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1 FACTORING QUADRATICS and Chapter 8 introduces students to quadratic equations. These equations can be written in the form of y = ax 2 + bx + c and, when graphed, produce a curve called a parabola. There are multiple methods that can be used to solve quadratic equations. One of them requires students to factor. Students have used algebra tiles to build rectangles of quadratic expressions. Later, they used the formula for the area of a rectangle, (base)(height) = area, to create generic rectangles and write equations expressing the area as both a sum and a product. In the figure below, the length and width of the rectangle are factors since they multiply together to produce the quadratic x 2 + x + 8. Notice the 4x and 2x are located diagonally from each other. They are like terms and can be combined and written as x. The dimensions of the rectangle are (x + 2) and (x + 4). The area is x 2 + x x 4x 8 x 2 2x x + 2 The factors of x 2 + x + 8 are (x + 2) (x + 4). The ax 2 term (1x 2 ) and the constant (8) are always diagonal to one another in a generic rectangle. In this example, the diagonal product is 8x 2 and the c term is the product: 2(4) = 8. The two x-terms are the other diagonal and can be combined into a sum when they are like terms. The b term of the quadratic is the sum of the factors: 2x + 4 x. Their diagonal product is (2x)(4x) = 8x 2, the same as the other diagonal. (See the Math Notes box on page 332 for why the products of the two diagonals are always equal.) To factor, students need to think about sums and products simultaneously. Students can Guess & Check or use a diamond problem from Chapter 1 to help organize their sums and products. See the Math Notes box on page Algebra Connections Parent Guide

2 Example 1 Factor x 2 + 7x Sketch a generic rectangle with 4 sections. x 2 12 Write the x 2 and the 12 along one diagonal. Find two terms whose product is 12x 2 and whose sum is 7x (3x and 4x). Students are familiar with this situation as a diamond problem from Chapter 1. Write these terms as the other diagonal. Find the base and height of the rectangle by using the partial areas. Write the complete equation. (x + 3)(x + 4) = x 2 + 7x x 3x 12 x 2 3x 12 x 2 4x 4x x + 4 Example 2 The same process works with negative values. Factor x 2 + 7x! 30. Sketch a generic rectangle with 4 sections. Write the x 2 and the -30 along one diagonal. Find two terms whose product is -30x 2 and whose sum is 7x. Write these terms as the other diagonal. 3x!30 x 2 10x 3 x 3x!30 x 2 10x x + 10 Find the base and height of the rectangle. Pay particular attention to the signs so the sum is 7x, not -7x. Write the complete equation. (x! 3)(x + 10) = x 2 + 7x! 30 Chapter 8: Quadratics 59

3 Example 3 Factor x 2! 15x + 5. Sketch a generic rectangle with 4 sections. Write the x 2 and the 5 along one diagonal. Find two terms whose product is 5x 2 and whose sum is -15x. 8x 5 x 2 7x 8 x 8x 5 x 2 7x x 7 Write these terms as the other diagonal. Find the base and height of the rectangle. Mentally multiply the factors to be sure they are correct. Pay close attention to the signs of the factors. Write the complete equation. (x! 7)(x! 8) = x 2! 15x + 5 Example 4 Factor 2x 2 + 7x +. Sketch a generic rectangle with 4 sections. Write the 2x 2 and the along one diagonal. Find two terms whose product is 12x 2 and whose sum is 7x. Write these terms as the other diagonal. 3x 2x 2 4x 3 + 2x 3x 2x 2 4x x + 2 Find the base and height of the rectangle. Write the complete equation. (2x + 3)(x + 2) = 2x 2 + 7x + 0 Algebra Connections Parent Guide

4 Example 5 Factor 12x 2! 19x + 5. Sketch a generic rectangle with 4 sections. Write the 12x 2 and the 5 along one diagonal. Find two terms whose product is 0x 2 and whose sum is -19x. 15x 5 12x 2 4x 5 4x 15x 5 12x 2 4x 3x 1 Write these terms as the other diagonal. Find the base and height of the rectangle. Check the signs of the factors. Write the complete equation. (3x! 1)(4x! 5) = 12x 2! 19x + 5 Problems 1. x 2 + 5x x 2 + 5x x 2 + 4x x x x x x 2 + 7x + 7. x x x 2 + 4x! x x x x! x 2 + x! x 2! 20x! x 2! 11x x x! 15. 2x 2 + 5x! 3 1. x 2! 3x! x 2! 12x x 2 + 2x! x 2! x! x 2! 18x + 8 Answers 1. (x + 2)(x + 3) 2. (x + 1)(2x + 3) 3. (3x + 1)(x + 1) 4. (x + 5)(x + 5) 5. (x + 11)(x + 4). (x + )(x + 1) 7. (x + 8)(x + 3) 8. (x + 8)(x! 4) 9. (2x + 3)(2x + 3) 10. (3x! 1)(4x + 5) 11. (x! 8)(x + 9) 12. (x! 7)(3x + 1) 13. (x! 4)(x! 7) 14. (x + )(2x! 1) 15. (x + 3)(2x! 1) 1. (x! 5)(x + 2) 17. (2x! 3)(2x! 3) 18. (3x + 5)(x! 1) 19. (2x + 1)(3x! 2) 20. (3x! 4)(3x! 2) Chapter 8: Quadratics 1

5 USING THE ZERO PRODUCT PROPERTY and A parabola is a symmetrical curve. Its highest or lowest point is called the vertex. It is formed using the equation y = ax 2 + bx + c. Students have been graphing parabolas by substituting values for x and solving for y. This can be a tedious process if the range of x values is unknown. One possible shortcut is to find the x-intercepts first, then find the vertex and other convenient points. To find the x-intercepts, substitute 0 for y and solve the quadratic equation, 0 = ax 2 + bx + c. Students will learn multiple methods to solve quadratic equations in this course. The method described below utilizes two ideas: (1) When the product of two or more numbers is zero, then one of the numbers must be zero. (2) Some quadratics can be factored into the product of two binomials, where coefficients and constants are integers. This procedure is called the Zero Product Method or Solving by Factoring. See the Math Notes boxes on pages 349 and 31. Example 1 Find the x-intercepts of y = x 2 + x + 8. The x-intercepts are located on the graph where y = 0, so write the quadratic expression equal to zero, then solve for x. x 2 + x + 8 = 0 Factor the quadratic. (x + 4)(x + 2) = 0 Set each factor equal to 0. (x + 4) = 0 or (x + 2) = 0 Solve each equation for x.!4 or!2 The x-intercepts are (-4, 0) and (-2, 0). You can check your answers by substituting them into the original equation. (!4) 2 + (!4) + 8 " 1! " 0 (!2) 2 + (!2) + 8 " 4! " 0 2 Algebra Connections Parent Guide

6 Example 2 Solve 2x 2 + 7x! 15 = 0. Factor the quadratic. (2x! 3)(x + 5) = 0 Set each factor equal to 0. (2x! 3) = 0 or (x + 5) = 0 Solve for each x or!5 Example 3 If the quadratic does not equal 0, rewrite it algebraically so that it does, then use the zero product property. Solve 2 = x 2! x. Set the equation equal to 0. Factor the quadratic. 2 = x 2! x 0 = x 2! x! 2 0 = (2x + 1)(3x! 2) Solve each equation for x. (2x + 1) = 0!!!!!!!or!!!!!!!(3x! 2) = 0 2!1!!!!!or!!!!!!!!!!!!!!!3 2! 1 2!!!!!!!!!!!!!!!!!!!!!!!!! 2 3 Example 4 Solve 9x 2! x + 1 = 0. Factor the quadratic. 9x 2! x + 1 = 0 (3x! 1)(3x! 1) = 0 Solve each equation for x. Notice the factors are the same so there will be only one x value. For this parabola, the x-intercept is the vertex. (3x! 1) = Chapter 8: Quadratics 3

7 Problems Solve for x. 1. x 2! x! 12 = x 2! 7x! = 0 3. x 2 + x! 20 = x x + 10 = 0 5. x 2 + 5!4. x! 9 = x 2 7. x 2 + 5x! 4 = 0 8. x 2! x + 8 = 0 9. x 2! x! 15 = x x + 9 = x 2! x 2 + 8x + = x x 2! x 2 = 45! 4x Answers 1. 4 or!3 2.! 2 3 or 3 3.!5 or 4 4.! 5 3 or!2 5.!4 or! ! 4 3 or or 2 9.! 3 2 or ! or!2 12.!1 or!3 13.! 1 5 or 2 14.! 1 or or Algebra Connections Parent Guide

9 Example 2 Solve 3x 2 + 5x + 1 = 0. Identify a, b, and c. a = 3,!b = 5,!c = 1 Write the quadratic formula.!b ± b2! 4ac Substitute a, b, and c in the formula and do the initial calculations.!(5) ± (5)2! 4(3)(1) 2(3) Simplify the.!5 ± 25! 12!5 ± 13!5 + 13!5! 13 The solution is or. These are the exact values of x. The x values are the x-intercepts of the parabola. To graph these points, use a calculator to find the decimal approximation of each one. The x-intercepts are (! "0.23, 0) and (! "1.43, 0). Example 3 Solve 25x 2! 20x + 4 = 0. Identify a, b, and c. a = 25,!b =!20,!c = 4 Write the quadratic formula.!b ± b2! 4ac Substitute a, b, and c in the formula and do the initial calculations.!(!20) ± (!20)2! 4(25)(4) 2(25) 20 ± 400! Simplify the. 20 ± 0 50 This quadratic has only one solution: 2 5. Algebra Connections Parent Guide

10 Example 4 Solve x 2 + 4x + 10 = 0. Identify a, b, and c. a = 1,!b = 4,!c = 10 Write the quadratic formula.!b ± b2! 4ac Substitute a, b, and c in the formula and do the initial calculations.!(4) ± (4)2! 4(1)(10) 2(1) Simplify the.!4 ± 1! 40 2!4 ±!24 2 It is impossible to take the square root of a negative number; therefore this quadratic has no real solution. The graph would still be a parabola, but there would be no x-intercepts. This parabola is above the x-axis. Example 5 Solve (3x + 1)(x + 2) = 1. Rewrite in standard form. (3x + 1)(x + 2) = 1 3x 2 + 7x + 2 = 1 3x 2 + 7x + 1 = 0 Identify a, b, and c. a = 3,!b = 7,!c = 1 Write the quadratic formula.!b ± b2! 4ac Substitute a, b, and c in the formula and do the initial calculations.!(7) ± (7)2! 4(3)(1) 2(3) Simplify. The x-intercepts are (! "0.15,!0) and (! "2.18,!0).!7 ± 49! 12!7 ± 37 Chapter 8: Quadratics 7

11 Example Solve 3x 2 + x + 1 = 0. Identify a, b, and c. a = 3,!b =,!c = 1 Write the quadratic formula.!b ± b2! 4ac Substitute a, b, and c in the formula and do the initial calculations.!() ± ()2! 4(3)(1) 2(3) Simplify.! ± 3! 12! ± 24 The x-intercepts are (! "1.82, 0) and (! "0.18, 0). Math Note describes another form of this expression that can be written by simplifying the square root. The result is equivalent to the exact values above. Factor the 24, then simplify by taking the square root of = 4 = 2 Simplify the fraction by dividing every term by 2.! ± 2!3 ± 3 8 Algebra Connections Parent Guide

12 Problems Solve each of these using the Quadratic Formula. 1. x 2! x! 2 = 0 2. x 2! x! 3 = 0 3.!3x 2 + 2x + 1 = 0 4.!2! 2x 2 = 4x ! 2x 2.!x 2! x + = 0 7.! 4x + 3x 2 = x 2 + x! 1 = 0 9. x 2! 5x + 3 = = 10x 2! 2x x(!3x + 5) = 7x! (5x + 5)(x! 5) = 7x Answers 1. 2 or!1 2. 1± 13 2! 2.30!or!" ! 1 3 or 1 4.!1 5.!7± ± 145!12! 1.09!or!"4.59! "1.09!or! ± 40 = 2± !1± ± 13 2! 1.72!or!"0.39! 0.39!or!"0.4! 4.30!or! No solution 11. 2± 124 =! 1± 31! ± ! "2.19!or!1.52!.21!or!"0.81 Chapter 8: Quadratics 9

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