Splitting the middle term once the tiles come away

0:00There is a picture where a quadratic is built out of tiles: one square, some strips, some units, pushed into a rectangle. It is a good picture. Now put it away. Not because it was wrong. Because there are quadratics it cannot draw, and we are about to meet one. So watch what survives when the tiles go. Two conditions survive. Add to the middle term, multiply to the constant. And it turns out they were never about tiles at all.

0:34When are two expressions the same expression? Answer: when their coefficients match, position by position. Same number in front of x squared, same in front of x, same on the end. Worth checking rather than believing, so I checked it. Fifteen thousand six hundred and twenty five pairs of quadratics, asked two ways. One way: do the coefficients match? The other way: do the two agree when you feed them three different numbers?

1:05The same hundred and twenty five pairs, both times. Three numbers is the honest minimum. Ask only two and four hundred and twenty five pairs sneak through, agreeing twice and differing everywhere else. Now use it. Here is x squared plus seven x plus twelve. And here, underneath, is what any pair of brackets multiplies out to: x squared, plus a plus b, times x, plus a b. We want these to be the same expression. So match them.

1:40The top coefficients are both one. Nothing to learn there. Match the middle: a plus b is seven. Match the end: a b is twelve. And now stop. There is nothing else to match - three places, all three used. So the two conditions are not rules somebody handed you. They are one comparison performed twice, and there is no room for a third. That turns factorising into a search that finishes.

2:11Taken one at a time, neither condition is any use. Two numbers adding to seven? In a window of forty-one whole numbers there are seventeen. Widen the window and there are thirty-seven. That search never ends. Two numbers multiplying to twelve? That one does end. Twelve has six factor pairs and that is all there will ever be. So run the finite one first. List the factor pairs of the constant, then test their sums.

2:44For twelve, exactly one pair survives both. Three and four. And three and four, or four and three, is the same factorisation with the brackets the other way round. One answer, not two. Now one where the wrong answers are worth more than the right one. x squared plus eleven x plus thirty. Factor pairs of thirty. One and thirty. Two and fifteen. Three and ten. Five and six. Four of them, then you run out.

3:18Now the sums. Thirty-one. Seventeen. Thirteen. Eleven. Try two and fifteen. The product is right, thirty exactly. The sum is seventeen. Rejected. Try three and ten. Product right again. Sum thirteen. Rejected. Five and six. Thirty, and eleven. Both conditions. So it is x plus five, times x plus six. Two candidates with the right product and the wrong sum is not a detour. It is the method working - the product narrows the field, the sum picks the winner.

3:58Now the one the tiles could not draw. x squared minus five x plus six. There is no negative tile in a box of tiles, so the picture stops here. The algebra does not. We need two numbers with sum minus five and product six. Look at the product first. It is positive, so the two numbers share a sign. Both plus, or both minus. Now the sum. It is negative. So the sign they share is minus.

4:30Minus two and minus three. x minus two, times x minus three. And notice what did not happen. Neither is negative because of the minus in the middle. Both are negative because a positive product and a negative sum leave nothing else available. That is a rule you can run before any arithmetic at all. The constant decides first. The middle term decides second. Positive constant, positive middle: both parts are positive.

5:07Positive constant, negative middle: both parts are negative - the last one. Negative constant: one of each, always, whatever the middle term does. In that third case the middle term still has a job: the part with the larger size carries its sign. I ran every pair from minus twelve to twelve through that - five hundred and seventy-six of them. Right every time, and all three cases get used.

5:39Three quick ones, to make the search automatic. s squared minus eleven s plus twenty-four. Positive constant, negative middle, so both parts are minus. Minus three and minus eight. s minus three, times s minus eight. r squared minus r minus forty-two. Negative constant, so one of each. Minus seven and six. r minus seven, times r plus six. Now one half given. Something, times x plus one, makes three x squared minus four x minus seven.

6:19The three x squared must come from somewhere, so the bracket starts three x. The minus seven must too, so it ends minus seven. Three x minus seven. And that is a warning shot: the front of that one was not a single x. Ten x squared minus eleven x minus six. Try the recipe as it stands and it fails. Factor pairs of minus six: not one of them sums to minus eleven. The search comes back empty.

6:54Here is the repair, and it is the one thing nobody tells you. Do not split the constant. Split the leading coefficient times the constant. Ten times minus six is minus sixty. Two numbers with product minus sixty and sum minus eleven: minus fifteen and four. Now write the middle term as those two. Ten x squared, minus fifteen x, plus four x, minus six. Four terms. Take them in pairs. Out of the first two comes five x, leaving two x minus three. Out of the last two comes two, leaving two x minus three.

7:36The same bracket, twice. That is the trick, and it is why the split had to be exactly that one. So: five x plus two, times two x minus three. I built nineteen hundred and forty-four quadratics from brackets and put every one through that routine. Every one came back as what it was built from. The same identity will multiply numbers for you. Twenty-three times seventeen. Both are twenty, give or take three. So it is twenty plus three, times twenty minus three.

8:14The two parts cancel in the middle, and you are left with four hundred minus nine. Three hundred and ninety-one. Now eighteen times twenty-nine, and the trick seems to break: no whole number sits midway between them. It does not break. Call it twenty minus two, times twenty plus nine. Four hundred, plus twenty times seven, minus eighteen. Five hundred and twenty-two. The difference of two squares was never a separate identity. It is this one, in the special case where the two parts cancel.

8:53Last, a use for it all. A rectangular pool. Its breadth is four metres less than its length, its area ninety-six square metres. How big is it? Call the length x. Then the breadth is x minus four, and the area is x times x minus four. Set that equal to ninety-six and move everything to one side. x squared minus four x minus ninety-six equals nothing. Negative constant, so one part of each sign. Product minus ninety-six, sum minus four. Minus twelve and eight.

9:33Split, pair, and take out the bracket. x minus twelve, times x plus eight. A product is nothing only when one of the pieces is nothing. So x is twelve, or x is minus eight. Two roots. One pool. Twelve gives a length of twelve, a breadth of eight, and twelve eights is ninety-six. That works. Minus eight would give a breadth of minus twelve. There is no such pool.

10:04So we throw that root away — and I want to be careful about why. Not because the algebra says so - the algebra is perfectly happy with it. Because a length cannot be negative, and that is a fact about pools, not equations. Here is the proof it matters. A number plus its reciprocal is ten thirds. Clear the fraction: three x squared minus ten x plus three, which factorises as three x minus one, times x minus three.

10:39Two roots again: three, and a third. This time you keep both, because each is the other's reciprocal. Same method, same two roots, opposite decision. Solving the equation was never the step where a root gets discarded.