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

Power Functions Do Now: express in exponential form: Students will be able to determine the constant of variation and the power of power functions and recognize their graphs. Do Now: express in exponential form:

Power Functions Do Now: express in exponential form: Students will be able to determine the constant of variation and the power of power functions and recognize their graphs. Do Now: express in exponential form:

The parts of power functions State the constant of variation of each power function, and the power. Ex: f(x) = 3x constant of variation is 3, power is 1. In general, given K is the constant of variation, n is the power. If n>0, the function is a direct variation, if n<0, the function is an inverse variation

Examples State the constant of variation, the power, and whether the function is a direct or an inverse variation Give an example of a function that is not a power function

Examples State the constant of variation, the power, and whether the function is a direct or an inverse variation Give an example of a function that is not a power function Y=2x

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 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

General shapes of nth degree polynomials Limits of end behavior: Odd degree, positive Odd degree, negative Even degree, positive Even degree, negative

Odd degree polynomials rise on the left and fall on the right hand sides of the graph (like x3) if the coefficient is negative. turning points in the middle left hand behaviour: rises right hand behaviour: falls

Odd degree polynomials fall on the left and rise on the right hand sides of the graph (like x3) if the coefficient is positive. turning Points in the middle right hand behaviour: rises left hand behaviour: falls

Even degree polynomials fall on both the left and right hand sides of the graph (like - x4) if the coefficient is negative. turning points in the middle left hand behaviour: falls right hand behaviour: falls

Even degree polynomials rise on both the left and right hand sides of the graph (like x4) if the coefficient is positive. The additional terms may cause the graph to have some turns near the center but will always have the same left and right hand behaviour determined by the highest powered term. left hand behaviour: rises right hand behaviour: rises

End behavior Give the degree of the polynomial, and give the end behavior

End behavior Give the degree of the polynomial, and give the end behavior Degree = 3, neg:

Multiplicity The multiplicity of roots means how many repeated times that root is in the equation. Do Now: copy diagram 2.28 and it’s equation from page 198 of your textbook. Student will be able to determine the multiplicity of a root from it’s equation and it’s graph.

Determining Multiplicity One A root with muliplicity one occurs when the graph goes straight through the x- axis. In this graph, the roots x=2 and x=-2 have multiplicity one.

Determining Multiplicity Two A root with multiplicity two occurs when the graph touches the x- axis and “turns around”. Like the vertex of a parabola. In this graph the root x=1 has multiplicity 2.

Determining Multiplicity Three A root with multiplicity three has an inflection point on the x-axis. (It has a slight curve) In this graph the root x=-1 has a multiplicity of three.

Even and odd multiplicities If a multiplicity is 2, 4, 6,( even) the graph tangent to the x axis. Ex: 1 state the degree, the roots and their multiplicities If the multiplicity is 3, 5, 7, (odd) the graph flattens through the x axis. Ex: 2

Even and odd multiplicities If a multiplicity is 2, 4, 6, (even) the graph tangent to the x axis. Deg: 5 1 has mult. Of 1 6 has mult. Of 4 If the multiplicity is 3, 5, 7, (odd) the graph flattens through the x axis. Deg = 6 4 has mult of 1 -8 has mult of 5

Even and odd multiplicities If a multiplicity is 2, 4, 6, (even) the graph tangent to the x axis. Deg: 5 1 has mult. Of 1 6 has mult. Of 4 If the multiplicity is 3, 5, 7, (odd) the graph flattens through the x axis. Deg = 6 4 has mult of 1 -8 has mult of 5

Today we will find the equation of a polynomial function from its graph. Locate the x-intercepts (roots). Determine the multiplicity of each root. Write the polynomial in factored form. Solve for a, the constant of proportionality, by using another point on the graph. (don’t use a root)

Example – 6th degree polynomial function: Roots Mult: Factored form: f(x)=

Example – 6th degree polynomial function: Roots: -3, -1, 2 Mult: 1, 2, 3 Factored form: (use a) y=a(x+3)1(x+1)2(x-2)3 Use the point (0,-2): f(x)= a (x+3)(x+1)2(x-2)3

Example – 6th degree polynomial function: Roots: -3, -1, 2 Mult: 1, 2, 3 Factored form: (use a) y=a(x+3)1(x+1)2(x-2)3 Use the point (0,-2): -2=a(3)(1)2(-2)3 -2=-24a, so a= f(x)= (x+3)(x+1)2(x-2)3

Note that we needed a W shape Sketching a function: Note: we know it’s a positive, degree 4 function First, find the roots by graphing for now: -3, -1, 2, 5 We need a smooth curve through the intercepts that has the correct left and right hand behavior. Note that we needed a W shape (0,30)

x and y intercepts would be useful and we know how to find those x and y intercepts would be useful and we know how to find those. To find the y intercept we put 0 in for x. To find the x intercept we put 0 in for y. Finally we need a smooth curve through the intercepts that has the correct left and right hand behavior. To pass through these points, it will have 3 turns (one less than the degree so that’s okay) (0,30)

Synthetic division Set up for Summary Statement, for: Is (x-4) a factor of The root on the outside and the coefficients, including a zero placeholder: 1 2 0 -1 1

Synthetic division We multiply the bottom number to the root and add down….. 1 2 0 -1 4 24 96 1 6 24 95 The remainder is 95

Synthetic division Summary Statement, for (x-4) a factor of 1 2 0 -1 4 24 96 1 6 24 95 This is the quotient 1X2+6x=24

3 10 9 -19 6 10 Synthetic Division: Determine if 3 is a root of F(x)= 10 9 -19 6 Set Up with coefficients 10

Determine if 3 is a root 3 10 9 -19 6 Don’t forget your remainder should be zero 30 117 294 10 39 98 300 Remainder is 300, so 3 is not a root! The remainder is 300

Determining if -2 is a root-Synthetic Division: 10 9 -19 6 10

Determining if -2 is a root-Synthetic Division: 10 9 -19 6 Don’t forget your remainder should be zero YES! -20 22 -6 (No remainder) so -2 is a root 10 -11 3 0

Determining if -2 is a root-Synthetic Division: 10 9 -19 6 We can write a factorization here! -20 22 -6 10 -11 3 0

Summary Statement:is a factor 3 1 -4 0 -1 3 1

Summary Statement: 3 1 -4 0 -1 3 3 -3 -9 -30 1 -1 -3 -10 -27 1 -4 0 -1 3 Notice that the first term of the quotient is a 3rd degree-one less than the original function 3 -3 -9 -30 1 -1 -3 -10 -27

Rational Zero Test To use the Rational Zero Test, you should list all rational numbers whose numerators are factors of the constant term and whose denominators are factors of the 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) It only test for rational numbers

EXAMPLE: Using the Rational Zero Theorem List all possible rational zeros of f (x) = 15x3 + 14x2 - 3x – 2. Solution The constant term is –2 and the leading coefficient is 15. Divide 1 and 2 by 1. 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) = 15x3 + 14x2 - 3x – 2 = 0 is {-1, -1/3, 2/5}, which contains 3 of the 16 possible solutions.

Student will be able to list the potential roots of a polynomial function List the possible roots, and find one that works!

Rational Zeros Students will be able to find all zeros of higher degree polynomial functions by using synthetic division. Do Now: Find the rational zeros. You may use your calculator.

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 *Never start synthetic division over, do one after another!

Find all the zeros of each polynomial function Example 1: Find all the zeros of each polynomial function 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 -2 10 9 -19 6 -20 22 -6 10 -11 3 0 Second, use the zero you found from the graph and do synthetic division to find a smaller polynomial -2 10 9 -19 6 Don’t forget your remainder should be zero -20 22 -6 10 -11 3 0 The new, smaller polynomial is:

This quadratic can be factored into: (5x – 3)(2x – 1) Example 1 Continued: Third, factor or use the quadratic formula to find the remaining zeros. This quadratic can be factored into: (5x – 3)(2x – 1) Therefore, the zeros to the problem are:

Rational Zeros Find the rational zeros.

Rational Zeros Find the rational zeros. None -1,2 5 1 (1 has multiplicity of 2)

Find all the real zeros (Hint: start by finding the rational zeros)

Find all the real zeros (Hint: start by finding the rational zeros)

Writing a Factorization 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 P is a polynomial function and a + bi is a root, then a – bi is also a root. ***If P is a polynomial function and is a root, then is also a root

Given roots are 2 (mult of 2) and -5 f(x)=(x – 2)(x – 2)(x+5) First FOIL or box two of the factors Second, box your answer from above with your remaining factor to get your polynomial:

Given roots are 2 (mult of 2) and -5 f(x)= f(x)=(x – 2)(x – 2)(x+5) First FOIL or box two of the factors Second, box your answer from above with your remaining factors to get your polynomial: X 5 ANSWER checks

Example 1: The zeros of a fourth-degree polynomial are 1 (multiplicity 2) and Write the factorization as a product of irreducible linear and quadratic factors First, write the zeros in factored form f(x)=(x -1)(x -1)(x –2i)(x+2i) Second, multiply the factors out- two at a time

f(x)=(x2-2x+1)(x2+4) X2 4 First FOIL or box two of the factors Second, box your answer from above with your remaining factors to get your polynomial: X2 4

X2 4 First FOIL or box two of the factors Second, box your answer from above with your remaining factors to get your polynomial: X2 4 ANSWER checks

Example 2: The zeros of a third-degree polynomial are 3 (multiplicity 2) and 1/3. Write a polynomial. First, write the zeros in factored form (x – 3)(x – 3)(x – (1/3)) = (x – 3)(x – 3)(3x-1) Second, multiply the factors out to check your polynomial

If x = the root then x - the root is the factor form. 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. Multiply the last two factors together. All i terms should disappear when simplified. -1 Now multiply the x – 2 through Here is a 3rd degree polynomial with roots 2, 1 - 3i and 1 + 3i

The last problem has an alternate method… 2, 1+3i,1-3i Add the conjugate roots and multiply them. Then use: Sum = 2 Product = 10

f(x) =(x-2)(x2 - 2x + 10) The last problem has an alternate method… Add the conjugate roots and multiply them. Then use: f(x) =(x-2)(x2 - 2x + 10)

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 (with multiplicity of 2) and 3i, -3i 2, 4 + i,

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 (with multiplicity of 2) and 3i, -3i 2, 4 + i, 4 – I

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 (with multiplicity of 2) and 3i, -3i 2, 4 + i, 4 – i Sum = 8,pr=17

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

Find Roots/Zeros of a Polynomial If the known root is imaginary, we can use the Complex Conjugates Thm. Ex: Find all the roots of 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 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: X -4 -i

Example (con’t) Ex: Find all the roots of If one root is 4 - i. - - + If the product of the two non-real factors is then the third factor (that gives us the real root) is the quotient of P(x) divided by - + - + + - + - - + The third root is x = -3 So, all of the zeros are: 4 – i, 4 + i, and -3

FIND ALL THE ZEROS (Given that 1 + 3i is a zero of f)

FIND ALL THE ZEROS (Given that 1 + 3i is a zero of f) X= 3, -2, 1-3i, 1+3i (Given that 5 + 2i is a zero of f) X = 5+2i, 5 – 2i,-3

All complex roots…. Given -1 + 2i is a root, Find the rest…

All complex roots…. Given -1 + 2i is a root, -1 - 2i is also a root

All complex roots…. add the roots, Multiply the roots Using -2 and 5, we find the quadratic factor, x2 +2x +5 Use long division and divide it into the function…

All complex roots…. Given -1 + 2i is a root, Find the rest…

More Finding of Zeros

Descartes’s Rule of Signs Let be a polynomial with real coefficients and 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) = x3 + 2x2 + 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 + 2(-x)2 + 5(-x) + 4 f (x) = x3 + 2x2 + 5x + 4 This is the given polynomial function. Replace x with -x. = -x3 + 2x2 - 5x + 4

EXAMPLE: Using Descartes’ Rule of Signs Determine the possible number of positive and negative real zeros of f (x) = x3 + 2x2 + 5x + 4. Solution Now count the sign changes. f (-x) = -x3 + 2x2 - 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

Potential and actual Roots Let f(x) be a polynomial with real coefficients and a positive leading coefficient. Suppose f(x) is divided by x – c, using synthetic division Find all the potential roots of the function Use the factor theorem with the rational root to express as irreducible linear and quadratic factors find all the actual rational and complex zeros:

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

h(x) = x4 + 6x3 + 10x2 + 6x + 9 1 6 10 6 9 1 2 4 6 1 4 6 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?

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?