 So let's try to tackle factoring in the hardest way possible. Actually, if you want to do yourself a real favor, just skip this video and go on to the next one. So far, we found that we can factor quadratic second-degree polynomials by using the quadratic formula to find a root, and then the root theorem to find the factors, or by rewriting the quadratic expression as a difference of squares. But these methods are too easy, and they work on all quadratics. And maybe you don't want to do things the easy way. Maybe you'd much rather do things the hard way, and so we'll introduce a hard method that doesn't always work. And again, if you want to do yourself a favor, skip this video and go on to the next one. Now, this problem is so hard we have to consider two cases. The first concern what are called monic polynomials, where the coefficient of x squared is 1, and an even harder case is when we have non-monic polynomials, where the coefficient of x squared is, wait for it, not 1. So let's consider this factoring a monic polynomial. If a quadratic polynomial is factorable, it must be the product of two linear factors. And so we observe that if I multiply x plus a times x plus b, I get x squared plus bx plus ax plus ab. And the thing to notice is that these two terms here in the middle can be rewritten as a plus b times x. And so this means if a monic polynomial can be written as a product of two binomials, the coefficient of x has to be the sum of the constant terms, and the constant itself is the product of the constant terms. And this suggests the following approach, which is known as trial and error factorization. First, make a list of every possible pair of integers that can multiply to the constant, and then find a pair that adds to the coefficient of x. Now, fair warning, trial and error factorization is the slowest, least efficient, and worst method of trying to factor. So if you don't want to use the quadratic formula, if you don't want to rewrite things as a difference of squares, what you're going to have to do is a lot of extra work. How bad is it? Well, let's try it out. So let's say we want to factor x squared plus 2x minus 24. So we want to find two numbers that multiply to negative 24 and add to 2. Well, since the numbers multiply to negative 24, we know one of them has to be positive and the other one has to be negative. And we now just need to go through our list of things that multiply to negative 24 and see which one adds to 2. So let's make a list of all the numbers that multiply to negative 24, and we have to check through our entire list to see if we can find a pair that adds to 2. So 1 and negative 24 don't add to 2. Negative 1 and 24 don't add to 2. 2 and negative 12 don't add to 2. Negative 2 and 12, nope, 3 and negative 8, nope, negative 3 and 8, nope, 4, negative 6, nope, negative 4 and 6, no, oh wait, wait, we found it. And so that says that x minus 4 and x plus 6 are our two factors. So for example, consider x squared minus 14x plus 13. We want to find two numbers that multiply to 13 and add to negative 14. Since they multiply to a positive, both of them have to have the same sign. Since they add to a negative, both of them have to be negative. And here's why this trial and error factorization is helpful. Since they multiply to 13, we only have one choice, negative 1 and negative 13. It's vitally important to understand that we have not yet found the factors, but we have found what might work. We still have to check to see if these add to negative 14. And it turns out they do, and so we have our factorization x minus 13 times x minus 1. How about x squared plus 12x minus 24? We need to find two numbers that multiply to negative 24 and add to 12. So we list the numbers that multiply to negative 24, then check to see if they add to 12. So the numbers that multiply to negative 24, and now we have to check every one of them. But none of them work. And so since there are no numbers that multiply to negative 24 and add to 12, this is not factorable over the rationals. And what's important to remember here is that if you choose to use trial and error factorization, you may have to try every possible set of factors before concluding that nothing works. And so this is part of the reason why trial and error factorization should be your last choice of how to factor a quadratic polynomial. What if we have non-monic polynomials? In that case, one or both factors must have a coefficient on x. So ax plus b times cx plus d will be equal to acx squared plus ad plus bx times x plus bd. And this means we need to find four numbers a, b, c, and d, where the product ac is the coefficient of x squared. The product bd is the constant term, and the sum ad plus bc is the coefficient of x. Good luck. So let's try to factor 2x squared minus 5x minus 12. We want to find four numbers where ac equals 2, bd equals negative 12, and ad plus bc is negative 5. Now we can simplify our problem slightly. Since ac equals 2, then a and c have to have the same signs. But if both are negative, we can remove two factors of negative 1 and make both positive. So we can assume that a and c are both positive. Since the only two numbers that multiply the 2 are 1 and 2, we can simplify and assume that if a factorization exists, 2x squared minus 5x minus 12 must be 2x plus c times x plus d. And this means we need c and d to multiply to negative 12, and 2d plus c has to be negative 5. So now we go through our trial and error process. We list every pair of numbers that multiply to negative 12, and check to see if 2c plus d is equal to minus 5. And sometime next week we find that d equals negative 4, c equals 3 work, and we have our factorization. So you've gotten all the way to the end of this video, which means you didn't take my advice to skip it, and that's okay, because there is value to learning trial and error factorization, and it comes from the following idea. If the only tool you have is a hammer, then every problem must be treated like a nail. And that's great if you're trying to drive a nail, but not so great if you're trying to work with a screw, and if you're trying to open a pickle jar, you really don't want to do that. The key idea is that the different methods are better in different situations. So remember we can factor by using the quadratic formula, or we can rewrite the expression as a difference of squares, or we can use trial and error. You might consider the problems that we looked at. So it's worth taking a moment and think about applying the different methods to these factorization problems and deciding whether you want to use a hammer to open that pickle jar.