7/9/2013 Quadratic Equations 38  Solve x 2 – 8x + 12 = 0 by factoring x 2 + ( a + b )x + ab = (x + a )(x + b ) Trinomials with one negative term – the middle term Not-So-Special Trinomials Terms a, b of same sign Sign of both terms  Find a and b such that Possible factors of 12 : –1, –12 –2, –6 –3, –4 Since –2 – 6 = –8, then a = –2 and b = –6
{"Image DocumentName": "Not-So-Special Trinomials", "Image ImageDescription": "Solving Quadratic Equations. Trinomials with one negative term. \u2013 the middle term. Sign of both terms. Terms a , b of same sign. Solve x2 \u2013 8x + 12 = 0 by factoring. Find a and b such that. x2 + (a + b)x + ab = (x + a)(x + b) Quadratic Functions: Not-So-Special Forms. Here we simply set out to factor a quadratic polynomial, possibly in connection with solving a quadratic equation. The sign of the constant term tells about the signs of the constants a and b, and the sign of the linear term tells which of these two is larger. As shown, the possible pairs of factors for the constant term can be tested quickly against the coefficient of the linear term, while skipping all the pairs that don\u2019t work. Once the quadratic expression is factored, we can use it to solve a related quadratic equation as shown. The factoring allows us to use the zero product property to find the solutions of the equation \u2013 IF we have one. In this case there is no equation to solve. We are merely changing the form of the given quadratic expression \u2013 not its value ! As the illustration shows, the sign of the constant term in the trinomial tells us, in this case, that the signs of a and b are different (producing a negative product) and the sign of the middle term tells us which is larger. Knowing these two things tells us all the possible combinations of a and b, which will produce these results. We then list and test these combinations. Note that without the sign analysis, looking only at the product 24, we would have four times as many pairs of a and b values to test. Possible factors of 12 : \u20131, \u201312 \u20132, \u20136 \u20133, \u20134. Since \u20132 \u2013 6 = \u20138 , then a = \u20132 and b = \u20136. 7/9/2013. Quadratic Equations. Quadratic Equations. 7/9/2013.", "Image XResolution": "37", "Image YResolution": "37", "Image ResolutionUnit": "Pixels/Centimeter", "Image Software": "SlidePlayer", "Image YCbCrPositioning": "Centered"}