Chapter 2 notes from powerpoints

Synthetic division – and basic definitions Sections 1 and 2

No! No! Definition of a Polynomial Function: Let n be a nonnegative integer and let an, an-1, …, a2, a1, a0 be real numbers. The following function is called a polynomial function of x with degree n. No! No! In polynomial functions, THE EXPONENTS ON THE VARIABLE CANNOT BE FRACTIONS AND CANNOT BE NEGATIVE.

2.1 Polynomials Constant Term Linear Term Quadratic Term n-1 Term

Name of Polynomial Function Classifying Polynomials Polynomials are often classified by their degree. The degree of a polynomial is the highest degree of its terms. Degree Name of Polynomial Function Example Zero f(x) = -3 First f(x) = 2x + 5 Second f(x) = 3x2 –5x + 2 Third f(x) = x3 – 2x –1 Fourth f(x) = x4 –3x3 +7x-6 Fifth f(x) = 2x5 + 3x4 – x3+x2 Constant Linear Quadratic Cubic Quartic Quintic

Any value of x for which 𝑃 𝑥 =0 is a root of the equation and a zero of the function. 0=𝑥 2 −4=(𝑥+2)(𝑥−2) 𝑃 𝑥 = 𝑥 2 −4 ±2 are the zeros of the function

Main Ideas: Synthetic Division

Synthetic Division Main Ideas 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑥−2 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥) Is 𝑥−2 a factor? The factor theorem states that if a is a constant, then 𝑥−𝑎 is a factor of polynomial 𝑃 𝑥 if and only if 𝑃 𝑎 =0 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑥−2 𝑖𝑠 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑃(𝑥)

Synthetic Division Main Ideas Is 𝑥+2 a factor? 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒, 𝑥+2 𝑖𝑠 𝑛𝑜𝑡 𝑎 𝑓𝑎𝑐𝑡𝑜𝑟 𝑜𝑓 𝑆(𝑥) The factor theorem states that if a is a constant, then 𝑥−𝑎 is a factor of polynomial 𝑃 𝑥 if and only if 𝑃 𝑎 =0

Section 2.2 Synthetic Division; The Remainder and Factor Theorems Objective: To use synthetic division and to apply the remainder and factor theorems.

Vocabulary 4 is not a factor of 23, because if when 23 is divided by 4 the remainder is not zero. When 23 is divided by 4, the quotient is 5 and the remainder is 3.

The Remainder Theorem When a polynomial P(x) is divided by x – a, the remainder is P(a) Remainder =P(a)

The Factor Theorem   ⇔ if and only if

SECTion 3- graphing polynomials

Starts Up, Ends Up Starts Down, Ends Down

Starts Down, Ends Up Starts Up, Ends Down

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions Zeros: -1, 1, 2

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions

2.3 Graphing Polynomial Functions

Effect of A Squared Factor

Effect of A Squared Factor

Homework Section 2.2 Page 61 #1-25 odds

Effect of A Squared Factor

Effect of A Squared Factor

Effect of A Cubed Factor

Effect of A Cubed Factor

Effect of A Cubed Factor

Effect of A Cubed Factor

Graphing review For test review: 9/26

examples Determine end behavior and orientation I can graph polynomials examples Determine end behavior and orientation 𝐟 𝐱 = 𝟔𝒙 𝟕 −𝟑 𝒙 𝟑 + 𝒙 𝟐 −𝟏𝟐 𝐟 𝐱 =−𝟑 𝒙 𝟔 −𝒙+𝟏 Find the x- and y-intercepts 𝐲=𝟒 𝒙−𝟓 𝟐 (𝒙+𝟏)(𝟐𝒙−𝟓)

Warm Up 𝑥→−∞, 𝑓 𝑥 →−∞ 𝑥→∞, 𝑓 𝑥 →∞ 𝑥→−∞, 𝑓 𝑥 →−∞ 𝑥→∞, 𝑓 𝑥 →−∞ Determine end behavior and orientation 𝒇 𝒙 =𝟔𝒙 𝟕 −𝟑 𝒙 𝟑 + 𝒙 𝟐 −𝟏𝟐 𝐟 𝐱 =−𝟑 𝒙 𝟔 −𝒙+𝟏 𝑥→−∞, 𝑓 𝑥 →−∞ 𝑥→∞, 𝑓 𝑥 →∞ 𝑥→−∞, 𝑓 𝑥 →−∞ 𝑥→∞, 𝑓 𝑥 →−∞

Use Zero Product Property! Warm Up Find the x- and y-intercepts y=4 𝑥−5 2 (𝑥+1)(2𝑥−5) X-intercepts Use Zero Product Property! 𝑥−5=0 𝑥+1=0 2𝑥−5=0 𝑥=5 𝑥=−1 𝑥=5/2 y-intercept Plug in 0 for x y=4 0−5 2 0+1 2 0 −5 y=4 −5 2 (1)(−5) 𝑦=−500

Graphing Polynomials 𝑦= 𝑥−1 2 (𝑥+1) 𝒙 𝒚 −2 −1 1 2

Graphing Polynomials 𝑦= (𝑥−1) 2 (𝑥+1) 2 𝒙 𝒚 −2 −1 1 2

Graphing a Polynomial Steps Find the roots (x-intercepts) Example: 𝒚=𝟐 𝒙+𝟏 𝒙−𝟐 (𝒙+𝟖) Find the roots (x-intercepts) 2) Find the y-intercept 𝑥+1=0 𝑥−2=0 𝑥+8=0 𝑥=−1 𝑥=2 𝑥=−8 (−1,0) (2,0) (−8,0) 𝑦=2(0+1)(0−2)(0+8) 𝑦=2 1 −2 8 𝑦=−32 (0,−32)

Graphing a Polynomial As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, 𝑓 𝑥 →∞ Steps Example: 𝒚=𝟐 𝒙+𝟏 𝒙−𝟐 (𝒙+𝟖) 3) Determine orientation and end behavior Odd degree, positive leading term As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, 𝑓 𝑥 →∞

Graphing a Polynomial As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, 𝑓 𝑥 →∞ Steps 4) Graph Example: 𝒚=𝟐 𝒙+𝟏 𝒙−𝟐 (𝒙+𝟖) 4) Graph −1,0 (2,0) (−8,0) (0,−32) As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, 𝑓 𝑥 →∞

Graphing a Polynomial Steps Find the roots (x-intercepts) Example 2: 𝒚= −𝒙+𝟓 (𝒙−𝟏) 𝟐 𝒙+𝟐 Find the roots (x-intercepts) 2) Find the y-intercept −𝑥+5=0 𝑥−1=0 𝑥+2=0 𝑥=5 𝑥=1 𝑥=−2 (5,0) (1,0) (−2,0) 𝑦= −0+5 (0−1) 2 0+2 𝑦= 5 (−1) 2 2 𝑦=10 (0,10)

Even degree, negative leading term Graphing a Polynomial Steps Example 2: 𝒚= −𝒙+𝟓 (𝒙−𝟏) 𝟐 𝒙+𝟐 3) Determine orientation and end behavior Even degree, negative leading term As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, 𝑓 𝑥 →−∞

Graphing a Polynomial As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞, Steps 4) Graph 5,0 Example: 𝒚= −𝒙+𝟓 (𝒙−𝟏) 𝟐 𝒙+𝟐 4) Graph 5,0 (1,0) (−2,0) (0,10) As 𝑥→−∞, 𝑓 𝑥 →−∞ As 𝑥→∞,

Try on your own: Graphing a Polynomial Steps Example 3: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) Find the roots (x-intercepts) 2) Find the y-intercept −𝑥=0 𝑥+1=0 𝑥−3=0 𝑥=0 𝑥=−1 𝑥=3 (0,0) (−1,0) (3,0) 𝑦=0 0+1 0−3 𝑦= 0 (1) −3 𝑦=0 (0,0)

Graphing a Polynomial As 𝑥→−∞, 𝑓 𝑥 →∞ As 𝑥→∞, 𝑓 𝑥 →−∞ Steps Example: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) 3) Determine orientation and end behavior Odd degree, negative leading term As 𝑥→−∞, 𝑓 𝑥 →∞ As 𝑥→∞, 𝑓 𝑥 →−∞

Warm Up Graph: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) x – intercepts: y – intercept: I can graph a polynomial Warm Up Graph: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) x – intercepts: y – intercept: End Behavior:

Try on your own: Graphing a Polynomial Steps Example 3: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) Find the roots (x-intercepts) 2) Find the y-intercept −𝑥=0 𝑥+1=0 𝑥−3=0 𝑥=0 𝑥=−1 𝑥=3 (0,0) (−1,0) (3,0) 𝑦=0 0+1 0−3 𝑦= 0 (1) −3 𝑦=0 (0,0)

Graphing a Polynomial As 𝑥→−∞, 𝑓 𝑥 →∞ As 𝑥→∞, 𝑓 𝑥 →−∞ Steps Example: 𝒚=−𝒙 𝒙+𝟏 (𝒙−𝟑) 3) Determine orientation and end behavior Odd degree, negative leading term As 𝑥→−∞, 𝑓 𝑥 →∞ As 𝑥→∞, 𝑓 𝑥 →−∞

X – Intercepts: (0,0) (−1,0) 3,0 Y – Intercept:(0,0) End Behavior: Odd degree, negative leading term As 𝑥→−∞, 𝑓 𝑥 →∞ As 𝑥→∞, 𝑓 𝑥 →−∞

2.6 Solving Polynomial Functions by Factoring

Grouping best to use if: P(x) is a cubic. P(x) has 4 terms. Pair up terms. Factor within the terms Factor again.

Try it: 𝑥 3 +6 𝑥 2 −4𝑥−24=0 𝑥=−6, ±2

𝑥 3 +6 𝑥 2 −4𝑥−24=0 𝑥 3 +6 𝑥 2 + −4𝑥−24 =0 𝑥 2 𝑥+6 −4 𝑥+6 =0 𝑥 3 +6 𝑥 2 + −4𝑥−24 =0 𝑥 2 𝑥+6 −4 𝑥+6 =0 𝑥+6 𝑥 2 −4 =0 𝑥+6 𝑥+2 𝑥−2 =0 𝑥+6=0, 𝑥−2=0, 𝑥+2=0 𝑥=−6, 𝑥=2, 𝑥=−2

2 𝑥 4 − 𝑥 2 −3=0 Let 𝑢= 𝑥 2 Then: 2 𝑥 4 − 𝑥 2 −3=0, becomes: 2 𝑢 2 −𝑢−3=0 2𝑢−3 𝑢+1 =0 𝑢= 3 2 , 𝑢=−1 Remember 𝑢= 𝑥 2 , so solving for x: 3 2 = 𝑥 2 , 𝑥=± 3 2 = 3 2 = 3 2 2 2 =± 6 2 , and −1= 𝑥 2 𝑠𝑜 𝑥=±𝑖

When to use quadratic form The polynomial has 3 terms The degree is an even number (though this is not required) The exponent of the middle term is half the degree (necessary) The 3rd term is constant Examples: 𝑥 8 + 𝑥 4 +20, 𝑥 22 + 𝑥 11 +15, 𝑥 13 + 𝑥 6.5 −215

Try it: 2𝑥 4 =−7 𝑥 2 +15 𝑥=± 6 2 , ±𝑖 5

Classwork Class Exercises Page 60 #1-5

Homework Page 61 #5,7,13,19,23 Page 66 #3,9,11,13,15,21,23,27,29

Homework Page 83 #1, 3, 9, 10, and 13 – 34 every 3rd

Classwork Page 83 #1-6