holt algebra 1 7-8 special products of binomials 7-8 special products of binomials holt algebra 1...

Holt Algebra 1

7-8 Special Products of Binomials7-8 Special Products of Binomials

Holt Algebra 1

Warm UpWarm Up

Lesson PresentationLesson Presentation

Lesson QuizLesson Quiz

Holt Algebra 1

7-8 Special Products of Binomials

Warm UpSimplify.

1. 42

3. (–2)2 4. (x)2

5. –(5y2)

16 49

4 x2

2. 72

6. (m2)2 m4

7. 2(6xy) 2(8x2)8. 16x2

–25y2

12xy

Holt Algebra 1


Find special products of binomials.

Objective

Holt Algebra 1


Vocabularyperfect-square trinomialdifference of two squares

Holt Algebra 1


Imagine a square with sides of length (a + b):

The area of this square is (a + b)(a + b) or (a + b)2. The area of this square can also be found by adding the areas of the smaller squares and the rectangles inside. The sum of the areas inside is a2 + ab + ab + b2.

Holt Algebra 1


This means that (a + b)2 = a2+ 2ab + b2.You can use the FOIL method to verify this:

(a + b)2 = (a + b)(a + b) = a2 + ab + ab + b2

F L

I

O = a2 + 2ab + b2

A trinomial of the form a2 + 2ab + b2 is called a perfect-square trinomial. A perfect-square trinomial is a trinomial that is the result of squaring a binomial.

Holt Algebra 1


Multiply.

Example 1: Finding Products in the Form (a + b)2

A. (x +3)2

(a + b)2 = a2 + 2ab + b2

Use the rule for (a + b)2.

(x + 3)2 = x2 + 2(x)(3) + 32

= x2 + 6x + 9

Identify a and b: a = x and b = 3.

Simplify.

B. (4s + 3t)2

(a + b)2 = a2 + 2ab + b2

(4s + 3t)2 = (4s)2 + 2(4s)(3t) + (3t)2


= 16s2 + 24st + 9t2

Identify a and b: a = 4s and b = 3t.

Simplify.

Holt Algebra 1


Multiply.

Example 1C: Finding Products in the Form (a + b)2

C. (5 + m2)2

(a + b)2 = a2 + 2ab + b2

(5 + m2)2 = 52 + 2(5)(m2) + (m2)2

= 25 + 10m2 + m4


Identify a and b: a = 5 and b = m2.

Simplify.

Holt Algebra 1


Check It Out! Example 1C

Multiply.

(1 + c3)2 Use the rule for (a + b)2.

(a + b)2 = a2 + 2ab + b2 Identify a and b: a = 1 and b = c3.

(1 + c3)2 = 12 + 2(1)(c3) + (c3)2

= 1 + 2c3 + c6 Simplify.

Holt Algebra 1


You can use the FOIL method to find products in the form of (a – b)2.

(a – b)2 = (a – b)(a – b) = a2 – ab – ab + b2

F L

IO = a2 – 2ab + b2

A trinomial of the form a2 – ab + b2 is also a perfect-square trinomial because it is the result of squaring the binomial (a – b).

Holt Algebra 1


Multiply.

Example 2: Finding Products in the Form (a – b)2

A. (x – 6)2

(a – b) = a2 – 2ab + b2

(x – 6) = x2 – 2x(6) + (6)2

= x2 – 12x + 36

Use the rule for (a – b)2.


Simplify.

B. (4m – 10)2

(a – b) = a2 – 2ab + b2

(4m – 10) = (4m)2 – 2(4m)(10) + 102

= 16m2 – 80m + 100


Identify a and b: a = 4m and b = 10.

Simplify.

Holt Algebra 1


Multiply.

Example 2: Finding Products in the Form (a – b)2

C. (2x – 5y )2

(a – b) = a2 – 2ab + b2

(2x – 5y)2 = (2x)2 – 2(2x)(5y) + (5y)2

= 2x2 – 20xy +25y2


Identify a and b: a = 2x and b = 5y.

Simplify.

D. (7 – r3)2

(a – b) = a2 – 2ab + b2

(7 – r3) = 72 – 2(7)(r3) + (r3)2

= 49 – 14r3 + r6


Identify a and b: a = 7 and b = r3.

Simplify.

Holt Algebra 1


You can use an area model to see that (a + b) = a2 – b2.

Begin with a square with area a2. Remove a square with area b2. The area of the new figure is a2 – b2.

Then remove the smaller rectangle on the bottom. Turn it and slide it up next to the top rectangle.

The new arrange- ment is a rectangle with length a + b and width a – b. Its area is (a + b)(a – b).

So (a + b)(a – b) = a2 – b2. A binomial of the form a2 – b2 is called a difference of two squares.

Holt Algebra 1


Multiply.

Example 3: Finding Products in the Form (a + b)(a – b)

A. (x + 4)(x – 4)

(a + b)(a – b) = a2 – b2

(x + 4)(x – 4) = x2 – 42

= x2 – 16

Use the rule for (a + b)(a – b).


Simplify.

B. (p2 + 8q)(p2 – 8q)

(a + b)(a – b) = a2 – b2

(p2 + 8q)(p2 – 8q) = p4 – (8q)2

= p4 – 64q2


Identify a and b: a = p2 and b = 8q.

Simplify.

Holt Algebra 1


Multiply.

Example 3: Finding Products in the Form (a + b)(a – b)

C. (10 + b)(10 – b)

(a + b)(a – b) = a2 – b2

(10 + b)(10 – b) = 102 – b2

= 100 – b2


Identify a and b: a = 10 and b = b.

Simplify.

Holt Algebra 1


Check It Out! Example 3

Multiply.a. (x + 8)(x – 8)

(a + b)(a – b) = a2 – b2

(x + 8)(x – 8) = x2 – 82

= x2 – 64



Simplify.

b. (3 + 2y2)(3 – 2y2)

(a + b)(a – b) = a2 – b2

(3 + 2y2)(3 – 2y2) = 32 – (2y2)2

= 9 – 4y4


Identify a and b: a = 3 and b = 2y2.

Simplify.

Holt Algebra 1


Write a polynomial that represents the area of the yard around the pool shown below.

Example 4: Problem-Solving Application

Holt Algebra 1


List important information:• The yard is a square with a side length of x + 5.• The pool has side lengths of x + 2 and x – 2.

11 Understand the Problem

The answer will be an expression that shows the area of the yard less the area of the pool.

Example 4 Continued

Holt Algebra 1


22 Make a Plan

The area of the yard is (x + 5)2 less the area of the pool. The area of the pool is (x + 2)(x – 2). You can subtract the area of the pool from the yard to find the area of the yard surrounding the pool.

Example 4 Continued

Holt Algebra 1


Solve33

Step 1 Find the total area.

(x +5)2 = x2 + 2(x)(5) + 52 Use the rule for (a + b)2: a = x and b = 5.

Step 2 Find the area of the pool.

(x + 2)(x – 2) = x2 – 2x + 2x – 4 Use the rule for (a + b)(a – b): a = x and b = 2.

x2 + 10x + 25=

x2 – 4 =

Example 4 Continued

Holt Algebra 1


Step 3 Find the area of the yard.

Area of yard = total area – area of pool

a = x2 + 10x + 25 (x2 – 4) –

= x2 + 10x + 25 – x2 + 4

= (x2 – x2) + 10x + ( 25 + 4)

= 10x + 29

Identify like terms.

Group like terms together

The area of the yard is 10x + 29.

Example 4 Continued

Solve33

Holt Algebra 1


Look Back44

Suppose that x = 20. Then the total area in the back yard would be 252 or 625. The area of the pool would be 22 18 or 396. The area of the yard around the pool would be 625 – 396 = 229.

According to the solution, the area of the pool is 10x + 29. If x = 20, then 10x +29 = 10(20) + 29 = 229.

Example 4 Continued

Holt Algebra 1


To subtract a polynomial, add the opposite of each term.

Remember!

Holt Algebra 1


Check It Out! Example 4

Write an expression that represents the area of the swimming pool.

Holt Algebra 1


Check It Out! Example 4 Continued

11 Understand the Problem

The answer will be an expression that shows the area of the two rectangles combined.

List important information:• The upper rectangle has side lengths of 5 + x

and 5 – x .• The lower rectangle is a square with side

length of x.

Holt Algebra 1


22 Make a Plan

The area of the upper rectangle is (5 + x)(5 – x). The area of the lower square is x2. Added together they give the total area of the pool.


Holt Algebra 1


Solve33

Step 1 Find the area of the upper rectangle.

Step 2 Find the area lower square.

(5 + x)(5 – x) = 25 – 5x + 5x – x2 Use the rule for (a + b)

(a – b): a = 5 and b = x.–x2 + 25=

x2 =

= x x


Holt Algebra 1


Step 3 Find the area of the pool.

Area of pool = rectangle area + square area

a x2 +

= –x2 + 25 + x2

= (x2 – x2) + 25

= 25

–x2 + 25=

Identify like terms.

Group like terms together

The area of the pool is 25.

Solve33


Holt Algebra 1


Look Back44

Suppose that x = 2. Then the area of the upper rectangle would be 21. The area of the lower square would be 4. The area of the pool would be 21 + 4 = 25.

According to the solution, the area of the pool is 25.


Holt Algebra 1


Holt Algebra 1


Lesson Quiz: Part I

Multiply.

1. (x + 7)2

2. (x – 2)2

3. (5x + 2y)2

4. (2x – 9y)2

5. (4x + 5y)(4x – 5y)

6. (m2 + 2n)(m2 – 2n)

x2 – 4x + 4

x2 + 14x + 49

25x2 + 20xy + 4y2

4x2 – 36xy + 81y2

16x2 – 25y2

m4 – 4n2

Holt Algebra 1


Lesson Quiz: Part II

7. Write a polynomial that represents the shaded area of the figure below.

14x – 85

x + 6

x – 6x – 7

x – 7

holt algebra 1 7-8 special products of binomials 7-8 special products of binomials holt algebra 1...

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