Splitting the middle term, then setting each factor to zero

0:00Here is a piece of news about two numbers, and it is worth almost nothing. Their product is twelve. Three and four? Two and six? A half and twenty-four? On a grid of ordinary numbers, twenty pairs multiply to twelve, and not one contains a nought. Try a hundred and twenty-four: twelve pairs, and again no nought. Now the news that is worth something: their product is nothing. Across that same grid, three hundred and eighty-five pairs multiply to nothing, and every one has a nought in it. There is no other way for it to happen.

0:41That asymmetry is the engine of the whole topic: an ordinary product tells you almost nothing, and a product of nothing tells you almost everything. And it is a fact about the real numbers, not about multiplying. Take the numbers on a clock face, where twelve is back to nought. Three fours are nothing there — and three is not nothing, and four is not nothing. Seventeen pairs there multiply to nothing with neither of them being nothing. So the word REAL is doing work, and the word ONLY has been earned.

1:20Here is the step, and everything that follows uses it. If two real numbers multiply to nothing, then at least one of them is nothing. At least one. Not both. That will matter later. So get an equation into the shape SOMETHING times SOMETHING equals nought, and one question about a squared unknown becomes two about a plain one. Now the trap. Suppose a situation hands you x minus five, times forty minus x, equals a hundred and twenty-four.

1:55It is tempting to set each bracket to a hundred and twenty-four. That gives a hundred and twenty-nine, and minus eighty-four. Tidied, the equation is x squared minus forty-five x plus three hundred and twenty-four equals nought. At a hundred and twenty-nine it comes out at eleven thousand, one hundred and sixty — and at minus eighty-four, exactly the same. Both wrong, by the same enormous amount. The step is licensed by nought and nothing else.

2:30So the first move is always: gather everything onto one side, and get a nought on the other. Watch the machinery run once, with the factorising handed to you. Two x minus three, times x minus one, equals nought. The product is nothing, so at least one factor is nothing. First case: two x minus three is nought, so x is three halves. Second case: x minus one is nought, so x is one. Two candidate answers, and no formula anywhere.

3:07Now check the words AT LEAST. Put three halves into the second factor: three halves minus one is a half. Not nothing — so exactly one factor vanished, and the other is perfectly alive. Across every example here, thirty checks of that kind were made, and in twenty-four of them exactly one factor was nothing. In the other six both were, and those are the special case we meet later.

3:37So the rule is not that both factors are nothing. It is that they cannot both be alive. That leaves the real work: reaching that shape. It is called splitting the middle term, and the recipe is: find two numbers whose sum is b and whose product is a times c. Where does that come from? Work backwards. Suppose the quadratic really is a product of two linear factors: p x plus q, times r x plus s.

4:10Multiply out. The x squared term is p r, so a is p r. The constant is q s, so c is q s. And the middle term is p s plus q r, times x. Look at those two pieces, p s and q r. They add to b, by construction. And multiplied, p s times q r is p r times q s — which is a times c.

4:38So the two numbers are not conjured. They are the two halves of the middle term a factorisation would have left behind. Which is why both conditions are needed: the right sum with the wrong product is a different equation. First one. Two x squared minus five x plus three equals nought. a times c is six, and b is minus five: we want two numbers multiplying to six and adding to minus five.

5:10Six is positive, so they share a sign; they add to something negative, so both are negative. Minus two and minus three. Split the middle term with those: two x squared, minus two x, minus three x, plus three. Group in pairs. The first gives two x, times x minus one; the second gives minus three, times the same bracket. The same bracket twice, which is what tells you the split was right.

5:42So the equation is two x minus three, times x minus one, equals nought — which we already solved. Three halves, and one. And verify — the cheapest step there is. At three halves: nine halves, minus fifteen halves, plus three. Nothing. At one: two, minus five, plus three. Nothing. Second one, and the only new thing in it is a sign. Six x squared minus x minus two equals nought.

6:15a times c is minus twelve, and b is minus one. A negative product forces the two numbers to have opposite signs — not a rule to memorise, just what a negative product means. Three and minus four. Product minus twelve, sum minus one. Split: six x squared, plus three x, minus four x, minus two. Group, and both pairs leave two x plus one. So the factors are three x minus two, and two x plus one, and the roots are two thirds and minus one half.

6:53Both check out on substitution: eight thirds, minus two thirds, minus two is nothing, and three halves, plus one half, minus two is nothing. Third one, and this is where students decide the method has stopped working. It has not — only the numbers have changed. Three x squared minus two root six x plus two equals nought. a times c is six, and b is minus two root six. We want two numbers multiplying to six and adding to minus two root six.

7:30Nothing says they have to be whole. Root six times root six is six, and two of them add to minus two root six. So the split is minus root six, twice. For the grouping, three is root three squared and two is root two squared — so each pair leaves the same bracket, root three x minus root two. One factor, appearing twice. Setting it to nought gives x equals root two over root three — the same number as root six over three.

8:04Verify it exactly, with no decimals. That number squared is two thirds, so the first term is two; the middle term is minus four; the last term is two. Two, minus four, two. Nothing. It passes. Now watch what happens when only one condition is met: a wrong answer can look right for four lines. Back to two x squared minus five x plus three. Try minus one and minus four. They add to minus five, correctly. Their product is four, and it needed to be six.

8:44Split anyway: two x squared, minus x, minus four x, plus three. The first pair leaves two x minus one; the second leaves two x minus one and a half. The brackets do not match, and nothing can be taken out. Push on regardless and what you have factorised is two x minus one, times x minus two — which is two x squared minus five x plus TWO. The right middle term, the wrong constant.

9:15So when the brackets do not match, you have not slipped in the arithmetic. You picked the wrong pair. Now the case that makes people think they have lost an answer. A hundred x squared minus twenty x plus one equals nought. a times c is a hundred, b is minus twenty. Minus ten and minus ten. The factorisation is ten x minus one, all squared. Set it to nought: x is one tenth. And the other factor is the same factor, saying the same thing.

9:52So there is ONE distinct number that solves this equation, and TWO roots, because the factor supplying it appears twice. The first counts numbers; the second counts factors. This is also the case where both factors really are nothing at once — the six checks out of thirty from earlier. Now finish something left open: a hall with a carpet covering exactly three hundred square metres, whose length is one metre more than twice its breadth.

10:27Calling the breadth x, that became two x squared plus x minus three hundred equals nought. a times c is minus six hundred, and b is one. Opposite signs, nearly equal in size: minus twenty-four and twenty-five. Split, group, and both pairs leave x minus twelve. So the factors are x minus twelve, and two x plus twenty-five. The roots are twelve, and minus twelve point five. Twelve is a breadth. The length is then twenty-five, and twelve times twenty-five is three hundred. It fits.

11:07And minus twelve point five is a perfectly good root. It is not wrong. It is simply not a breadth. So the hall is twelve metres by twenty-five — and saying WHY the other root goes is part of the answer. That last step is where marks go quietly missing. An equation knows nothing about where it came from. Seven situations were modelled for this video, and their equations produce fourteen roots between them. Ten are answers a situation can use. Four are not.

11:44A breadth of minus twelve point five metres. A base of minus five centimetres. A workshop that made minus seven and a half articles. A negative integer, when the question asked for a positive one. Each is a correct root, thrown out for a reason that lives in the story and never in the algebra. And the test is not always simply POSITIVE: a count of articles has to be whole, and a triangle's base has to exceed seven or the altitude does not exist.

12:21So write the conditions beside the equation, at the moment you form it, while you still remember what the letter meant. And sometimes nothing is thrown out. Two people sharing forty-five marbles give nine and thirty-six, and both are usable — the same pair, the other way about. One last thing, and it is the most important thing to know about this method: it usually does not work. Every example here was chosen so that two convenient numbers exist, and most quadratics have no such pair.

13:00That was measured rather than guessed. Sixteen hundred and eighty quadratics with small whole coefficients were swept, and each was searched for a split. Two hundred and thirty-one of them have one. One thousand four hundred and forty-nine have none at all. Roughly one in seven. Sit down with a quadratic you did not choose, and you are usually hunting for something that is not there. That is not a flaw in the method. It is the reason a method that always works comes next — not an alternative to this one, but the thing that has to exist because this one runs out.

13:43A word on the checking, because this claim confirms itself if you are not careful. 'A product is nothing only when a factor is nothing' is TRUE — so a sweep that only looks at products which ARE nothing agrees for free, having tested nothing. So every pair was filed by two things at once: what the product came to, and whether either factor was nought. The cell that had to stay empty sits beside cells that had to fill.

14:14Then the identical filing ran over the numbers modulo twelve, through the same line of code, where that empty cell fills seventeen times. That occupied cell is what makes the empty one evidence. The roots were reached three ways sharing no machinery: read off the factors, with nothing evaluated; found by substitution, with nothing factorised; and a third, wrong on purpose, whose actual answers were asserted rather than merely required to differ.

14:50Each admissibility test was put to a value it has to refuse — a breadth of nought, a base of seven, half an article — because a test only shown things it accepts has not been tested. And the surd work was exact: no decimal appears anywhere, because root six over three is not a decimal. Factorising does not solve a quadratic equation. Nought does. Among real numbers a product is nothing only when something multiplied was nothing — which is why everything is gathered to one side first.

15:31At least one factor, not both. Take the cases in turn. Splitting the middle term hunts two numbers with sum b and product a c — exactly the two pieces a pair of linear factors would have left behind, which is why both conditions have to hold. Surds are allowed, and a repeated factor supplies its root twice — so one distinct number and two roots is not a contradiction.

16:00A root the situation cannot use is still a root. Throw it out, and say why. And most quadratics have no convenient split at all — not a gap in what you have learned, but the reason for what comes next.