4.1 Polynomial Functions

1. 4.1 Polynomial Functions Objectives: Define a polynomial. Divide polynomials. Apply the remainder and factor theorems, and the connections between remainders and factors. Determine the maximum number of zeros of a polynomial.

2. Polynomials Characteristics of Polynomials: A sum or difference of monomials All exponents are whole numbers No variable contained in the denominator No variable is under a radical

3. Degree of a Polynomial

4. Example #1Polynomial Division Divide: Expand the dividend and fill in missing terms, then follow the same steps as when dividing real numbers: divide, multiply, subtract, bring down. Divide 2x3 (2x4 − 8x3) Bring down Multiply 8x3 Subtract – 25x2 Once this cycle is made once, it is repeated until the degree of the remainder is less than the degree of the dividend.

5. Example #1Polynomial Division Divide: + 8x2 2x3 (2x4 − 8x3) Divide Bring down 8x3 – 25x2 (8x3 – 32x2) Multiply Subtract 7x2 – 30x

6. Example #1Polynomial Division Divide: + 8x2 2x3 + 7x (2x4 − 8x3) Bring down Divide 8x3 – 25x2 (8x3 – 32x2) Multiply 7x2 – 30x (7x2 – 28x) Subtract −2x + 5

7. Example #1Polynomial Division Divide: + 8x2 2x3 + 7x − 2 (2x4 − 8x3) Divide 8x3 – 25x2 (8x3 – 32x2) Multiply 7x2 – 30x (7x2 – 28x) The degree of the remainder is less than that of the divisor so the cycle is completed. −2x + 5 (−2x+ 8) Subtract −3 Remainder

8. Example #1Polynomial Division Divide: + 8x2 2x3 + 7x − 2 (2x4 − 8x3) Finally the remainder is put over the divisor and added to the quotient. 8x3 – 25x2 (8x3 – 32x2) 7x2 – 30x (7x2 – 28x) −2x + 5 (−2x+ 8) −3

9. Example #2Synthetic Division Divide using synthetic division: Caution! Synthetic division only works when the divisor is a first-degree polynomial of the form (x−a). Set the divisor equal to 0 and solve. Place it in the left corner in a half-box Write the leading coefficients of the dividend in order of degree next to the half-box. Include 0’s for missing terms if necessary. 1 5 2 6 −17 −5 −5 1 0 Bring down the first digit. Add the column and repeat the process. Multiply it by the -5. Write the answer under the next digit.

10. Example #2Synthetic Division Divide using synthetic division: 1 5 2 6 −17 −5 −5 0 2 1 0 Multiply the sum again by the -5. Write the answer under the next digit. Add the column and repeat the process.

11. Example #2Synthetic Division Divide using synthetic division: 1 5 2 6 −17 −5 −5 0 −10 2 −4 1 0 Multiply the sum again by the -5. Write the answer under the next digit. Add the column and repeat the process.

12. Example #2Synthetic Division Divide using synthetic division: 1 5 2 6 −17 −5 −5 0 −10 20 2 −4 3 1 0 Multiply the sum again by the -5. Write the answer under the next digit. Add the column. The last digit forms the remainder.

13. Example #2Synthetic Division Divide using synthetic division: After using up all terms, the quotient needs the variables added back into the final answer. The first term is always one degree less than the dividend. Remember to place the remainder over the original divisor. 1 5 2 6 −17 −5 −5 0 −10 20 2 −4 3 1 0

14. Example #3Factors Determined by Division x Divide (3x3 + 0x2 – 2x) Bring down Subtract 15x2 + 0x −10 Multiply

15. Example #3Factors Determined by Division x + 5 Divide (3x3 + 0x2 – 2x) 15x2 + 0x −10 (15x2 + 0x – 10) Multiply Subtract 0

16. Example #3Factors Determined by Division x + 5 (3x3 + 0x2 – 2x) 15x2 + 0x −10 (15x2 + 0x – 10) 0 Because the remainder is 0, the divisor divides evenly into the dividend and is then said to be a factor of the dividend.

17. Remainder Theorem If a polynomial f(x) is divided by x – c, then the remainder is f(c).

18. Example #4The remainder when dividing by x – c: Using the remainder theorem, this problem can be performed without actually dividing the dividend by the divisor. First find the value of c in the divisor x – c. Next evaluate f(c). The remainder is therefore 7.

19. Example #5The remainder when dividing by x + k Find the remainder when 2x4 + 4x3 + 5x – 6 is divided by x + 3.

20. Factor Theorem A polynomial function f(x) has a linear factor x – aif and only if f(a) = 0. The idea in this theorem was actually introduced in Example #3. The difference is that this theorem only works for linear factors, in which case division is not necessary, otherwise you must use division to check if they are factors.

21. Example #6The Factor Theorem Since f (2) = 0, then it is a factor of the polynomial. Next we’ll find the actual quotient using synthetic division. 2 1 5 −2 −24 The answer can be checked by multiplication. 2 14 24 1 7 12 0

22. Fundamental Polynomial Connections If one of the following are true, all are true: r is a zero of the function f r is an x-intercept of the graph of the function f x = r is a solution, or root, of the function f(x) = 0 x – ris a factor of the polynomial f(x)

23. Example #7Fundamental Polynomial Connections **Note**: From the graph it is clear that the x-intercepts are at −2.5, 1, & 5. The zeros of f are also −2.5, 1, & 5. The solutions are x = −2.5, x = 1, & x = 5. Setting each solution equal to 0 we obtain the linear factors of (2x + 5), (x – 1), & (x – 5).

24. Example #8A Polynomial with Specific Zeros Find three polynomials of different degrees that have -2, 1, and 3 as zeros. The simplest polynomial must be of degree 3 to have all the zeros requested. For a degree of 4, one (not necessarily the factor shown) of the factors must repeat twice, but highest exponent (if it were multiplied out) would still only be 4. Finally, for a degree of 5, two of the factors must repeat.

25. Example #8A Polynomial with Specific Zeros Find three polynomials of different degrees that have -2, 1, and 3 as zeros. Here, an alternative polynomial of degree 5 is shown that retains the same zeros as the one above it. First of all, the 7 out front is a constant and has NO effect on where the graph crosses the x-axis. This could be any real number, even a negative. Secondly, the final group of (x2 + 1) has imaginary roots at ±i. When everything is multiplied out, the resulting polynomial will still have a degree of 5, but the imaginary roots will have no effect on where the graph crosses the x-axis either.

26. Number of Zeros A polynomial of degree n has at most ndistinct real zeros.

27. Example #8Determining the Number of Zeros Find the maximum number of zeros that the polynomial could contain. The degree of this polynomial is 5. At most, this polynomial can have only 5 distinct real zeros. This means that at most, this polynomial will cross the x-axis 5 times.