Splash Screen.

Slides:



Advertisements
Similar presentations
Aim: How do we solve polynomial equations using factoring?
Advertisements

Splash Screen. Lesson Menu Five-Minute Check (over Lesson 9–4) CCSS Then/Now New Vocabulary Key Concept: The Quadratic Formula Example 1:Use the Quadratic.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 5–4) CCSS Then/Now New Vocabulary Key Concept: Sum and Difference of Cubes Example 1:Sum and.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–4) CCSS Then/Now New Vocabulary Example 1:Use the Distributive Property Key Concept: Factoring.
Using the Distributive Property Lesson 8-5 Splash Screen.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–2) CCSS Then/Now New Vocabulary Example 1: The Distributive Property Key Concept: FOIL Method.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 5–5) CCSS Then/Now New Vocabulary Key Concept: Remainder Theorem Example 1:Synthetic Substitution.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 6–5) Then/Now New Vocabulary Key Concept: Remainder Theorem Example 1:Synthetic Substitution.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 4–6) CCSS Then/Now New Vocabulary Example 1:Write Functions in Vertex Form Example 2:Standardized.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 5–7) CCSS Then/Now Key Concept: Rational Zero Theorem Example 1:Identify Possible Zeros Example.
Splash Screen.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–3) Then/Now New Vocabulary Example 1: Solve a Logarithmic Equation Key Concept: Property of.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 10–5) CCSS Then/Now New Vocabulary Example 1:Real-World Example: Use Pascal’s Triangle Key Concept:
Splash Screen. Concept Example 1 Identify Possible Zeros A. List all of the possible rational zeros of f(x) = 3x 4 – x Answer:
The Remainder and Factor Theorems
7.4 THE REMAINDER & FACTOR THEOREMS Objectives: The student will be able to… 1)evaluate functions using synthetic substitution 2)determine whether a binomial.
Ch. 6.3 Dividing Polynomials. Divide x 2 + 2x – 30 by x – 5. ALGEBRA 2 LESSON 6-3 Dividing Polynomials – 30Subtract: (x 2 + 2x) – (x 2 – 5x) = 7x. Bring.
Five-Minute Check (over Lesson 2-2) Then/Now New Vocabulary
Splash Screen. Over Lesson 8–4 5-Minute Check 1 A.16x B. 16x x + 25 C. 16x x + 25 D. 4x x + 5 Find (4x + 5) 2.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 11–2) CCSS Then/Now New Vocabulary Example 1:Find Excluded Values Example 2:Real-World Example:
Splash Screen.
5.6 The Remainder and Factor Theorems. [If you are dividing by (x - 6), the remainder will be the same as if you were evaluating the polynomial using.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 5–2) CCSS Then/Now New Vocabulary Example 1:Degrees and Leading Coefficients Example 2:Real-World.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 8–2) CCSS Then/Now New Vocabulary Example 1: The Distributive Property Key Concept: FOIL Method.
Then/Now You factored quadratic expressions to solve equations. (Lesson 0–3) Divide polynomials using long division and synthetic division. Use the Remainder.
Splash Screen Unit 8 Quadratic Expressions and Equations EQ: How do you use addition, subtraction, multiplication, and factoring of polynomials in order.
Algebra 2 Divide x 2 + 2x – 30 by x – 5. Lesson 6-3 Dividing Polynomials – 30Subtract: (x 2 + 2x) – (x 2 – 5x) = 7x. Bring down –30. xDivide = x. x – 5.
Splash Screen Unit 8 Quadratic Expressions and Equations EQ: How do you use addition, subtraction, multiplication, and factoring of polynomials in order.
Then/Now You used the Distributive Property and factoring to simplify algebraic expressions. Evaluate functions by using synthetic substitution. Determine.
Splash Screen Unit 8 Quadratic Expressions and Equations EQ: How do you use addition, subtraction, multiplication, and factoring of polynomials in order.
Splash Screen. Lesson Menu Five-Minute Check (over Lesson 4–4) CCSS Then/Now New Vocabulary Example 1:Equation with Rational Roots Example 2:Equation.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Name:__________ warm-up 5-6
LESSON 8–6 Solving x2 + bx + c = 0.
Solve 25x3 – 9x = 0 by factoring.
Splash Screen.
Five-Minute Check (over Lesson 3–4) Mathematical Practices Then/Now
Splash Screen.
Splash Screen.
Splash Screen.
The Remainder and Factor Theorems
Splash Screen.
Splash Screen.
Chapter 7.4 The Remainder and Factor Theorems Standard & Honors
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
Splash Screen.
The Remainder and Factor Theorems
Splash Screen.
Splash Screen.
Splash Screen.
The Remainder and Factor Theorems
Splash Screen.
Presentation transcript:

Splash Screen

Five-Minute Check (over Lesson 5–5) CCSS Then/Now New Vocabulary Key Concept: Remainder Theorem Example 1: Synthetic Substitution Example 2: Real-World Example: Find Function Values Key Concept: Factor Theorem Example 3: Use the Factor Theorem Lesson Menu

Factor 8c3 – g3. If the polynomial is not factorable, write prime. A. (2c)(4c2 + cg + g2) B. (2c – g)(4c2 + 2cg + g2) C. (c – g)(2c + g + g2) D. prime 5-Minute Check 1

Factor 8c3 – g3. If the polynomial is not factorable, write prime. A. (2c)(4c2 + cg + g2) B. (2c – g)(4c2 + 2cg + g2) C. (c – g)(2c + g + g2) D. prime 5-Minute Check 1

Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx. If the polynomial is not factorable, write prime. A. (2a – 3z)(5x – c) B. (2a – 6z)(2a + b + c) C. (5x + 6z)(2a – b – c) D. prime 5-Minute Check 2

Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx Factor 12az – 6bz – 6cz + 10ax – 5bx – 5cx. If the polynomial is not factorable, write prime. A. (2a – 3z)(5x – c) B. (2a – 6z)(2a + b + c) C. (5x + 6z)(2a – b – c) D. prime 5-Minute Check 2

Factor 8x3m2 – 8x3n2 + y3m2 – y3n2. If the polynomial is not factorable, write prime. A. (8x + y)(m + n)(m – n) B. (4x + y)(2x – y2)(m + n)(m – n) C. (2x + y)(4x2 – 2xy + y2)(m + n)(m – n) D. prime 5-Minute Check 3

Factor 8x3m2 – 8x3n2 + y3m2 – y3n2. If the polynomial is not factorable, write prime. A. (8x + y)(m + n)(m – n) B. (4x + y)(2x – y2)(m + n)(m – n) C. (2x + y)(4x2 – 2xy + y2)(m + n)(m – n) D. prime 5-Minute Check 3

Solve 16d4 – 48d2 + 32 = 0. A. B. C. D. 5-Minute Check 4

Solve 16d4 – 48d2 + 32 = 0. A. B. C. D. 5-Minute Check 4

Solve k3 + 64 = 0. A. 16 B. 8 C. –2 D. –4 5-Minute Check 5

Solve k3 + 64 = 0. A. 16 B. 8 C. –2 D. –4 5-Minute Check 5

The width of a box is 3 feet less than the length The width of a box is 3 feet less than the length. The height is 4 feet less than the length. The volume of the box is 36 cubic feet. Find the length of the box. A. 2 ft B. 3 ft C. 4 ft D. 6 ft 5-Minute Check 6

The width of a box is 3 feet less than the length The width of a box is 3 feet less than the length. The height is 4 feet less than the length. The volume of the box is 36 cubic feet. Find the length of the box. A. 2 ft B. 3 ft C. 4 ft D. 6 ft 5-Minute Check 6

Mathematical Practices 7 Look for and make use of structure. Content Standards A.APR.2 Know and apply the Remainder Theorem: For a polynomial p(x) and a number a, the remainder on division by x – a is p(a), so p(a) = 0 if and only if (x – a) is a factor of p(x). F.IF.7.c Graph polynomial functions, identifying zeros when suitable factorizations are available, and showing end behavior. Mathematical Practices 7 Look for and make use of structure. CCSS

Evaluate functions by using synthetic substitution. You used the Distributive Property and factoring to simplify algebraic expressions. Evaluate functions by using synthetic substitution. Determine whether a binomial is a factor of a polynomial by using synthetic substitution. Then/Now

synthetic substitution depressed polynomial Vocabulary

Concept

If f(x) = 2x4 – 5x2 + 8x – 7, find f(6). Synthetic Substitution If f(x) = 2x4 – 5x2 + 8x – 7, find f(6). Method 1 Synthetic Substitution By the Remainder Theorem, f(6) should be the remainder when you divide the polynomial by x – 6. 2 12 67 410 2453 Notice that there is no x3 term. A zero is placed in this position as a placeholder. 2 0 –5 8 –7 12 72 402 2460 Answer: Example 1

If f(x) = 2x4 – 5x2 + 8x – 7, find f(6). Synthetic Substitution If f(x) = 2x4 – 5x2 + 8x – 7, find f(6). Method 1 Synthetic Substitution By the Remainder Theorem, f(6) should be the remainder when you divide the polynomial by x – 6. 2 12 67 410 2453 Notice that there is no x3 term. A zero is placed in this position as a placeholder. 2 0 –5 8 –7 12 72 402 2460 Answer: The remainder is 2453. Thus, by using synthetic substitution, f(6) = 2453. Example 1

Method 2 Direct Substitution Synthetic Substitution Method 2 Direct Substitution Replace x with 6. Original function Replace x with 6. Simplify. Answer: Example 1

Method 2 Direct Substitution Synthetic Substitution Method 2 Direct Substitution Replace x with 6. Original function Replace x with 6. Simplify. Answer: By using direct substitution, f(6) = 2453. Example 1

If f(x) = 2x3 – 3x2 + 7, find f(3). A. 20 B. 34 C. 88 D. 142 Example 1

If f(x) = 2x3 – 3x2 + 7, find f(3). A. 20 B. 34 C. 88 D. 142 Example 1

Find Function Values COLLEGE The number of college students from the United States who study abroad can be modeled by the function S(x) = 0.02x 4 – 0.52x 3 + 4.03x 2 + 0.09x + 77.54, where x is the number of years since 1993 and S(x) is the number of students in thousands. How many U.S. college students will study abroad in 2011? Answer: Example 2

Find Function Values COLLEGE The number of college students from the United States who study abroad can be modeled by the function S(x) = 0.02x 4 – 0.52x 3 + 4.03x 2 + 0.09x + 77.54, where x is the number of years since 1993 and S(x) is the number of students in thousands. How many U.S. college students will study abroad in 2011? Answer: In 2011, there will be about 451,760 U.S. college students studying abroad. Example 2

HIGH SCHOOL The number of high school students in the United States who hosted foreign exchange students can be modeled by the function F(x) = 0.02x 4 – 0.05x 3 + 0.04x 2 – 0.02x, where x is the number of years since 1999 and F(x) is the number of students in thousands. How many U.S. students will host foreign exchange students in 2013? A. 616,230 students B. 638,680 students C. 646,720 students D. 659,910 students Example 2

HIGH SCHOOL The number of high school students in the United States who hosted foreign exchange students can be modeled by the function F(x) = 0.02x 4 – 0.05x 3 + 0.04x 2 – 0.02x, where x is the number of years since 1999 and F(x) is the number of students in thousands. How many U.S. students will host foreign exchange students in 2013? A. 616,230 students B. 638,680 students C. 646,720 students D. 659,910 students Example 2

Concept

Use the Factor Theorem Determine whether x – 3 is a factor of x3 + 4x2 – 15x – 18. Then find the remaining factors of the polynomial. The binomial x – 3 is a factor of the polynomial if 3 is a zero of the related polynomial function. Use the factor theorem and synthetic division. 1 7 6 0 1 4 –15 –18 3 21 18 Example 3

x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Use the Factor Theorem Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Answer: Example 3

x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Use the Factor Theorem Since the remainder is 0, (x – 3) is a factor of the polynomial. The polynomial x3 + 4x2 – 15x –18 can be factored as (x – 3)(x2 + 7x + 6). The polynomial x2 + 7x + 6 is the depressed polynomial. Check to see if this polynomial can be factored. x2 + 7x + 6 = (x + 6)(x + 1) Factor the trinomial. Answer: So, x3 + 4x2 – 15x – 18 = (x – 3)(x + 6)(x + 1). Example 3

Use the Factor Theorem Check You can see that the graph of the related function f(x) = x3 + 4x2 – 15x – 18 crosses the x-axis at 3, –6, and –1. Thus, f(x) = (x – 3)[x – (–6)][x – (–1)].  Example 3

Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10 Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10. If so, find the remaining factors of the polynomial. A. yes; (x + 5)(x + 1) B. yes; (x + 5) C. yes; (x + 2)(x + 3) D. x + 2 is not a factor. Example 3

Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10 Determine whether x + 2 is a factor of x3 + 8x2 + 17x + 10. If so, find the remaining factors of the polynomial. A. yes; (x + 5)(x + 1) B. yes; (x + 5) C. yes; (x + 2)(x + 3) D. x + 2 is not a factor. Example 3

End of the Lesson