# 4.1 polynomial functions

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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. Polynomials. Characteristics of Polynomials : - PowerPoint PPT Presentation

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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.4.1 Polynomial FunctionsPolynomialsCharacteristics of Polynomials:A sum or difference of monomialsAll exponents are whole numbersNo variable contained in the denominatorNo variable is under a radical

PolynomialsNot Polynomials

Degree of a Polynomial

Example #1Polynomial DivisionDivide:

Expand the dividend and fill in missing terms, then follow the same steps as when dividing real numbers: divide, multiply, subtract, bring down.

2x38x3

Divide

Multiply(2x4 8x3)SubtractBring down 25x2Once this cycle is made once, it is repeated until the degree of the remainder is less than the degree of the dividend.Example #1Polynomial DivisionDivide:

2x38x3

Divide

Multiply(2x4 8x3)SubtractBringdown 25x2 + 8x2(8x3 32x2)7x2 30xExample #1Polynomial DivisionDivide:

2x38x3

Divide

Multiply(2x4 8x3)SubtractBringdown 25x2 + 8x2(8x3 32x2)7x2 30x + 7x(7x2 28x)2x+ 5Example #1Polynomial DivisionDivide:

2x38x3

Divide

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

2x38x3

(2x4 8x3) 25x2 + 8x2(8x3 32x2)7x2 30x + 7x(7x2 28x)2x+ 5 2(2x + 8)3Finally the remainder is put over the divisor and added to the quotient.Example #2Synthetic DivisionDivide using synthetic division:

Caution! Synthetic division only works when the divisor is a first-degree polynomial of the form (x a).

51 5 2 6 17Set the divisor equal to 0 and solve. Place it in the left corner in a half-boxWrite the leading coefficients of the dividend in order of degree next to the half-box. Include 0s for missing terms if necessary.1Bring down the first digit.5Add the column and repeat the process.0Multiply it by the -5.Write the answer under the next digit.

Example #2Synthetic DivisionDivide using synthetic division:

51 5 2 6 1715Add the column and repeat the process.0Multiply the sum again by the -5.Write the answer under the next digit.

02Example #2Synthetic DivisionDivide using synthetic division:

51 5 2 6 1715Add the column and repeat the process.0Multiply the sum again by the -5.Write the answer under the next digit.

02410Example #2Synthetic DivisionDivide using synthetic division:

51 5 2 6 1715Add the column. The last digit forms the remainder.0Multiply the sum again by the -5.Write the answer under the next digit.

02410320Example #2Synthetic DivisionDivide using synthetic division:

51 5 2 6 1715002410320After 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.

Example #3Factors Determined by Division

x(3x3 + 0x2 2x)DivideMultiplySubtract15x2 + 0x10Bring downExample #3Factors Determined by Division

x(3x3 + 0x2 2x)DivideMultiplySubtract15x2 + 0x10+ 5(15x2 + 0x 10)0Example #3Factors Determined by Division

x(3x3 + 0x2 2x)15x2 + 0x10+ 5(15x2 + 0x 10)0Because the remainder is 0, the divisor divides evenly into the dividend and is then said to be a factor of the dividend.Remainder TheoremIf a polynomial f(x) is divided by x c, then the remainder is f(c).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.Example #5The remainder when dividing by x + k

Find the remainder when 2x4 + 4x3 + 5x 6 is divided by x + 3.Factor TheoremA polynomial function f(x) has a linear factor x a if 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.Example #6The Factor Theorem

Since f (2) = 0, then it is a factor of the polynomial. Next well find the actual quotient using synthetic division.2 1 5 2 241 7 12 02 14 24

The answer can be checked by multiplication.

Fundamental Polynomial ConnectionsIf one of the following are true, all are true:

r is a zero of the function fr is an x-intercept of the graph of the function fx = r is a solution, or root, of the function f(x) = 0x r is a factor of the polynomial f(x)Example #7Fundamental Polynomial Connections

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).

**Note**:23Example #8A Polynomial with Specific ZerosFind 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. Example #8A Polynomial with Specific ZerosFind 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.Number of ZerosA polynomial of degree n has at most n distinct real zeros.

Example #8Determining the Number of ZerosFind 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.

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