 # 1. 2 Polynomial Function A polynomial function is a function of the form where n is a nonnegative integer and each a i (i = 0,1,…, n) is a real number.

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2 Polynomial Function A polynomial function is a function of the form where n is a nonnegative integer and each a i (i = 0,1,…, n) is a real number. The polynomial function has a leading coefficient a n and degree n. Def.:

3 + 2 Dividing Polynomials Ex1: Divide x 2 + 3x – 2 by x + 1 and check the answer. x x 2 + x 2x2x– 2 2x + 2 – 4– 4 remainder Check: 1. 2. 3. 4. 5. 6. (x + 2) quotient (x + 1) divisor + (– 4) remainder = x 2 + 3x – 2 dividend Answer: x + 2 + – 4– 4 Dividing Polynomials

4 Notes: P(x ): Dividend D(x): Divisor Q(x): Quotient R(x): Remainder 1) 2) The degree of R(x) <the degree of D(x)

5 Ex2 Ex3(HW)

6 16 Synthetic Division Synthetic division is a shorter method of dividing polynomials. This method can be used only when the divisor is of the form x – c. It uses the coefficients of each term in the dividend. Ex4: Divide 3x 2 + 2x – 1 by x – 2 using synthetic division. 3 2 – 1 2 Since the divisor is x – 2, c = 2. 3 1. Bring down 3 2. (2 3) = 6 6 815 3. (2 + 6) = 8 4. (2 8) = 16 5. (–1 + 16) = 15 coefficients of quotient remainder value of c coefficients of the dividend 3x + 8Answer: 15

7 Ex5: Use synthetic division to divide by Ex6: Use synthetic division to divide by

8 Remainder Theorem If a polynomial P (x) is divided by x –c, then the remainder equals P(c). Ex7: Using the remainder theorem, evaluate P(x) = x 4 – 4x – 1 when x = 3. 9 1 0 0 – 4 – 1 3 1 3 39 6927 2368 The remainder is 68 at x = 3, so P(3) = 68. You can check this using substitution: P(3) = (3) 4 – 4(3) – 1 = 68. value of x Note:

9 Ex8: If P(x)=211x 4 -212x 3 +212x 2 +210x-3, then find P(1/211). Q69/289: Find the remainder of 5x 48 +6x 10 -5x+7 divided by x-1. Ex9:(HW) Find the remainder of P(x)=x 103 +x 102 +x 101 + x 100 divided by x+i.

10 Factor Theorem A polynomial P(x) has a factor (x – c) if and only if P(c) = 0. Ex10: Show that (x + 2) and (x – 1) are factors of P(x) = 2x 3 + x 2 – 5x + 2. 6 2 1 – 5 2 – 2 2 – 4 – 31 – 2 0 The remainders of 0 indicate that (x + 2) and (x – 1) are factors. – 1 2 – 3 1 1 2 2 – 10 The complete factorization of P is (x + 2)(x – 1)(2x – 1).

11 Zeros of a Function A real number c is a zero of P (x) if and only if P(c) = 0. A polynomial function of degree n has at most n real zeros. Real Zeros of Polynomial Functions If P(x) is a polynomial and c is a real number then the following statements are equivalent. 1. x = c is a zero of P. 2. x = c is a solution of the polynomial equation P (x) =0. 3. (x – c) is a factor of the polynomial P (x). 4. (c, 0) is an x-intercept of the graph of P (x). and we have P(x)=(x-c) Q(x), where Q(x) is a polynomial of degree< degree of P(x) by 1 called the reduced polynomial

12 Ex11: Verify that (x+4) is a factor of P(x) = 3x 4 +11x 3 -6x 2 - 6x+8. and write P(x) as the product of (x+4) and the reduced polynomial Q(x). Ex12:Prove that for any positive odd integer n, P(x) = x n +1 has x-1 as a factor. Ex13: If x-i is a factor of the polynomial P(x) = 7x 171 -8x 172 - 9x 173 +kx 174, then find the value of k.

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