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McGraw Hill Integrated II, 2012

MH

McGraw Hill Integrated II, 2012 View details

Study Guide and Review

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Exercise 34 Page 82

To find the area of a shaded region, subtract the area of the white region from the area of the whole.

A=3x^2 - 21

Practice makes perfect

When we want to find the area of a shaded region, we need to subtract the area of the smaller white region from the area of the whole region. Let's start with the area of the whole region. It's length is 2x+5 and width is 2x-5.

A = l * w

l= 2x+5, w= 2x-5

A = ( 2x+5)( 2x-5)

▼

Simplify right-hand side

ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

A = (2x)^2 - (5)^2

Calculate power

A = 4x^2 - 25

Now, let's find the area of the white region with its length x+2 and width x-2.

A = l * w

l= x+2, w= x-2

A = ( x+2)( x-2)

▼

Simplify right-hand side

ExpandDiffSquares

(a+b)(a-b)=a^2-b^2

A = (x)^2 - (2)^2

Calculate power

A = x^2 - 4

From here, we can subtract the white area from the whole area. cc A_(Shaded Region)&= &A_(Whole)&-& A_(White) & & ⇓ && ⇓ &=& (4x^2 - 25) &-& (x^2 - 4) Let's simplify that expression.

A = (4x^2 - 25) - (x^2 - 4)

▼

Simplify right-hand side

Distribute - 1

A = 4x^2-25-x^2+4

CommutativePropAdd

Commutative Property of Addition

A = 4x^2-x^2 -25+4

Subtract terms

A = 3x^2 - 25+4

Add terms

A = 3x^2 - 21

Subchapter links

1 Adding and Subtracting Polynomials p.81

2 Multiplying a Polynomial by a Monomial p.81

3 Multiplying Polynomials p.81

4 Special Products p.82

5 Using the Distributive Property p.82

6 Solving x^2+bx+c=0 p.83

7 Solving ax^2+bx+c=0 p.83

8 Differences of Squares p.84

9 Perfect Squares p.84

10 Roots and Zeros p.85