Algebra 10.5 Factoring Trinomials With a Lead Coefficient of 1 to Solve Factoring to solve… • a quadratic expression can be solved by factoring and then using the zeroproduct property. Solve: x2 + 10x + 21 = 0 …and whose sum is 10. Find two numbers whose product is 21... (x + 7)(x + 3) = 0 x = -7 and -3 They should check! Figuring out the signs! Frame it with the signs. x2 + bx + c = 0 (x + Frame it with the signs. )(x + )=0 x2 – bx + c = 0 (x - )(x - Frame it with the signs. )=0 x2 – bx – c = 0 (x - )(x + )=0 The larger # goes here. Frame it with the signs. x2 + bx – c = 0 (x - )(x + )=0 The larger # goes here. Methods Method 1: List out all the factors of the constant in the trinomial. List factors of -36 x2 + 5x – 36 = 0 Factor. Frame it. Solve. -12 and 3 -18 and 2 12 and -3 18 and -2 9 and -4 36 and -1 -9 and 4 -36 and 1 (x + 9)(x - 4 ) = 0 Which set of factors add to +5? x = 4 and -9 Method 2: Do this process in your head!!! x2 – 14x = -48 Factor. Put in standard form. x2 – 14x + 48 = 0 Frame it with signs. 6 and -6 (x - 6 )(x - 8 ) = 0 Solve. x = 6 and 8 Solve. x2 – 15x – 7 = -61 + 61 +61 Put in standard form! x2 – 15x + 54 = 0 (x - 9 )(x - 6 ) = 0 x = 9 and 6 Solve. 1) x2 + 3x – 18 = 0 (x – 3)(x + 6) = 0 2) m2 + 11m = -10 m2 + 11m + 10 = 0 (m + 10)(m + 1) = 0 m = -10 and -1 3) x2 – 2x – 40 = 8 x2 – 2x – 48 = 0 (x + 6)(x – 8) = 0 x = -6 and 8 4) a2 – 33a = 280 a2 – 33a – 280 = 0 (a – 40)(a + 7) = 0 a = 40 and -7 x = 3 and -6 Solve. 5) x2 + 3x = 6 x2 + 3x – 6 = 0 (x – )(x + ) = 0 Then how do you solve the equation? x = -3 +√33 2 x = -3 - √33 2 If you think the quadratic equation cannot be factored, check the discriminant. If the discriminant is a perfect square: The equation can be factored. If the discriminant is not a perfect square: The equation cannot be factored. b2 – 4ac 32 – 4(1)(-6) 9 + 24 33 Not a perfect square, the trinomial cannot be factored. HW • P. 607-609 #15-47, 52-56

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# 10.5 Solving Quadratics with a Lead Coefficient of 1 by Factoring