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Transcript of Holt McDougal Algebra 1 6-6 Special Products of Binomials 6-6 Special Products of Binomials Holt...
Holt McDougal Algebra 1
6-6 Special Products of Binomials6-6 Special Products of Binomials
Holt Algebra 1
Warm UpWarm Up
Lesson PresentationLesson Presentation
Lesson QuizLesson Quiz
Holt McDougal Algebra 1
Holt McDougal Algebra 1
6-6 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 McDougal Algebra 1
6-6 Special Products of Binomials
Find special products of binomials.
Objective
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Vocabularyperfect-square trinomialdifference of two squares
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
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
Use the rule for (a + b)2.
= 16s2 + 24st + 9t2
Identify a and b: a = 4s and b = 3t.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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
Use the rule for (a + b)2.
Identify a and b: a = 5 and b = m2.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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.
Use the rule for (a + b)2.
(x + 6)2 = x2 + 2(x)(6) + 62
= x2 + 12x + 36 Simplify.
b. (5a + b)2
(a + b)2 = a2 + 2ab + b2
Use the rule for (a + b)2.
(5a + b)2 = (5a)2 + 2(5a)(b) + b2
Identify a and b: a = 5a and b = b.
= 25a2 + 10ab + b2 Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Check It Out! Example 1C
Multiply.
c. (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 McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
Multiply.
Example 2: Finding Products in the Form (a – b)2
A. (x – 6)2
(a – b)2 = a2 – 2ab + b2
(x – 6)2 = x2 – 2x(6) + (6)2
= x2 – 12x + 36
Use the rule for (a – b)2.
Identify a and b: a = x and b = 6.
Simplify.
B. (4m – 10)2
(a – b)2 = a2 – 2ab + b2
(4m – 10)2 = (4m)2 – 2(4m)(10) + (10)2
= 16m2 – 80m + 100
Use the rule for (a – b)2.
Identify a and b: a = 4m and b = 10.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Multiply.
Example 2: Finding Products in the Form (a – b)2
C. (2x – 5y)2
(a – b)2 = a2 – 2ab + b2
(2x – 5y)2 = (2x)2 – 2(2x)(5y) + (5y)2
= 4x2 – 20xy +25y2
Use the rule for (a – b)2.
Identify a and b: a = 2x and b = 5y.
Simplify.
D. (7 – r3)2
(a – b)2 = a2 – 2ab + b2
(7 – r3)2 = 72 – 2(7)(r3) + (r3)2
= 49 – 14r3 + r6
Use the rule for (a – b)2.
Identify a and b: a = 7 and b = r3.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Check It Out! Example 2
Multiply.
a. (x – 7)2
(a – b)2 = a2 – 2ab + b2
(x – 7)2 = x2 – 2(x)(7) + (7)2
= x2 – 14x + 49
Use the rule for (a – b)2.
Identify a and b: a = x and b = 7.
Simplify.
b. (3b – 2c)2
(a – b)2 = a2 – 2ab + b2
(3b – 2c)2 = (3b)2 – 2(3b)(2c) + (2c)2
= 9b2 – 12bc + 4c2
Use the rule for (a – b)2.
Identify a and b: a = 3b and b = 2c.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Check It Out! Example 2c
Multiply.
(a2 – 4)2
(a – b)2 = a2 – 2ab + b2
(a2 – 4)2 = (a2)2 – 2(a2)(4) + (4)2
= a4 – 8a2 + 16
Use the rule for (a – b)2.
Identify a and b: a = a2 and b = 4.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
You can use an area model to see that: (a + b)(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.
Remove the 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 McDougal Algebra 1
6-6 Special Products of Binomials
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).
Identify a and b: a = x and b = 4.
Simplify.
B. (p2 + 8q)(p2 – 8q)
(a + b)(a – b) = a2 – b2
(p2 + 8q)(p2 – 8q) = (p2)2 – (8q)2
= p4 – 64q2
Use the rule for (a + b)(a – b).
Identify a and b: a = p2 and b = 8q.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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
Use the rule for (a + b)(a – b).
Identify a and b: a = 10 and b = b.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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
Use the rule for (a + b)(a – b).
Identify a and b: a = x and b = 8.
Simplify.
b. (3 + 2y2)(3 – 2y2)
(a + b)(a – b) = a2 – b2
(3 + 2y2)(3 – 2y2) = 32 – (2y2)2
= 9 – 4y4
Use the rule for (a + b)(a – b).
Identify a and b: a = 3 and b = 2y2.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Check It Out! Example 3
Multiply.
c. (9 + r)(9 – r)
(a + b)(a – b) = a2 – b2
(9 + r)(9 – r) = 92 – r2
= 81 – r2
Use the rule for (a + b)(a – b).
Identify a and b: a = 9 and b = r.
Simplify.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Write a polynomial that represents the area of the yard around the pool shown.
Example 4: Problem-Solving Application
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
22 Make a Plan
The area of the yard is (x + 5)2. 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 McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
Step 3 Find the area of the yard around the pool.
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 around the pool is 10x + 29.
Example 4 Continued
Solve33
Holt McDougal Algebra 1
6-6 Special Products of Binomials
To subtract a polynomial, add the opposite of each term.
Remember!
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Check It Out! Example 4
Write an expression that represents the area of the swimming pool.
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
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.
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Solve33
Step 1 Find the area of the upper rectangle.
Step 2 Find the area of the 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
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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
Check It Out! Example 4 Continued
Holt McDougal Algebra 1
6-6 Special Products of Binomials
Holt McDougal Algebra 1
6-6 Special Products of Binomials
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 McDougal Algebra 1
6-6 Special Products of Binomials
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