Zeros of Polynomial Functions

Zeros of Polynomial Functions Objectives: 1.Use the Fundamental Theorem of Algebra to determine the number of zeros of polynomial functions 2.Find rational zeros of polynomial functions 3.Find conjugate pairs of complex zeros 4.Find zeros of polynomials by factoring 5.Use Descartes s Rule of Signs and the Upper and Lower Bound Rules to find zeros of polynomials

In the complex number system, every nth-degree polynomial has precisely n zeros. Fundamental Theorem of Algebra If f(x) is a polynomial of degree n, where n > 0, then f has at least one zero in the complex number system Linear Factorization Theorem If f(x) is a polynomial of degree n, where n > 0, then f has precisely n linear factors f (x) a n (x c 1 )(x c 2 )...(x c n ) c 1,c 2,...,c n where are complex numbers

Zeros of Polynomial Functions Give the degree of the polynomial, tell how many zeros there are, and find all the zeros f (x) x 2 f (x) x 2 6x 9 f (x) x 3 4 x f (x) x 4 1

Rational Zero (Root) Test To use the Rational Zero Test, you should list all rational numbers whose numerators are factors of the constant term & whose denominators are factors of the leading coefficient Possible Rational Zeros factors of the constant term factors of leading coefficient Once you have all the possible zeros test them using substitution or synthetic division to see if they work and indeed are a zero of the function (Also, use a graph to help determine zeros to test) NOTE: It only tests for rational numbers.

EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f (x) 15x 3 2 14x 3x 2. Solution The constant term is 2 & the leading coefficient is 15. Possible rational zeros Divide 1 and 2 by 1. Factors of the constant term, 2 Factors of the leading coefficient, 15 1, 2 1, 3, 5, 15 1 2 1 2 1 2 3 3 5 5 15 15 1, 2,,,,,, Divide 1 and 2 by 3. Divide 1 and 2 by 5. Divide 1 and 2 by 15. There are 16 possible rational zeros. The actual solution set to f (x) 15x 3 14x 2 3x 2 = 0 is {-1, 1 /3, 2 /5}, which contains 3 of the 16 possible solutions.

You Try: Using the Rational Zero Theorem List all possible rational zeros of f (x) x 3 2 5x 2x + 8. Solution The constant term is 8 & the leading coefficient is 1. Possible rational zeros Factors of the constant term, 8 Factors of the leading coefficient, 1 1, 2, 4, 8 1 There are 8 possible rational zeros. The actual solution set to f (x) x 3 5x 2 2x + 8 is {-1, 2, 4}, which contains 3 of the 8 possible solutions.

Roots & Zeros of Polynomials II Finding the Roots/Zeros of Polynomials (Degree 3 or higher): Graph the polynomial to find your first zero/root Use synthetic division to find a smaller polynomial If the polynomial is not a quadratic follow the 2 steps above using the smaller polynomial until you get a quadratic. Factor or use the quadratic formula to find your remaining zeros/roots

Example 1: Find all the zeros of each polynomial function 32 109196 xxx First, graph the equation to find the first zero ZERO From looking at the graph you can see that there is a zero at -2

Example 1 Continued 32 109196 xxx -2 10 9-19 6-20 22-6 Second, use the zero you found from the graph and do synthetic division to find a smaller polynomial Don t forget your remainder should be zero 10-11 3 0 The new, smaller polynomial is: 2 10 xx 13

Example 1 Continued: 2 10 xx 13 Third, factor or use the quadratic formula to find the remaining zeros. This quadratic can be factored into: Therefore, the zeros to the problem (5x 3)(2x 1) x 32 109196 xxx are: 31 2,, 52

Using the Rational Zero Theorem: Find all Zeros! 3 2 f ( x) x 4x 21x 34 Try listing all possible rational zeros of Solution The constant term is 34 & the leading coefficient is 1. Possible rational zeros Factors of the constant term, 34 Factors of the leading coefficient, 1 1, 2, 17, 34 1 There are 8 possible rational zeros. The actual solution set to f (x) x 3 4x 2 21x 34 is {2, 1 + 4i, 1 4i}, which contains 3 of the 8 possible solutions.

Find the rational zeros. Rational Zeros f x x x x x 4 3 2 ( ) 3 6 f x x x x 3 2 ( ) 8 40 525 f x x x x 3 2 ( ) 2 3 8 3

Find all the real zeros (Hint: start by finding the rational zeros) f (x) 10x 3 15x 2 16x 12 f (x) 3x 3 19x 2 33x 9

Writing a Polynomial given the zeros. To write a polynomial you must write the zeros out in factored form. Then you multiply the factors together to get your polynomial. Factored Form: (x zero)(x zero)... ***If it is a polynomial function and a + bi is a root, then a bi is also a root. ***If it is a polynomial function and a bis a root, then a bis also a root

Example 1: The zeros of a third-degree polynomial are 2 (multiplicity 2) and -5. Write a polynomial. (x 2)(x 2)(x (-5)) = (x 2)(x 2)(x+5) Second, multiply the factors out to find your polynomial 2 (2)(2)44 xxxx 2 ( x 5)( x 4x 4) First, write the zeros in factored form xxx 32 1620

Example 1 Continued (x 2)(x 2)(x+5) 2 (2)(2)44 xxxx First FOIL or box two of the factors X 5 xx 2 44 3 x 2 2 5x 4x 4x 20x 20 Second, box your answer from above with your remaining factors to get your polynomial: xxx 32 1620 ANSWER

Conjugate Pairs Complex Zeros Occur in Conjugate Pairs = If a + bi is a zero of the function, the conjugate a bi is also a zero of the function (the polynomial function must have real coefficients) EXAMPLES: Find a polynomial with the given zeros -1, -1, 3i, -3i 2, 4 + i, 4 i

So if asked to find a polynomial that has zeros, 2 and 1 3i, you would know another root would be 1 + 3i. Let s find such a polynomial by putting the roots in factor form and multiplying them together. If x = the root then (x - the root) is the factor form. xix 31 x 2xix 31 31 i Multiply the last two factors together. All i terms should disappear when simplified. xx 2 3i xi931 2 2 iixi x 2x 2x 10 Now multiply the x 2 through xx 3 4 2 14 x 20 Here is a 3 rd degree polynomial with roots 2, 1-3i and 1 + 3i i x2 31-1

Example 3: 4 3 2 Find ALL the zeros of Given that 1+3i is a zero! x 3x 6x 2 60

You Try! Find ALL the zeros of Given that (7i) is a zero! 3 2 x 2x 49x 98

STEPS For Finding the Zeros given a Solution 1) Find a polynomial with the given solutions (FOIL) 2) Use long division to divide your polynomial you found in step 1 with your polynomial from the problem 3) Factor or use the quadratic formula on the answer you found from long division. 4) Write all of your answers out

Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Theorem. Ex: Find all the roots of f(x) x 3 5x 2 7x 5 If one root is 4 - i. Because of the Complex Conjugate Thm., we know that another root must be 4 + i.

Example (con t) Ex: Find all the roots of f(x) x 3 5x 2 7x 51 If one root is 4 - i. If one root is 4 - i, then one factor is [x - (4 - i)], and Another root is 4 + i, & another factor is [x - (4 + i)]. Multiply these factors: [(4)][(4)](4)(4) xixixixi X -4 -i x x 4i 2 4x ix 4x 16 4i ix 4i 2 i xx 2 817

Example (con t) Ex: Find all the roots of f(x) x 3 5x 2 7x 51 If one root is 4 - i. If the product of the two non-real factors is x 2 8x 17 then the third factor (that gives us the real root) is the quotient of P(x) divided by x 2 8x 17 x 3 x 2 8x 17x 3 5x 2 7x 51 x 3 5x 2 7x 51 0 The third root is x = -3 So, all of the zeros are: 4 i, 4 + i, and -3

FIND ALL THE ZEROS f (x) x 4 3x 3 6x 2 2x 60 (Given that 1 + 3i is a zero of f) f (x) x 3 7x 2 x 87 (Given that 5 + 2i is a zero of f)

More Finding of Zeros f (x) x 5 x 3 2x 2 12x 8 f (x) 3x 3 4x 2 8x 8

Descartes s Rule of Signs Let f (x) a n x n a n 1 x n 1... a 2 x 2 a 1 x a 0 with real coefficients and a 0 0 be a polynomial The number of positive real zeros of f is either equal to the number of variations in sign of f(x) or less than that number by an even integer The number of negative real zeros of f is either equal to the number of variations in sign of f(-x) or less than that number by an even integer Variation in sign = two consecutive coefficients have opposite signs

EXAMPLE: Using Descartes Rule of Signs Determine the possible number of positive and negative real zeros of f (x) x 3 2x 2 5x + 4. Solution 1. To find possibilities for positive real zeros, count the number of sign changes in the equation for f (x). Because all the terms are positive, there are no variations in sign. Thus, there are no positive real zeros. 2. To find possibilities for negative real zeros, count the number of sign changes in the equation for f ( x). We obtain this equation by replacing x with x in the given function. f (x) x 3 2x 2 5x + 4 This is the given polynomial function. Replace x with x. f ( x) ( x) 3 2( x) 2 5 x 4 x 3 2x 2 5x + 4

EXAMPLE: Using Descartes Rule of Signs Determine the possible number of positive and negative real zeros of f (x) x 3 2x 2 5x + 4. Solution Now count the sign changes. f ( x) x 3 2x 2 5x + 4 1 2 3 There are three variations in sign. # of negative real zeros of f is either equal to 3, or is less than this number by an even integer. This means that there are either 3 negative real zeros or 3 2 1 negative real zero.

Descartes s Rule of Signs EXAMPLES: describe the possible real zeros f (x) 3x 3 5x 2 6x 4 f (x) 3x 3 2x 2 x 3

Upper & Lower Bound Rules Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x c, using synthetic didvision If c > 0 and each number in the last row is either positive or zero, c is an upper bound for the real zeros of f If c < 0 and the numbers in the last row are alternately positive and negative (zero entries count as positive or negative), c is a lower bound for the real zeros of f EXAMPLE: find the real zeros f (x) 6x 3 4x 2 3x 2

h(x) = x 4 + 6x 3 + 10x 2 + 6x + 9 Factors of9 1, 3, 9 Factors of 1 1 1 1 6 10 6 9 2 4 6 0 1 4 6 0 9 Signs are all positive, therefore 1 is an upper bound.

EXAMPLE You are designing candle-making kits. Each kit contains 25 cubic inches of candle wax and a model for making a pyramid-shaped candle. You want the height of the candle to be 2 inches less than the length of each side of the candle s square base. What should the dimensions of your candle mold be?