Algebra unit 8.4

8.4 MULTIPLYING SPECIAL CASES8.4 MULTIPLYING SPECIAL CASES

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

Find special products of binomials.

Objective

Vocabularyperfect-square trinomialdifference of two squares

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.

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.

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.

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.

Check It Out! Example 1

Multiply.A. (x + 6)2

(a + b)2 = a2 + 2ab + b2 Identify a and b: a = x and b = 6.


(x + 6)2 = x2 + 2(x)(6) + 62

= x2 + 12x + 36 Simplify.

B. (5a + b)2

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


(5a + b)2 = (5a)2 + 2(5a)(b) + b2

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

= 25a2 + 10ab + b2 Simplify.

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.

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).

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

= x – 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.

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.


Multiply.a. (x – 7)2

(a – b) = a2 – 2ab + b2

(x – 7)2 = x2 – 2(x)(7) + (7)2

= x2 – 14x + 49



Simplify.

b. (3b – 2c)2

(a – b) = a2 – 2ab + b2

(3b – 2c)2 = (3b)2 – 2(3b)(2c) + (2c)2

= 9b2 – 12bc + 4c2


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

Simplify.

Check It Out! Example 2c

Multiply. (a2 – 4)2

(a – b) = a2 – 2ab + b2

(a2 – 4)2 = a4 – 2(a2)(4) + (4)2

= a4 – 8a2 + 16


Identify a and b: a = a2 and b = 4.

Simplify.

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.

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.

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.


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.


Multiply.

c. (9 + r)(9 – r)

(a + b)(a – b) = a2 – b2

(9 + r)(9 – r) = 92 – r2

= 81 – r2


Identify a and b: a = 9 and b = r.

Simplify.

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

Example 4: Problem-Solving Application

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

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

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

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

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

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

Remember!


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

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.

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.


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•


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


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.


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

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

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Algebra unit 8.4

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Transcript of Algebra unit 8.4