# SOLUTION: 2. In order to fully understand what I'm doing here, I need to have it explained to me how to factor a polynomial of the form ax^2 + bx + c when a is not equal to 1. I know there i

Algebra ->  -> SOLUTION: 2. In order to fully understand what I'm doing here, I need to have it explained to me how to factor a polynomial of the form ax^2 + bx + c when a is not equal to 1. I know there i      Log On

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 Question 157297: 2. In order to fully understand what I'm doing here, I need to have it explained to me how to factor a polynomial of the form ax^2 + bx + c when a is not equal to 1. I know there is the grouping approach as well as reversing FOIL and I want to be able to see an example. Which approach would be the best? Could somebody explain this to me as well? Thank you so very much, I truly need guidance on this. Thank you.Answer by gonzo(654)   (Show Source): You can put this solution on YOUR website!to create an example, you can work backwards. take (7x+5) * (5x-3) for example. the quadratic equation would then be -3*5 -21x + 25x + 35x^2 which would then equal 35x^2 + 25x - 21x - 15 which would then equal 35x^2 + 4x - 15 how would you solve this equation? if you couldn't figure out the combinations that would work fairly quickly, then the best approach would be to use the quadratic equation of and looking at the equation yields the following conclusions. it's been reduced to its simplest form. there are no common factors that can be eliminated. the c factors would be a combination of (1 and 15) or (3 and 5) the a factors would be a combination of (35 and 1) or (5 and 7) try (35 and 1) for the a factors first combined with (1 and 15) for the c factors. cross multiplying the a factors with the c factors yields the following taking into account that one of the c factors has to be minus: 35x * 1 = -35x or 35x 35x * 15 = 525x or -525x possibilities are 35x, -35x, 525x, -525x. none of these can be added together to get a b factor of 4. conclusion is (35 and 1) combined with (1 and 15) won't work. intuition might tell you to try (5 and 7) for the a factors and (3 and 5) for the c factors next. this will also shorten the time required to explain this. remember one of the c factors has to be minus, either 3 or 5. 5x * 3 15x or -15x 5x * 5 = 25x or -25x 7x * 3 = 21x or -21x 7x * 5 = 35x or -35x possibilities are 15x, -15x, 25x, -25x, 21x, -21x, 35x, -35x. out of all these, the combination that will yield a b factor of 4 is 25x - 21x. you answer has to be some combination of (5x * 5) and (7x * 3) in order to get that cross multiplication you would need to have factors of either (5x-3) or (5x+3) and (7x-5) or (7x+5). intuition will tell you that the 25 must be positive to get a positive b factors, so you would choose (7x+5) * (5x-3) because your cross multiplication then gets you 25x - 21x. cross multiplying those factors will get you -15 -21x + 25x + 35x^2. combining and moving terms gets you 35x^2 +4x -15 which is the equation you want.