# Factor `6x^2-x-12`The leading coefficient is giving me trouble and I just don't know what method to use. Thank you!

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In order to do this, we need to first consider possible combinations of the coefficients of x and the constant terms. We know that our result will take the following form:

`(ax+m)(bx+n)`

where a and b are factors of the leading coefficient and m and n are factors of the constant term. When FOIL is used to find the resulting trinomial, we get the following expression:

`abx^2 + (bm+an)x + mn`

So, we know the following information based on the trinomial:

1) a and b must be either 1 and 6 or 2 and 3 (assuming we have integer coefficients).

2) m and n are 1 and 12, 2 and 6, or 3 and 4 with one negative value and one positive value (because they need to multiply to -12).

In order to factor this without a calculator, you basically just need to take the shotgun approach to see what combination of numbers will give the following result:

`(bm+an) = -1`

This result is what you're looking for because, if you look at the given trinomial, this is the coefficient of the `x` term.

This might take a few trials for you to find, but you'll eventually see that the following result will come from the possible factors we looked at above:

`(2*4 + 3*(-3)) = -1`

This gives us the following values:

`a = 3, b = 2, m = 4, n = -3`

Notice that you can switch a and b if you also switch m and n. This result then gives you the following factorization:

`(3x+4)(2x-3)`

If you FOIL this out again, you get the original trinomial, confirming we factored this correctly!

I hope this helped!