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Lesson 8-3 Factoring Trinomials (Quadratic Equations x2+bx+c) Factoring x2+bx+c – find two integers m and p with a sum of b and a product of c, then write x2+bx+c as (x+m)(x+p). Example 1 b and c are Positive Factor x2 + 16x + 63. In this trinomial, b = 16 and c = 63. Since c is positive and b is positive, you need to find two positive factors with a sum of 16 and a product of 63. Make an organized list of the factors of 63, and look for the pair of factors with a sum of 16. Factors of 63 1, 63 3, 21 7, 9 Sum of Factors 64 24 16 x2 + 16x + 63 = (x + m)(x + p) = (x + 7)(x + 9) Check The correct factors are 7 and 9. Write the pattern. m = 7 and p = 9 You can check this result by multiplying the two factors. The product should be equal to the original equation. (x + 7)(x + 9) = x2 + 9x + 7x + 63 FOIL method = x2 + 16x + 63 Simplify. Example 2 b is Negative and c is Positive Factor x2 – 11x + 24. In this trinomial, b = –11 and c = 24. Since c is positive and b is negative, you need to find two negative factors with a sum of –11 and a product of 24. Make a list of the negative factors of 24, and look for the pair of factors with the sum of –11. Factors of 24 –1, –24 –2, –12 –3, –8 –4, –6 Sum of Factors –25 –14 –11 The correct factors are –3 and –8. –10 x2 – 11x + 24 = (x + m)(x + p) = (x – 3)(x – 8) Check Write the pattern. m = –3 and p = –8 You can check this result by using a graphing calculator. Graph y = x2 – 11x + 24 and y = (x – 3)(x – 8) on the same screen. Since only one graph appears, the two graphs must coincide. Therefore, the trinomial has been factored correctly. Example 3 c is Negative Factor each polynomial. a. x2 + x – 6 In this trinomial, b = 1 and c = -6. Since c is negative, the factors m and p have opposite signs. So either m or p is negative, but not both. Since b is positive, the factor with the greater absolute value is also positive. List the factors of –6, where one factor of each pair is negative. Look for the pair of factors with a sum of 1. Factors of –15 Sum of Factors –1, 6 5 1, –6 –5 –2, 3 1 The correct factors are –2 and 3. 2, –3 –1 x2 + x – 6 = (x + m)(x + p) = (x – 2)(x + 3) Write the pattern. m = –2 and p = 3 b. x2 – 4x – 21 In this trinomial, b = -4 and c = -21. Either m or p is negative, but not both. Since b is negative, the factor with the greater absolute value is also negative. List the factors of 21, where one factor of each pair is negative. Look for the pair of factors with a sum of 4. Factors of –21 Sum of Factors –1, 21 20 1, –21 –20 –3, 7 4 3, –7 –4 The correct factors are 3 and –7. x2 – 4x – 21 = (x + m)(x + p) = (x + 3)(x – 7) CHECK Write the pattern. m = 3 and p = –7 Graph y = x2 – 4x – 21 and y = (x + 3)(x 7) on the same screen. Since the graphs coincide, the trinomial has been factored correctly. Quadratic Equation – an equation of the form x2+bx+c =0. To solve: factor the quadratic and set the factors equal to zero. Example 4 Solve an Equation by Factoring Solve x2 – 8x + 7 = 0. Check your solutions. x2 – 8x + 7 = 0 (x – 1)(x – 7) = 0 x–1=0 or x – 7 = 0 x=1 x=7 Original equation Factor. Zero Product Property Solve each equation. The roots are 1 and 7. CHECK Substitute 1 and 7 for x in the original equation. x2 – 8x + 7 = 0 x2 – 8x + 7 = 0 ? (1)2 – 8(1) + 7 0 ? (7)2 – 8(7) + 7 0 ? 1–8+7 0 0=0 ? 49 56 + 7 0 0=0 Real-World Example 5 Solve a Problem by Factoring NUMBER THEORY Find two consecutive integers whose product is 56. Understand You are to find two consecutive integers that have a product of 56. Plan Write an equation representing the product of the two integers. Solve Let x = the first integer. Then x + 1 = the next consecutive integer. first integer x times x(x + 1) = 56 x2 + x = 56 2 x + x – 56 = 0 (x + 8)(x – 7) = 0 x + 8 = 0 or x – 7 = 0 x = –8 x=7 consecutive integer equals 56 (x + 1) = 56 Write the equation. Multiply. Subtract 56 from each side. Factor. Zero Product Property Solve each equation. One possible set of consecutive integers is –8 and –8 + 1 or –7. The other is 7 and 7 + 1 or 8. Check –8 (–7) = 56 and 7 8 = 56, so the solution checks.