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12-4 Factoring x² + bx + c (x + 2)(x + 3) = x² + 3x + 2x + 2·3 x² + 5x + 6 (x – 4)(x + 3) = x² + 3x - 4x - 4·3 x² - 1x - 12 (x + a)(x + b) = x² + bx + ax + a·b x² + (b + a)x + ab In this section you will be working backwards to create the 2 binomials, so you need to find sums & products that will give the middle and last terms. Factor the following trinomials. 1. x² + 50x + 96 (x + ?)(x + ?) (x + 2)(x + 48) x² + 48x + 2x + 96 x² + 50x + 96 All positive terms. Find the combination whose product is 96 and whose sum is 50. 1, 96 = 96, 97 no 2, 48 = 96, 50 Check mentally 2. y² - 3y - 54 (y + ?)(y + ?) (y + 6)(y - 9) Last term is negative so there must be 1 postive and 1 negative number. 1, -54 = -54, -53 no -1, 54 = -54, 53 no 2, -27 = -54, -25 no -2, 27 = -54, 25 no 3, -18 = -54, -15 no -3, 18 = -54, 15 no 6, -9 = -54, -3 yes 3. z² - 9z + 20 (z + ?)(z + ?) (z - 4)(z -5) 4. c² - 16 (c + ?)(c + ?) (c + 4)(c -4) Last term positive with middle term negative, so both must be negative. -1, -20 = 20, -21 no -2, -10 = 20, -12 no -4, -5 = 20, -9 yes This is the difference of squares that were introduced previously. 4, -4 = -16, 0 yes 5. d³ - 16d The terms have a gcf. Take care of it first. d(d² - 16) Factor now into 2 binomials d(d+4)(d-4) 6. Explain why c² + 16 is not factorable. Because there are no 2 numbers who’s product is positive and has a sum of 0.