Recognising an expression as an identity in disguise

0:00An identity has an equals sign in the middle of it, and an equals sign faces both ways. a plus b, all squared, equals a squared plus two a b plus b squared. Read left to right, that tells you how to expand a bracket. Read it right to left and it tells you something more useful. If you ever meet an expression of that shape, you are allowed to write it as a bracket squared.

0:29That is factorising by recognition, and it is not a search. Nothing gets tried and nothing gets guessed. It is a matching problem. Here is one. x squared, plus four x, plus four. Cover the middle term with your thumb and look only at the two ends. The first end is x squared, so if this is something all squared, that something starts with x. The last end is four, and four is two squared.

1:00So the candidate is x plus two. And notice what just happened. You had no freedom at all. The ends did not offer a choice between several answers. They named one candidate, and there was nothing to decide. Which leaves the middle term, and here is the idea that makes this quick. You do not work the middle term out. You check it. The candidate is x plus two. If it is right, the middle term has to be two, times x, times two. Four x.

1:35Look up at the expression. It says four x. The candidate passes, and you are finished. So the ends are a naming step and the middle is a testing step. Mixing up those two jobs is what makes this feel like guesswork when it is not. Now one where the naming step has a trap in it. Thirty six x squared, plus twelve x, plus one. The temptation is to read the candidate off as thirty six x.

2:07It is not. You want the thing that squares to give thirty six x squared, and that is six x, because six squared is thirty six. The other end is one, which is one squared, so the candidate is six x plus one. Test it. Two, times six x, times one, is twelve x. The expression says twelve x. It passes. So far every candidate has passed. Watch one fail.

2:38x squared, plus five x, plus four. The ends are exactly the same as before, so they name exactly the same candidate. x plus two. Test it. Two, times x, times two, is four x. But the expression says five x. Four is not five, so this is not a perfect square, and no amount of staring will make it one. But look at what has actually been ruled out. Not that it factorises. Only that it factorises into two identical brackets.

3:13In fact it is x plus one, times x plus four. It splits perfectly well. It just does not split into a matching pair. Now a harder failure, and a more interesting one. Fifty p squared, plus sixty p q, plus eighteen q squared. Start where you always start, at the ends. What squares to give fifty? Nothing you would want to write. Fifty sits between forty nine and sixty four, so its square root is not a whole number, and it is not a fraction either.

3:50So the naming step fails before the test ever runs, and that is a different kind of no. Last time the ends named a candidate and the middle rejected it. This time the ends cannot name one at all. Fifty, sixty and eighteen are all even. So take a two out of all three. What is left inside is twenty five p squared, plus thirty p q, plus nine q squared.

4:20Now the ends are twenty five and nine. Five and three. Candidate: five p plus three q. Test it. Two, times five p, times three q, is thirty p q. It passes. So the whole thing is two, times, five p plus three q, all squared. And here is why that worked, which is worth more than the answer. Fifty is two times twenty five, and the two is exactly the part of fifty that was stopping it being a square.

4:53Take that part away and a square is what is left. Checked on every number below four hundred, that always holds, and the ones that needed nothing taken out are exactly the twenty squares. The factor you take out does not have to be a whole number, and this is where most people stop too early. Three a squared, plus four a b, plus four thirds b squared. Neither end is a square.

5:23Take out a third. Inside is nine a squared, plus twelve a b, plus four b squared, and now the ends are nine and four. Three and two. Test: two, times three a, times two b, is twelve a b. It passes. A third, times, three a plus two b, all squared. The same move handles the one that looks worst. Nine fifths s squared, plus six s v, plus five v squared, is a fifth, times, three s plus five v, all squared.

6:00Two small things that stop this method dead if nobody mentions them. The first is a minus. Sixteen y squared, minus twenty four y, plus nine. The ends are still positive, and they still name four and three. The minus lives in the middle term, and the middle term is the test. So test with a minus three. Two, times four y, times minus three, is minus twenty four y. It passes. Four y minus three, all squared.

6:35The second is order. Sixteen s squared, plus twenty five t squared, minus forty s t. That is not written wrong, it is written in a different order, and the two ends of the expression are no longer the two ends of the identity. Put the two squares on the outside, and it is four s minus five t, all squared. It is worth knowing how special these expressions are, because the method recognises squares and nothing else.

7:08Take every three term expression whose two end coefficients run from one to twelve, and whose middle runs from minus forty to forty. Eleven thousand six hundred and sixty four of them. Eight hundred and eight split into two brackets with ordinary numbers in. And of those, eighteen are a single bracket squared. Eighteen, out of eleven and a half thousand. So the squares sit inside the ones that factorise, which sit inside the ones that exist, and each step in is a big step.

7:45Which tells you what to expect. This method will fail on most things, and that is not a failure of yours. One more distinction, because it is quietly everywhere. Saying x plus two is a factor of something is a weaker claim than saying the something is x plus two, all squared. x plus two is a factor of x squared plus four x plus four, and that one is the square.

8:15But x plus two is also a factor of x squared plus five x plus six, and that one is not a square at all. Its other factor is x plus three. So a factor tells you one bracket. A perfect square tells you both, and tells you they are the same bracket. So the method, in one breath. The two ends name the candidate and leave you no choice. The middle term tests it, and you either pass or you do not.

8:47If an end has no tidy square root, look for a common factor first, and remember the factor is allowed to be a fraction. And if the test fails you have learned something real. This is not a square. It may still factorise. Five thousand seven hundred and sixty expressions built to be a factor times a square were handed to the matcher cold. Eight hundred and sixty four went through as they stood, and the other four thousand eight hundred and ninety six all came good once the factor came out.

9:25But nudge the middle term by one, and every single one of them dies. The test really is a test. For everything that is not a square, you need a method that does search. That is the next lesson.