# What is the process of factoring a trinomial with a leading coefficient that is not equal to one and why is the process more difficult than when the leading coefficient is one?

degeneratecircle | High School Teacher | (Level 2) Associate Educator

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There isn't only one way to factor a trinomial, and the level of difficulty depends on the method used. I normally stick to just two methods, which I'll illustrate with an example. Suppose we wish to factorÂ `6x^2-7x-3.`

First, we multiply the coefficient of the second degree term and the constant term, in this case `6` and `-3,` respectively. This gives `-18.` Now we try to find two numbers that multiply to `-18` and add to the remaining coefficient, `-7.` These numbers are `-9` and `2.`

This leads us to break up the middle term and rewrite the trinomial as

`6x^2-9x+2x-3,` and now we pull out the greatest common factor of the first and second pairs of terms as follows:

`6x^2-9x+2x-3=3x(2x-3)+1(2x-3).` Since `2x-3` is now a common factor, we get the final result of

`6x^2-7x-3=(2x-3)(3x+1).`

If the leading coefficient is one, then this method becomes easier due to the fact that multiplication by 1 is simple and generally results in smaller numbers.

However, this method won't always work (not as easily as it just did, at least) and it is also possible to factor by using the quadratic formula. This is done by finding the roots of the trinomial. Doing this gives roots of `3/2` and `-1/3.` Since the leading coefficient is 6, we form two factors using the roots and stick a 6 in front to get

`6x^2-7x-3=6(x-3/2)(x-(-1/3))=(2x-3)(3x+1),` just as before. In my opinion, if we use this method then the problem is really no more difficult than if the leading coefficient were one, and it works every time and pretty quickly too.

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