In this case the coefficient on the first term is not just an invisible 1. So here when we multiply the first and last coefficients, we get something different than the last term. (2)(15) = 30, so we are looking for factors of 30 that add up to 13. Many students will be able to immediately recognize that 3 and 10 multiply up to 30 and add up to 13, but if the answer does not immediately pop into your head, there is something you can be doing with your hands while we wait for it to happen.
Write down all of the factorizations of 30 into a product of two numbers starting with 1 time 30.
When we divide 1, 2, and 3 into 30, we find that they go into it, and we get the first three results in our table. After 1, 2, and 3, we get to 4. 4 doesn't go into 30, so there is no entry for 4. After 4 comes 5 which does go into 30, and it appears in the table. After 5 comes 6, but we see a 6 in the left column of our table. As soon as you find yourself considering a number which appears in the left side of the table, you know that you have gotten all of the possibilities, because if we were to find a number bigger than 6 which went into 30, its quotient would be smaller than 5, and we were very thorough in already considering all of those possibilities.
Fortunately, we see there is a pair of numbers in the table that add up to the middle coefficient, namely -10 and -3. The (2)(15) = 30 is positive, and there are two ways two numbers can multiply to a positive: two positives or two negatives. Since they add up to -13, which is negative, they must both be negative. That gives us the coefficients of the O and I terms in the FOIL step.
F O I L
2x2 - 10x - 3x + 15
Factor out the greatest common factor in the F and O terms. That will be the first term in the first binomial.
We can now figure that the other First term must be an x.
and the other Outer term must b a -5.
There are two ways of looking at the final term. It is one of the Last terms, and it is one of the Inner terms. There is something which will satisfy both requirements, namely -3.
If the coefficient on the square term in the factoring problem is not an invisible 1, the problem becomes much more extensive. While the same techniques, which we have set forth here, will work with both types of problems, if the coefficient on the square term is an invisible 1, finding the coefficients of the O aand I terms amounts to simply looking for factors of the last term that add up to the middle term. These factors appear in the answer. If the coefficient of the square term is not an invisible 1, then when we multiply the first and last coefficients, we get something more extensive than just the last term. The factors of the product do not necessarily appear in the answer. One will need to run through the full technique to be able to get the answer.
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