First dive into factoring polynomials. This section covers factoring quadratics with leading coefficient of by factoring the coefficients.
Like many techniques in mathematics, the method of factoring coefficients is easier to understand by starting with what we hope to have at the end, and work our way backward to see what it looks like at the start. For example, let’s consider the following factored quadratic and it’s distributed form; As we mentioned, we want to think of this backward. So we would ask ourselves, “given , how could I figure out that it should factor into ? To answer this we want to make a couple clever observations that will ultimately lead to our method.
First we need to observe that our two zeros, and , both divide the last term in the expanded form. That is to say that and both divide . In fact, not only do they both divide , but their product is exactly (ie ).
The second observation we need to make is that the two zeros add up to the (negative) of the middle term’s coefficient in the expanded form; ie (or, an easier way to see it is that ).
With these observations in mind, we want to generalize the process to work for any zeros, not just and . To this end we will substitute and for the zeros and redo the above calculation (careful to note that we are using and so that if we plug in or we get zero!) expanding from the factored form to the expanded form:
Here we can verify that the zeros will multiply to give the constant coefficient in the expanded form and add to give the (negative of the) coefficient of the middle term. This is something we can use to figure out how to factor a general quadratic with leading coefficient of .
According to our observations we want two numbers that multiply to but add to . We can start by listing all the pairs of numbers that multiply to . This list will be the same as the list for (positive) , but we will need one of the numbers to be negative, so we will start by writing out pairs of numbers that multiply to . Those would be;
, ,
The above are all possible pairs for but one of them must be negative in order for the product to be . So we really want to see which combination of the above factors add to if one of those two factors is negative. For example, we might consider and check the sum if the is negative, which gets (not what we want) and then try again with the being negative, which gives (also not what we want). Since neither way worked, that pair of numbers doesn’t work and we move to the next pair. After some trial and error we will eventually figure out that the pair we want is and . This means we have the factorization;
If we weren’t sure, we could distribute the proposed roots to make sure we get the original polynomial back;
