## LARSON ALGEBRA 2 CHAPTER 5, LESSON 2, EXTRA ... - ClassZone

Algebra 2, Chapter 5, Lesson 2, Extra Examples. 1. LARSON ALGEBRA 2. CHAPTER 5, LESSON 2, EXTRA EXAMPLES. Extra Example 1 Factoring Trinomials ...

LARSON ALGEBRA 2 CHAPTER 5, LESSON 2, EXTRA EXAMPLES

Extra Example 1 Factoring Trinomials of the Form x2 + bx + c Factor x2 º 2x º 48. SOLUTION

You want x2 º 2x º 48 = (x + m)(x + n) where m+ n = º 2 and mn = º 48. Find factors of º48. Taking m = º8 and n = 6 gives x2 º 2x º 48 = (x º8)(x + 6). Extra Example 2 Factoring a Trinomial of the Form ax2 + bx + c Factor 4y2 º 4y º 3. SOLUTION

You want 4y2 º 4y º 3 = (ky + m)(ly + n) where k and l are factors of 4 and m and n are factors of 3. Check possible factorizations by multiplying. (4y º 3)(y + 1) = 4y2 + y º 3 (4y º 1)(y + 3) = 4y2 + 11y º 3 (2y º 1)(2y + 3) = 4y2 + 4y º 3 (2y º 3)(2y + 1) = 4y2 º 4y º 3

➧ The correct factorization is 4y2 º 4y º 3 = (2y º 3)(2y + 1).

Extra Example 3 Factoring with Special Patterns Factor the quadratic expression. a. 16y2 º 225

= (4y)2 º (15)2

Difference of two squares

= (4y º 15)(4y + 15) b.

4z2 º 12z + 9 = (2z)2 + 2(2z)(º3) + (º3)2

Perfect square trinomial

= (2z º 3)2 c. 36w2 + 60w + 25 = (6w)2 + 2(6w)(5) + 52

Perfect square trinomial

= (6w + 5)2

Algebra 2, Chapter 5, Lesson 2, Extra Examples

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Extra Example 4 Factoring Monomials First Factor the quadratic expression. a. 14x2 + 2x º12 = 2(7x2 + x º6) = 2(7x º 6)(x + 1) b. 3v2 º 18v = 3v(v º 6) c. 12x2 + 3x + 3 = 3(4x2 + x + 1) d.

4u2 º 36 = 4(u2 º 9) = 4(u + 3)(u º 3)

Extra Example 5 Solving Quadratic Equations Solve the equation. a. 9t2 º 12t + 4 = 0 SOLUTION a. 9t2 º 12t + 4

b. x2 º 10 = 3x º 6

=0

Write original equation.

(3t º 2)2 = 0

Factor.

3t º 2 = 0 t=

Use zero product property.

2 3

Solve for t.

2 ➧ The solution is . Check the solution in the original equation. 3 b.

x2 º 10 = 3x º 6 x2 º 3x º 4 = 0

Write original equation. Write in standard form.

(x º 4)(x + 1) = 0

Factor.

x º 4 = 0

Use zero product property.

or x + 1 = 0

x = 4 or

x =º 1

Solve for x.

➧ The solutions are 4 and º 1. Check the solutions in the original equation.

Algebra 2, Chapter 5, Lesson 2, Extra Examples

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Extra Example 6 Using a Quadratic Equation as a Model A painter is making a rectangular canvas for her next painting. She wants the length of the canvas to be 4 feet more than twice the width of the canvas. The area of the canvas must be 30 square feet. What should be the dimensions of the canvas? SOLUTION

Verbal Model Labels

Algebraic Model

Width of Length of Area of =  Canvas   •   canvas canvas Area of canvas = 30

(square feet)

Width of canvas = w

(feet)

Length of canvas = (2w + 4)

(feet)

30 = w(2w + 4) 30 = 2w2 + 4w

Simplify right side.

0 = 2w2 + 4w º 30

Write in standard form.

0 = (2w + 10)(w º 3)

Factor.

or

w º 3 = 0

w = º5 or

w=3

2w + 10 = 0

Use zero product property. Solve for w.

➧ Reject the solution w = º5, because the canvas’s width cannot be negative. The width of the canvas should be 3 feet, and the length should be 10 feet. Extra Example 7 Finding the Zeros of a Quadratic Function Find the zeros of y = 3x2 + 14x º 5. SOLUTION

Use factoring to write the function in intercept form. y = 3x2 + 14x º 5 = (3x º 1)(x + 5) ➧ The zeros of the function are

1 and º5. 3

✓ CHECK Graph 3x2 + 14x º 5. The graph crosses the x-axis at 1 (º5, 0) and ( , 0), which confirms 3 1 that the zeros are and º5. 3 Algebra 2, Chapter 5, Lesson 2, Extra Examples

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Extra Example 8 Writing a Quadratic Model You own an amusement park that averages 75,000 visitors per year who pay a \$12 admission charge. You plan to lower the admission price to attract new customers. It has been shown that each \$1 decrease in price results in 15,000 new visitors. What admission should you charge to maximize your annual revenue? What is the maximum revenue? SOLUTION

Revenue

= Number of visitors • Ticket price

Let R = your annual revenue and let x = the number of \$1 price decreases. R = (75,000 + 15,000x)(12 º x) = (15,000x + 75,000)(º x + 12) = 15,000(x + 5)(º 1)(x º 12) = º15,000(x + 5)(x º 12) The zeros of the revenue function are º 5 and 12. The value of x that maximizes R is º5 + 12 the average of the zeros, or x = = 3.5. To maximize revenue, you should 2 charge \$12.00 º \$3.50 = \$8.50 per visit. Your maximum revenue is R = º15,000(3.5 + 5)(3.5 º 12) = \$1,083,750.

Algebra 2, Chapter 5, Lesson 2, Extra Examples

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