Polynomial Functions

Polynomial Functions

Polynomials A Polynomial in one variable, x, is an expression of the form a n x 0 a 1 x n 1... a n 2 x 2 a n 1 x a n The coefficients represent complex numbers (real or imaginary), ao is not zero, and n represents a nonnegative integer.

The degree of a polynomial is a greatest exponent of the variable. The leading coefficient is a coefficient of the variable with the greatest exponent. Zeros of a polynomial function are values of x for which f(x) = 0

Example 1 Consider the polynomial function f(x) = x 3 - x 2-7x + 3. a. State the degree and leading coefficient of the polynomial. x 3 - x 2-7x + 3 has a degree of 3 and a leading coefficient of 1.

b. Determine whether 3 is a zero of f(x). Evaluate f(x) = x 3 - x 2-7x + 3 for x = 3. That is, find f(3). f(3) = (3) 3 - (3) 2-7(3) + 3 x = 3 f(3) = 27-9 - 21 + 3 f(3) = 0 Since f(3) = 0, 3 is a zero of f(x) = x 3 - x 2-7x + 3.

Fundamental Thm. Of Algebra Every Polynomial Equation with a degree higher than zero has at least one root in the set of Complex Numbers. COROLLARY: A Polynomial Equation of the form P(x) = 0 of degree n with complex coefficients has exactly n Roots in the set of Complex Numbers. P (x) = k(x -r1)(x - r2)(x - r3) (x - rn )

Example 2 a. Write a polynomial equation of least degree with roots 3, 2i, and -2i. b. Does the equation have an odd or even degree? How many times does the graph of the related function cross the x-axis?

a. If x = 3, then x - 3 is a factor of the polynomial. Likewise, if x = 2i and x = -2i, then x - 2i and x - (-2i) are factors of the polynomial. Therefore, the linear factors for the polynomial are x - 3, x - 2i, and x + 2i. Now find the products of these factors. (x - 3)(x - 2i)(x + 2i) = 0 (x - 3)(x 2-4i 2 ) = 0 (x - 3)(x 2 + 4) = 0-4i 2 = -4(-1) or 4 x 3-3x 2 + 4x - 12 = 0 A polynomial with roots 3, 2i, and -2i is x 3-3x 2 + 4x - 12 = 0.

b. The degree of this equation is 3. Thus, the equation has an odd degree since 3 is an odd number. Since two of the roots are imaginary, the graph will only cross the x-axis once.

Example 3 State the number of complex roots of the equation 4x 4-3x 2-1 = 0. Then find the roots and graph the related function. Factor the equation to find the roots. 4x 4-3x 2-1 = 0 (4x 2 + 1)(x 2-1) = 0 (4x 2 + 1)(x + 1)(x - 1) = 0 To find each root, set each factor equal to zero.

(4x 2 + 1)(x + 1)(x - 1) = 0 4x 2 + 1= 0 1 x 2 = - 4 x = 1 4 x = 1 i 2 x + 1= 0 x - 1 = 0 x= -1 x = 1 The roots are 1, -1, and 1. i 2

Lesson 4.2 Quadratic Equations.

Example 1. Solve the equation by completing the square. x 2-4x - 5= 0 x 2-4x = 5 x 2-4x + 4= 5 + 4 (x - 2) 2 = 9 x 2 = 3 b 4 4 ( ) 2 2 4 x - 2= 3 or x - 2= -3 x= 5 x= -1 Answer: The roots of the equation are -1 and 5.

Example 2. Solve 3x 2 + 4x + 4 = 0 by completing the square. b 4 ( 3 4 3 1 ) 2 2 16 36

Example 3. Find the discriminant of x 2 + 4x + 2 = 0 and describe the nature of the roots of the equation. Then solve the equation by using the Quadratic Formula. The discriminant b 2-4ac = 4 2-4(1)(2) = 8 Since the value of the discriminant is greater than zero, there are two real rational roots.

Example 4. Solve 2x 2 + 3x + 4 = 0.

Lesson 4.3 The Remainder and Factor Theorems

Example 1. Synthetic Substitution Divide f(x) = 2x 4 5x 2 + 8x 7 by x-6 using synthetic division 2 0 5 8 7 12 72 402 2460 2 12 67 410 2453 Notice that there is no x 3 term. A zero is placed in this position as a placeholder. Answer: The quotient is is R is 2453.

Example 2. Divide f(x) = 2x 4 5x 2 + 8x 7 by x-6 using direct substitution. State whether the binomial is a factor of the polynomial. Replace x with 6. Original function Replace x with 6. Simplify. Answer: f(6) is not a zero. So, x -6 is not a factor of the polynomial.

Example 3. Determine the binomial factors of x 3 + 4x 2 15x 18. One root is x = 3 and the factor is (x - 3). Use synthetic division to find the rest of the factors.

Example 3. Use the Factor Theorem Determine the binomial factors of x 3 + 4x 2 15x 18. Because f(3) = 0, x - 3 is a factor. Find the depressed polynomial. 1 4 15 18 3 21 18 1 7 6 0

The polynomial x 2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x 2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Answer: So, x 3 + 4x 2 15x 18 = (x 3)(x + 6)(x + 1).

Example 4. Find the value of k so that the remainder of (3x 4 + 8x 3-2x 2 - kx + 4) (x + 2) is 0. If the remainder is to be 0, x + 2 must be a factor of 3x 4 + 8x 3-2x 2 - kx + 4. So, f(-2) must equal 0. f(x)= 3x 4 + 8x 3-2x 2 - kx + 4 f(-2)= 3(-2) 4 + 8(-2) 3-2(-2) 2 - k(-2) + 4 0 = 48-64 - 8 + 2k + 4 Replace f(-2) with 0. 0 = 2k - 20 10= k

The value of k is 10. Check using synthetic division. -2 3 8-2 -10 4-6 -4 12-4 3 2-6 2 0

Lesson 4.4 The Rational Root Theorem

Example 1. List the possible rational roots of 1x 3 - x 2-10x - 8 = 0. Then determine the rational roots. possible values of p: 1, 2, 4, 8 possible values of q: 1 q p Answer: the possible rational roots p/q: 1, 2, 4, 8

You can use a graphing utility to narrow down the possibilities. You know that all possible rational roots fall in the domain -8 x 8. So, set your x-axis viewing window at [-9, 9]. Graph the related function f(x) = x 3 - x 2-10x - 8. A zero appears to occur at -2. Use synthetic division to check that -2 is a zero.

Solve by factoring x 2-3x - 4 =0 (x - 4)(x + 1)=0 x = 4, x = -1 Answer: -2, -1, and 4

Example 2. Find Numbers of Positive and Negative Zeros Find the number of possible positive real zeros and the number of possible negative real zeros for x 4-5x 2 +4. Then determine the rational zeros. To determine the number of possible positive real zeros, count the sign changes for the coefficients. There are two changes. So, there are 2 or 0 positive real zeros.

To determine the number of possible negative real zeros, find f(-x) and count the number of sign changes There are two changes. So, there are 2 or 0 negative real zeros.

Example 2. f(x) = x 4-5x 2 +4 possible values of p: 1, 2, 4 possible values of q: 1 Determine the rational zeros. Test the possible zeros using the Remainder Theorem. f(1) = 0, f(-1) = 0, f(2) = 0, and f(-2) = 0 Answer: 1, -1, 2, and -2

Suppose y = f(x) represents a polynomial function with real coefficients. If a and b are two numbers with f(a) negative and f(b) positive, the function has at least one real zero between a and b.

Example 1: Determine between which consecutive integers the real zeros of f(x) = x 3 + 3x 2-4x + 6 are located. Use the TABLE feature. The change in sign between -5 and -4 indicates that a zero exists between -5 and -4.

Example 2 Approximate the real zeros of f(x) = 5x 3-2x 2-4x + 1 to the nearest tenth. There are three complex zeros for this function. According to Descartes Rule of Signs, there are two or zero positive real roots and one negative real root. Use the TABLE feature of a graphing calculator. To find the zeros to the nearest tenth, use the TBLSET feature changing Tbl to 0.1. There are zeros between -0.9 and -0.8, between 0.2 and 0.3, and at 1.

Since 0.36 is closer to zero than -0.665, the zero is about -0.8. Since 0.16 is closer to zero than -0.245, the zero is about 0.2.

The third zero occurs at 1.

Upper Bound Theorem Suppose c is a positive number and P(x) is divided by x c. If the resulting quotient and remainder have no change in sign, then P(x) has no real zero greater than c. Thus c is an upper bound of the zeros of P(x). Lower Bound Theorem If c is an upper bound of the zeros of P(-x), then c is a lower bound of the zeros of P(x).

Example 3 Use the Upper Bound Theorem to find an integral upper bound and the Lower Bound Theorem to find an integral lower bound of the zeros of f(x) = x 3 + 5x 2-2x - 8. The Rational Root Theorem tells us that 1, 2, 4, and 8 might be roots of the polynomial equation x 3 + 5x 2-2x - 8 = 0. These possible zeros of the function are good starting places for finding an upper bound. f(x) = x 3 + 5x 2-2x - 8 f(-x) = -x 3 + 5x 2 + 2x - 8 r 1 5-2 -8 r -1 5 2-8 0-1 5 2-8 1 1 6 4-4 1-1 4 6-2 2 1 7 12 16 2-1 3 8 8 An upper bound is 2. 3-1 2 8 16 4-1 1 6 16 5-1 0 2 2 6-1 -1-4 -32 Since 6 is an upper bound of f(-x), -6 is a lower bound of f(x). This means that all real zeros of f(x) can be found in the interval -6 x 2.

Lesson 4.6

The best way to solve a rational equation: This can be done by multiplying each side of the equation by the LCD.

Example 1. Solve for x What is the LCD? (x 2)(x 5) 7 x 2 6 7 x 2 6 x 5 (x+2)(x-5) x 5 7(x 5) 6(x 2) 7x 35 6x 12 x 35 12 x 47 (x 2)(x 5)

The other method of solving rational equations is cross-multiplication. 7 x 2 6 x 5 7 (x 5) 6 (x 2) 7x 35 6x 12 x 35 12 x 47

Example 2. Solve for x Step 1: Find the LCD x 1 3x 6 5x 6 x 1 2 Hint: Factor the denominator x 1 3x 6 Therefore. This denominator can be factored into 3(x-2) LCD 6(x 2)

Step 2: Multiply both sides of equation by LCD. This eliminates the fraction. 6(x 2) x 1 3(x 2) 6(x 2) 5x 6 6(x 2) x 1 2 2(x 1) (x 2)5x 6

2(x 1) 5x(x 2) 1 6 2x 2 5x 2 10x 6 2x 2 2x 2 0 5x 2 12x 4 0 (5x 2)(x 2) x 2 5 x 2

Example 3 Decompose x 2 x 9 3x 10 into partial fractions. First factor the denominator. x 2-3x - 10= (x - 5)(x + 2) Express the factored form as the sum of two fractions using A and B as numerators and the factors as denominators. x 2 x 9 3x 10 x A 5 x B 2

Eliminate the denominators bymultiplying each side by the LCD, (x - 5)(x + 2). (x 5)(x 2) 2 x x 9 3x 10 (x 5)(x 2) x A 5 (x 5)(x 2) x B 2 x + 9= A(x + 2) + B(x - 5) Eliminate B by letting x = 5 so that x - 5 becomes 0. x + 9= A(x + 2) + B(x - 5) 5 + 9= A(5 + 2) + B(5-5) 14= 7A 2= A

x + 9= A(x + 2) + B(x - 5) Eliminate A by letting x = -2 so that x + 2 becomes 0. x + 9= A(x + 2) + B(x - 5) -2 + 9= A(-2 + 2) + B(-2-5) 7= -7B -1= B Now substitute the values for A and B to determine the partial fractions. A x 5 x B 2 x 2 5 1 x 2

x 2 x 9 3x 10 x 2 5 1 x 2

Example 4: Solve (x 1)(x 3) 0 2 (x 2)(x 4) 1. Solve the numerator (x +1)(x 3) = 0 x = -1 and x = 3 2. Solve the denominator (x - 2)(x +4)(x + 4) = 0 x = 2 and x = - 4 3. Make vertical dashed lines through -4, -1, 2, 3

4. Test a convenient value within each interval in the original rational inequality to see if the test value is a solution. For x < - 4, test x = -5: (x 1)(x 3) (x 2)(x 4) 2 0 ( 5 ( 5 1)( 5 2)( 5 3) 4) 2 ( )( ) ( )( ) So, x < -4 is not a solution.

For -4 < x < -1, test x = -2.5: (x 1)(x 3) (x 2)(x 4) 2 0 ( 2.5 ( 2.5 1)( 2.5 2)( 2.5 3) 4) 2 ( )( ) ( )( ) So, - 4 < x < -1 is not a solution.

For -1 < x < 2, test x = 0: (x 1)(x 3) (x 2)(x 4) 2 0 (0 1)(0 3) (0 2)(0 4) 2 ( )( ) ( )( ) So, -1 < x < 2 is a solution.

For 2 < x < 3, test x = 2.5: (x 1)(x 3) (x 2)(x 4) 2 0 (2.5 1)(2.5 3) (2.5 2)(2.5 4) 2 ( )( ) ( )( ) So, 2 < x < 3 is not a solution.

For x > 3, test x = 4: (x 1)(x 3) (x 2)(x 4) 2 0 (4 1)(4 3) (4 2)(4 4) 2 ( )( ) ( )( ) So, x > 3 is a solution.

This solution can be graphed on a number line.

Lesson 4.7

Equations in which radical expressions include variables are To solve radical equations 1. Isolate the radical on one side of the equation 2. Raise each side of the equation to the proper power to eliminate the radical expression.

Example 1. Solve

Example 2. Solve 50 7x 8 x 50 7x x 8 50 7x x 2 16x 64 0 x 2 16x 64 7x 50 0 x 2 9x 14 0 x ( x 2)( x 7 2 x ) 7

Check both solutions to make sure they are not extraneous.

Example 3. Solve

Check both solutions to make sure they are not extraneous.

Example 4. Solve 5x 4 8 1. Solve 5x + 4 = 0, x = -0.8 2. Solve 5x 4 8 5x 5x x 4 64 60 12

3. On a number line, mark -0.8 and 12 with vertical dashed lines. For x -0.8, test x = -1: This statement is meaningless. x -0.8 is not a solution.

For -0.8 x 12, test x = 0: -0.8 x 12 is a solution.

For x 12, test x = 13: x 12 is not a solution.