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My daughter is being taught the X method of factoring. In

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My daughter is being taught the "X" method of factoring. In her homework is the problem(3x squared - 5x - 2 = 0) The constant is -2, the co-efficient is -5. Since 2 is a prime, the only factors possible are 1 x -2 or -1 x 2, neither of which can be added together to achieve -5. How do we solve this problem, or what are we missing?

You need to look at the first coefficient and the last coefficient when factoring a trinomial. Please take a look at the attached Word document that illustrates the solution to this problem, thanks!

In the example, the first coefficient is 3, but the second is -5, the constant is -2. My daughter was taught to put the constant in the top of the X, and the second coefficient in the bottom of the X, then find the factors that would multiply to produce the number in the top of the X, and add to produce the number in the bottom. These two numbers are then entered on either side of the X. These then become the factors that solve the equation.

-on the left side of the X, place the two factors that make up the 1st coefficient -3. It's either going to be 1 and 3 or 3 and 1.

-on the right side of the X, place the two factors that make up the 3rd coefficient -2. It's going to be 2 and -1 or -2 and 1.

-Now is the time to select the correct factors out of the options you have. You do this by cross-multiplying and adding to get the 2nd coefficient, -5. To cross-multiply, take the number on the top-left of the X and multiply by the number on the bottom-right of the X. Also take the number on the top-right number of the X and multiply by the bottom-left of the X. Then add these two numbers. If you use the highlighted factors, you get 1*1 + -2*3 = 1 - 6 = -5. This is how you choose the correct factors.

-For the final step, all you need to do is fill in the factors into the equation (1x-2)*(3x+1). The top left factor and top right factor go into the first set (1x-2) and the bottom left factor and bottom right factor go into the 2nd set (3x+1).

Dear XXXXX: Apparently we are using two different versions of the technique. You refer to "top left" and "bottom right". Her teacher insists on her using the described technique of using the constant in the center top of the x, the second coefficient in the center bottom, cross-multiplying whatever factors produce the top number as well as add together to produce the bottom number. I can see that with the method you are using the factoring works and the problem is solved. But how can she achieve the same answer using the teacher's described technique? She has several problems that don't work this way, and several that do. So is this technique only applicable where the constant can be factored into numbers that add together to produce the 2nd coefficient? Other techniques must be used if this isn't possible?