section 5.1 multiplying polynomials

Name: _______________________ Period: ______________________

Reflection: 1

Section 5.1 Multiplying Polynomials Objective(s): Multiply polynomials.

Essential Question: What is your favorite method for multiply polynomials? Why?

Homework: Assignment 5.1. #1 – 20 in the homework packet.

Notes:

Vocabulary

A monomial is a polynomial with one term.

A binomial is a polynomial with two terms.

Multiplying Polynomials

Monomials

Ex: Find the product. 2 24 3 1x x x

Multiply each term of the trinomial by the monomial.

2 2 2 2

4 3 2

4 4 3 4 1

4 12 4

x x x x x

x x x

Binomials

Method 1: To multiply by a binomial, use the distributive property twice, then combine like terms.

Ex: Find the product. 23 5 2 8 3x x x

Distributive property:

2 2

2 2

3 2 2

3 (2 8 3) 5(2 8 3)

3 2 3 8 3 3 5 2 5 8 5 3

6 24 9 10 40 15

x x x x x

x x x x x x x

x x x x x

Combine like terms: 3 26 14 31 15x x x

Reflection: 2

Method 2: Box method.

Ex: Find the product. 23 5 2 8 3x x x

2x2 8x 3

3x

-5

Multiply

Example 1: (x + 2)(x – 2) Example 2: (2x + 1)(2x – 1)

Example 3: (5t – 3)(5t + 3) Example 4: (a + b)(a – b)

Sum and Difference Pattern 2 2a b a b a b

Multiply using the pattern.

Example 5: (3t + 4)(3t – 4) Example 6: (4x – 7)(4x + 7)

Example 7: (8x + 3)(8x – 3) Example 8: (7x – 5)(7x + 5)

Reflection: 3

Multiply

Example 9: (x + 3)2 Example 10: (2x + 7)2

x 3

x

3

2x 7

2x

7

Example 11: (x – 5)2 Example 12: (a + b)2

x -5

x

-5

a b

a

b

Square of a Binomial Pattern

2 2 2

2 2 2

2

2

a b a ab b

a b a ab b

Multiply using the pattern

Example 13: (8y – 3)2 Example 14: (3w + 5)2

Example 15: (5w – 1)2 Example 16: (4y + 3)2

Example 17: (8t + 5)2 Example 18: (3w – 7)2

Reflection: 4

Sample CCSD Common Exam Practice Question(s):

1. Multiply the binomials: 2 3 3 1x x

A. 25 7 3x x

B. 25 5 3x x

C. 26 7 3x x

D. 26 5 3x x

2. Multiply the polynomials: 25 2 3 4x x x

A. 3 22 7 11 20x x x

B. 3 22 7 19 20x x x

C. 3 22 13 11 20x x x

D. 3 22 13 19 20x x x

3. Expand the expression 2

3 5x

A. 29 25x

B. 29 25x

C. 29 30 25x x

D. 29 30 25x x

Reflection: 5

Section 5.2 Greatest Common Factor and Factoring by Grouping Objective(s): Factor polynomials

Essential Question: How can you find the GCF of a variable expression without writing out all of the

variables as factors?


Notes:

Vocabulary

A quadratic equation is one that can be written in the form ax2 + bx + c = 0 where a, b, and c are real

numbers and a is not 0.

Factoring is the process of writing a polynomial as a product.

Factoring Using the Distributive Property

Ex: Factor the polynomial 22 8x x .

Step One: Find the GCF of the terms. GCF = 2x

Step Two: Use the distributive property to factor the GCF out of the polynomial. 2 4x x

Ex: Factor the polynomial 3 2 2 214 21 7x y x y x y .

Step One: Find the GCF of the terms. GCF = 7x2y

Step Two: Use the distributive property to factor the GCF out of the polynomial. 27 2 3 1x y x y

Note: You can check your answers by multiplying using the distributive property.

Factor out the GCF from the polynomial.

Example 1: 40x – 20 Example 2: 14x + 7y – 7

Reflection: 6

Example 3: x(y + 15) – 2(y + 15) Example 4: 8a(a – b) + 1(a – b)

Example 5: 2r(2q + 7) + 5(2q + 7) Example 6: 4x(9y2 + 11) + 3(9y2 + 11)

Factoring by Grouping

Ex: Factor the polynomial 3 22 3 6x x x .

Step One: Group the first two terms and last two terms. 3 22 3 6x x x

Step Two: Factor the GCF from both sets of terms. 22 3 2x x x

Step Three: Factor the common factor using the distributive property. 22 3x x

Ex: Factor the polynomial 4 28 7x xy y .

Step One: Group the first two terms and last two terms. 4 28 7x xy y

Step Two: Factor the GCF from both sets of terms. 4 7 7x y x

Step Three: Factor the common factor using the distributive property. 7 4x y

Factor the four-term polynomial by grouping.

Example 7: xy + 11x – 6y – 66

Example 8: 3y – 36 + xy – 12x

Example 9: x3 + 2x2 + x + 2

Example 10: x3 + 7x + x2 + 7

Reflection: 7

Section 5.3 Factoring Trinomials Objective(s): Factor polynomials.

Essential Question: Are all quadratic trinomials factorable? If not, write a trinomial that cannot be

factored.


Notes:

Factoring a Quadratic ax2 + bx + c = 0 (a = 1)

Ex: Factor the polynomial x2 + 5x + 4

Step One: Recall from multiplying that the x2 goes in the upper left hand box and the constant goes in

the lower right hand box.

x2

4

Step Two: Since x times x is the only way to get x2, put an x above the x2 and another x to the left.

x

x x2

4

Step Three: Recall that the two rectangles are filled with x terms.

Reflection: 8

The 5x from the original problem needs to be split between the two rectangles. You have a couple of

choices. You could put 1x in one rectangle and 4x in the other, or you could put 2x in one rectangle and

3x in the other. Keep in mind that the coefficients have to multiply to get 4, the constant. Since 2 times 3

does NOT equal 4, the correct choice would be 1x and 4x.

x

x x2 1x

4x 4

Filling in the rest of the box, you would get the following.

x +1

x x2 1x

+4 4x 4

Answer: (x + 1)(x + 4)

Quick check: Multiply the diagonals. What is x2 times 4? What is 4x times 1x?

Factor the polynomial completely.

Example 1: x2 + 11x + 24

Answer:

Quick Check:

Example 2: x2 + 7x + 10

Answer:

Quick Check:

Reflection: 9

Example 3: x2 + 7x + 12

Example 4: x2 – 10x + 9 (How is this one

different?)

Example 5: x2 – 14x + 40 Example 6: x2 – 8x + 15

Example 7: x2 + 6x – 27 (How is this one different?)

Example 8: x2 – 3x – 10 Example 9: x2 + 7x – 8

Reflection: 10

Factoring a Quadratic ax2 + bx + c = 0 (a 1)

Multiply (2x + 3)(x + 4)

2x 3

x 2x2 3x

4 8x 12

If you do a quick check, both diagonals equal 24x2.


Example 10: 3x2 + 17x + 10 Example 11: 2x2 + 9x + 10

Example 12: 5x2 + 14x + 8 Example 13: 7w2 + 9w – 10


8x2 – 18x + 9

What do you do with the 8? Do you put 1x and 8x or 2x and 4x?

Reflection: 11

Factoring a Quadratic - The a c Method

8x2 – 18x + 9

Step One: Multiply a c . 8 9 72

Step Two: Find two integers such that their product is 72a c and their sum is 18b .

6 and -12

Step Three: Rewrite (“split”) the middle term as a sum of two terms using the numbers from Step Two.

28 6 12 9x x x (order does not matter when splitting the middle term)

Step Four: Factor by grouping. Group the first 2 terms and last 2 terms and factor out the GCF from each

pair.

28 6 12 9 2 (4 3) 3(4 3)x x x x x x

Step Five: If Step Four was done correctly, there should be a common binomial factor. Factor this

binomial out and write what remains from each term as the second binomial factor.

4 3 2 3x x


Example 14: 6x2 + 11x + 3 Example 15: 12x2 + x – 6

Example 16: 8x2 – 11x + 3 Example 17: 8x2 – 18x – 5

Reflection: 12

Special Factoring Patterns: Memorize this!

Difference of Two Squares: 2 2a b a b a b

Ex: Factor 2 29 16x y .

This appears to be a difference of two squares, since each term is a perfect square. Rewrite each term as

a monomial squared then use the pattern to factor.

2 2

3 4 3 4 3 4x y x y x y


Example 18: z2 – 36 Example 19: 81y2 – 100

Example 20: 121 – w2 Example 21: 4x2 – 25

Factoring a GCF Monomial

Ex: Factor 272 50y completely.

Step One: Factor out the GCF of 2. 22 36 25y

Step Two: Factor the remaining polynomial. 2 6 5 6 5y y


Example 22: 16x2 – 56x – 32 Example 23: 8y2 + 36y - 20

Reflection: 13

Section 5.4 Solve by Factoring Objective(s): Solve quadratic equations by factoring.

Essential Question: What must be true about a quadratic equation before you can solve it using the zero

product property?


Notes:

Vocabulary

Standard Form of a Quadratic Equation: 20ax bx c

Zero Product Property: If the product of two factors is 0, then one or both of the factors must equal 0.

If a and b are real numbers and a ∙ b = 0, then a = 0 or b = 0.

Solving a Quadratic Equation by Factoring

Ex: Solve the equation 2 22 4 8x x x x .

Step One: Write the equation in standard form. 23 5 8 0x x

Step Two: Factor the quadratic using the “ac method”.

24

5

8 and 3

a c

b

23 8 3 8 0

3 8 1 3 8 0

3 8 1 0

x x x

x x x

x x

Step Three: Set each factor equal to zero and solve.

3 8 0 1 0

81

3

x x

x x

The solutions can be written in set notation: 8

1,3

Reflection: 14

Solve the equation.

Example 1: (x + 2)(x – 8) = 0 Example 2: (2x – 3)(x – 12) = 0

Example 3: x(x – 23) = 0 Example 4: (7x + 5)(9x – 4) = 0

Example 5: x2 + 8x – 20 = 0 Example 6: x2 – x = 30

Example 7: 5x2 – 3x – 8 = 0 Example 8: x2 + 3x = 0

Example 9: 9x2 – 25 = 0 Example 10: x2 – 18x + 81 = 0

section 5.1 multiplying polynomials

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