Factoring Polynomials of Degree 2Date: 4/8/96 at 13:39:19 From: Lorrane Chung Subject: Factoring polynomials Hello, I am having a lot of difficulties factoring polynomials of the type x^2 + 6x +8 and 3x^2 + 10x +8. I have exhausted all the methods - they don't seem to work for me. I really need an easy method to factor other than the quadratic formula or for finding 2 factors that multiply to give last term but add to give the middle term. Please help. Desperate, Lorrane Date: 4/16/96 at 19:33:19 From: Doctor Patrick Subject: Re: Factoring polynomials Hi Lorrane! I hope we can help you. How did you try factoring these problems? When I have to factor, I figure out the possible factors for the first and last terms and then play with them a little to see what might work, and what is definitely going to be out of the question. In the first problem, all we need to factor is the 8 - the x^2 is going to have to be x*x. This gives us two choices: 8*1, or 2*4. Since all of the numbers in the problem are positive, we can ignore the other two possibilities of -8*-1 and -2*-4. Working with the 8*1 first, we get (x+8)*(x+1), which doesn't give us the middle term we are looking for. Since there are no other combinations using 8*1 (do you see why?) we can move on to the other possibility of 2*4. If we try factoring the equation with these numbers we get (x+2)(x+4). When we multiply this back out we get x*x +2x + 4x +8 which equals x^2+6x+8 after we combine like terms and multiply out the x*x. The second problem is more complicated since we have one additional term (the 3 in front of the x^2). Again, we have two ways to factor 8, but we also need to try multiplying the 3 by both of the factors to find out which combination works. If we start with 8*1 as factors of 8, again we get two possibilities: (3x+8)(x+1) or (3x+1)(x*8). Do you understand why there are two possibilities this time? The first possibility would end up giving us (3x*1) + (x*8), or 3x+8X, for the middle term. Since this would equal 11x it is not the combination we are looking for. Likewise (3x+1)(x*8) will not work. (Why not?) So now have to move on to the other set of factors (4*2). Why don't you try the remaining two possibilities on your own using the same methods we used above. Good luck! Write us back if you need more help! -Doctor Patrick, The Math Forum |
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