + 1) + (– 4) = x2 + 3x – 2 divisor remainder dividend x + 2 x2+ x 2x – 2 2x + 2 –4 Check:
Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – a. It uses the coefficients of each term in the dividend. coefficients of the dividend value of a coefficients of quotient remainder 15 Answer: 3x + 8 Example: Divide 3x2 + 2x – 1 by x – 2 using synthetic division. Since the divisor is x – 2, a = 2. 2 3 2 –1 1. Bring down 3 6 16 2. (2 • 3) = 6 3. (2 + 6) = 8 3 8 15 4. (2 • 8) = 16 5. (–1 + 16) = 15
Remainder Theorem: The remainder of the division of a polynomial f(x) by x – k is f(k). value of x Example: Using the remainder theorem, evaluate f(x) = x4 – 4x – 1 when x = 3. 3 1 0 0 –4 –1 3 9 27 69 68 1 3 9 23 The remainder is 68 at x = 3, so f(3) = 68. f(3) = (3)4 – 4(3)– 1 = 68. You can check this using substitution:
Factor Theorem: A polynomial f(x) has a factor (x – k) if and only if f(k) = 0. Example: Show that (x + 2) and (x – 1) are factors of f(x) = 2x3 + x2– 5x + 2. – 2 2 1 –5 2 2 – 3 1 1 – 4 6 – 2 2 – 1 2 – 3 1 0 2 – 1 0 The remainders of 0 indicate that (x + 2) and (x – 1) are factors. The complete factorization of f is (x + 2)(x – 1)(2x – 1).
