Unit 4 Polynomials

Polynomial Graphs: Domain, Range, Zeros and Extrema

Vocabulary Domain: the x-values of the function Range: they y-values of the function Zeros: where the graph crosses the x- axis Extrema: Relative Maximum – the highest point in a particular section of the graph (like a hill) Relative Minimum - the lowest point in a particular section of the graph (like a valley) Vocabulary

f(x)=x2 + 2x Domain: (-∞, ∞) Range: [-1, ∞) Zeros: Relative Minimum: Relative Maximum: Domain: (-∞, ∞) Range: [-1, ∞) {-2, 0} (-1,-1) none

g(x) = -2x2 + x Zeros: Relative Minimum: Relative Maximum: {0, ½} none Domain: (-∞, ∞) Range: (-∞, ¼] {0, ½} none (¼, ¼)

h(x)= x3 - x Domain: (-∞, ∞) Range: Zeros: Relative Minimum: Relative Maximum: Domain: (-∞, ∞) Range: {-1, 0, 1} (½ ,- ½) (-½ , ½)

j(x) = -x3 +2x2 +3x Domain: (-∞, ∞) Range: Zeros: Relative Minimum: Relative Maximum: Domain: (-∞, ∞) Range: {-1, 0, 3} (-½ ,- 1) (2 , 6)

k(x)= x4 -5x2 + 4 Domain: (-∞, ∞) Range: [-2, ∞) Zeros: Relative Minimum: Relative Maximum: Domain: (-∞, ∞) Range: [-2, ∞) {-2, -1, 1, 2} (-1½ ,-2), (1½ ,-2) (0 , 4)

l(x) = -(x4 -5x2 + 4) Domain: (-∞, ∞) Range: (-∞, 2] Zeros: Relative Minimum: Relative Maximum: Domain: (-∞, ∞) Range: (-∞, 2] {-2, -1, 1, 2} (0,-4) (-1½ , 2), (1½ , 2)

Intervals: Vocabulary Increase: Where on the graph is the function increasing (going up) **Use interval notation** Decrease: Where on the graph is the function decreasing (going down) Vocabulary

Symmetry: A function is considered EVEN if it has symmetry about the y-axis. Only even degree polynomials with ALL EVEN exponents can have even symmetry. Ex: 𝑥 4 +3 𝑥 2 +6 Non-Ex: 𝑥 6 −3𝑥 A function is considered ODD is it has symmetry about the origin. Only odd degree polynomials with ALL ODD exponents can have odd symmetry. Ex: 𝑥 5 +3 𝑥 3 +𝑥 Non-Ex: 𝑥 3 −3𝑥+2 Vocabulary

f(x)=x2 + 2x Odd/Even/Neither Interval of increase: Interval of decrease: Odd/Even/Neither [-1,∞) ( -∞, -1]

g(x) = -2x2 + x Odd/Even/Neither Interval of increase: Interval of decrease: Odd/Even/Neither (-∞, ¼] [¼, ∞)

h(x)= x3 - x Intervals of increase: Interval of decrease: Odd/Even/Neither (- ∞, -½] , [½ , ∞) [- ½ , ½ ]

j(x) = -x3 +2x2 +3x Odd/Even/Neither Intervals of increase: Interval of decrease: Odd/Even/Neither [- ½ , 2] (- ∞, -½] , [ 2 ,∞)

k(x)= x4 -5x2 + 4 Odd/Even/Neither Intervals of increase: Interval of decrease: Odd/Even/Neither [- 1½ , 0] , [ 1½, ∞) (- ∞, -1½] , [ 0 ,1½)

l(x) = -(x4 -5x2 + 4) Intervals of increase: Interval of decrease: Odd/Even/Neither (- ∞, -1½] , [ 0 ,1½) [- 1½ , 0] , [ 1½, ∞)

Write the standard form of the equation for the polynomial function with the given zeroes 1, 3(multiplicity 2) f(x) = (x-1)(x-3)(x-3) f(x) = (x2 -3x – x +3)(x-3) f(x) = (x2 - 4x + 3)(x–3) f(x) = x3 - 3x2 - 4x2 + 12x +3x– 9 f(x) = x3 - 7x2+15x-9 Multiplicity means that a zero is used as a factor more than once

Long Division

Algebraic long division Divide 2x³ + 3x² - x + 1 by x + 2 x + 2 is the divisor 2x³ + 3x² - x + 1 is the dividend The quotient will be here. Algebraic long division

Algebraic long division First divide the first term of the dividend, 2x³, by x (the first term of the divisor). This gives 2x². This will be the first term of the quotient. Algebraic long division

Algebraic long division Now multiply 2x² by x + 2 and subtract Algebraic long division

Algebraic long division Bring down the next term, -x. Algebraic long division

Algebraic long division Now divide –x², the first term of –x² - x, by x, the first term of the divisor which gives –x. Algebraic long division

Algebraic long division Multiply –x by x + 2 Algebraic long division and subtract

Algebraic long division Bring down the next term, 1 Algebraic long division

Algebraic long division Divide x, the first term of x + 1, by x, the first term of the divisor Algebraic long division which gives 1

Algebraic long division Multiply x + 2 by 1 and subtract

Algebraic long division The quotient is 2x² - x + 1 The remainder is –1. Algebraic long division

Algebraic long division Write the quotient as:

Synthetic Division

Synthetic Division Another way to divide certain polynomials The divisor must be a linear (exponent=1) binomial with leading coefficient 1 For example, x-5 Synthetic Division

Synthetic Division Example: Think about how you would set up the long division problem Make sure that the dividend is in descending order and that no terms are missing You may have to include some 0 coefficients if terms are missing Synthetic Division

Remove the variables from the dividend leaving the coefficients Don’t forget any 1’s or 0’s that may be there Set up the multiplier Use the divisor and solve it for zero 2 3 -1 1 -2 Synthetic Division

Synthetic Division -2 -4 2 -2 -1 1 -1 Leave a blank line below the coefficients, and underline it. Bring down the first coefficient Use the multiplier, and place the product below the next coefficient. Add Repeat 2 3 -1 1 2 -2 -4 2 -2 -1 1 -1 Synthetic Division

Synthetic Division -2 -4 2 -2 -1 1 -1 Now, use the bottom line to write the quotient as a polynomial Place the variables back into the problem The first coefficient has degree that is one less than in the dividend The last number is the remainder 2 3 -1 1 2 -2 -4 2 -2 -1 1 -1 Synthetic Division

2 -3 4 -1 -1 -2 5 -9 2 -5 9 -10 Example 2:

1 0 0 -3 5 4 4 16 64 244 1 4 16 61 249 Example 3:

1 0 0 0 0 -1 1 1 1 1 1 1 1 1 1 1 1 Example 4:

Rational Root Theorem

To find the zeroes of a Polynomial: The Rational Root Theorem: If f(x) = anxn + . . . + a1x + a0 has integer coefficients, then every rational zero of f(x) has the following form: p = factor of constant term a0 q factor of leading coefficient an To find the zeroes of a Polynomial:

What is the constant? -10 This is going to be the numerator What is the leading coefficient? 2 This is going to be the denominator Possible roots: Use the rational root theorem to determine all possible rational roots:

The degree of the polynomial will tell you the max number of zeroes Some roots may be imaginary Use factors of the constant term (a0) as your numerator Use factors of the leading coefficient (an) as your denominator Use your calculator and synthetic division to determine if the possible ratios are actual zeroes Things to consider:

Find the rational zeroes of: Possible: Use your calculator to graph the function and determine which possible zeroes to try. Which ones look like they may work? Use synthetic division to check them Find the rational zeroes of:

Actual Solutions 2 -3/2 1/4 Try 2 2 8 -6 -23 6 2 8 -6 -23 6 16 20 -6 8 10 -3 0 now use, the smaller polynomial: Try -3/2 -3/2 8 10 -3 -12 3 8 -2 0

Now, you can give the factored form of the equation The zeroes for are { 2, -3/2, ¼ } So, the factored form of the equation is: Now, you can give the factored form of the equation

Find the rational zeroes of: Possible: Use your calculator to graph the function and determine which possible zeroes to try. Which ones look like they may work? Use synthetic division to check them Find the rational zeroes of:

Actual Solutions 3 -3/2 1/3 Try 3 3 6 -11 -24 9 3 6 -11 -24 9 18 21 -9 6 7 -3 0 now use, the smaller polynomial: Try -3/2 -3/2 6 7 -3 -9 3 6 -2 0

Now, you can give the factored form of the equation The zeroes for are { 3, -3/2, 1/3 } So, the factored form of the equation is: Now, you can give the factored form of the equation

Solving Polynomial Equations

The degree of the polynomial determines the number of zeros/roots/solutions the polynomial has. These zeros may be real or complex. Quadratic Equation: 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Things to Remember…

Steps Example Use the Rational Root Theorem to find all the rational roots of the polynomial. 𝑥 4 −14 𝑥 2 +45=0 Possible Roots: ±1, ±3, ±5, ±9,±15, ±45 Try: 3, −3 3 1 0 -14 0 45 Example 1

Steps Example Use the remaining polynomial and the quadratic formula to find the last two roots. 𝑥 2 −5=0 𝑎=1 𝑏=0 𝑐=−5 𝑥= −0± 0 2 −4(1)(−5) 2(1) 𝑥= ± 20 2 = ±2 5 2 =± 5 𝑥= −𝑏± 𝑏 2 −4𝑎𝑐 2𝑎 Example 1

The solutions of 𝑥 4 −14 𝑥 2 +45=0 are {−3, − 5 , 5 , 3}

Solving Polynomial Inequalities

Use the Rational Root Theorem and the Quadratic Formula to solve a polynomial. With inequalities, we can use a “test point” to determine the validity of a value. Things to remember…

Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0 Steps Example Set the polynomial equal to zero. Solve the polynomial. 𝑥 3 +3 𝑥 2 −𝑥−3=0 𝑥={−3, 1, −1}

Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0 Use the solutions to break up the number line. -3 -1 1

Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0 Plug a “test point” from each section of the number in to the original inequality to test the validity of the value. -5 -3 -2 -1 1 2

Example 1: Solve 𝑥 3 +3 𝑥 2 −𝑥−3>0 Use the information from step 4 to write the solution to the inequality in interval notation. −3, −1 𝑎𝑛𝑑 (1, ∞) -5 -3 -2 -1 1 2