Trinomials by Factorising

Transcript

0:00Introduction

Hello everyone! Okay, a scary question to start. Find the coefficient of x to the power fifteen in one minus x, to the power ten, times one plus x plus x squared, to the power ten. Ten copies of a three term bracket! Multiply it all out? Please don't! One small factorisation does it in one line. This comes up often in JEE, so let's begin.

0:26Factorise first

So, the whole lesson rests on one small multiplication. Take one minus x, and multiply it by one plus x plus x squared. What do you get? Do it in your head. One minus x cubed, yes, exactly! Everything in the middle cancels. So one minus x cubed equals one minus x, times one plus x plus x squared. The small raised number is the power, okay? Now its twin, one plus x cubed. It is one plus x, times a three term bracket. Is that bracket one plus x plus x squared again? No, no, careful! The middle sign flips. One plus x cubed equals one plus x, times one minus x plus x squared. Students write a plus there in a hurry. Don't! Got it? Third, the general version. One plus x plus, and so on up to x to the power k minus one, equals one minus x to the power k, over one minus x. The three dots mean the powers keep going. Put k equal to three: our first line, just rearranged! And here is why it matters. Raise the first line to the power n. One minus x to the power n, times one plus x plus x squared to the power n, is just one minus x cubed, to the power n. The trinomial is gone!

1:49Coefficients of (1+x+x²)ⁿ: symmetry

Clear? Now the trinomial itself, with n a positive integer. We call its coefficients a sub r: a, with a small r below. The big Greek letter sigma means add up the terms a sub r times x to the power r, from r equals zero up to two n. It stops at two n, because each of the n brackets gives at most x squared. Okay? Let's see a small case, n equal to two. We get one plus two x plus three x squared plus two x cubed plus x to the power four. Read the coefficients: one, two, three, two, one. Notice anything? Yes! Same forwards and backwards. And it holds for every n: a sub r equals a sub two n minus r. First equals last, second equals second last, and so on. With me? The reason: replace x by one over x, and multiply by x to the power two n. On the left, the coefficients now come in reverse order. On the right, x squared plus x plus one, to the power n: the same trinomial! So the list reads the same both ways. Clear?

2:58Sums of the coefficients

Now, sums of these coefficients. Suppose you want to add a sub zero, then a sub one, and so on, all the way up to a sub two n. Which value of x should you put? x equals one, correct! Every power of x becomes one, and one plus one plus one is three. So all the coefficients add up to three, to the power n. For n equal to two: one plus two plus three plus two plus one is nine. Three squared, right? Now put x equal to minus one. The signs alternate: a sub zero minus a sub one plus a sub two, and so on. And the trinomial gives one minus one plus one, which is one; one to the power n is one. Got it? Now, what if they ask for only the even ones: a sub zero plus a sub two plus a sub four, and so on? You have two equations on screen. How will you use them? Add them! The odd ones cancel, and each even one appears twice. So halve: the even sum is three to the power n plus one, over two. For n equal to two, one plus three plus one is five, and nine plus one, over two, is five. Perfect! One more, a topper's trick. For the sum of r times a sub r, differentiate both sides, then put x equal to one. The bracket becomes three, to the power n minus one, and the derivative of the inside, one plus two x, becomes three. So we get n times three to the power n minus one, times three; the dot means times. That is n times three to the power n. An r in front? Differentiate!

4:38Example 1: coefficient of x^15 in (1-x)^10 (1+x+x^2)^10

Now, our scary question from the start, here on screen. The coefficient of x to the power fifteen. Hint: look at the powers of the two brackets. Pause the video and try it. Done? Okay, let's check. Both brackets have the same power, ten. So pair them: the whole product is just one minus x cubed, to the power ten. See? No trinomial left! Now the general term, T k plus one, with k counted from zero: capital T, with a small k plus one below. It is ten C k, the capital C with ten written at its top left, times minus x cubed, to the power k. Sign out: minus one to the power k, times ten C k, times x to the power three k. We want x to the power fifteen. So which k do we need? Three k equals fifteen, so k equals five. The double arrow means implies, okay? Now, ten C five is two hundred and fifty two. So the coefficient is two hundred and fifty two, right? Oh no, careful! The sign! Minus one to the power five is minus one, so the coefficient is minus two hundred and fifty two. Dropping that minus is the classic mistake here. See how fast that was?

5:57Example 2: coefficient of x^16 in (1-x)^11 (1+x+x^2)^10

Now a slightly harder one. Find the coefficient of x to the power sixteen in one minus x to the power eleven, times one plus x plus x squared to the power ten. Careful, eleven and ten this time. Pause the video and try it. Done? Okay, let's check. Eleven is ten plus one. So ten copies of one minus x pair with the trinomial to give one minus x cubed, to the power ten, and one single one minus x is left outside. Okay? Now think. In one minus x cubed, to the power ten, which powers of x can appear? Only multiples of three, exactly! x to the power zero, x cubed, x to the power six, and so on, up to x to the power thirty. Right? From the single one minus x, we pick either the one or the minus x. If we pick the one, we need x to the power sixteen from the other bracket. Is that possible? No! Sixteen is three times five, plus one: not a multiple of three. So we must pick minus x, with x to the power fifteen, the term with k equal to five: minus one to the power five, times ten C five, times x to the power fifteen. Clear? So the coefficient is minus one, times minus two hundred and fifty two. Two minus signs make a plus: two hundred and fifty two! Same ten C five as Example one, but the sign flips. Watch the signs!

7:27Quick revision: 1 - x^3 factor, symmetry, coefficient sums

So, quick revision. One: spot the factor. One minus x, times one plus x plus x squared, is one minus x cubed; and match the powers. Two: a sub r equals a sub two n minus r; all the coefficients add to three to the power n; with signs minus one to the power r, they add to one. Three: the even ones give three to the power n plus one, over two, and the sum of r times a sub r is n times three to the power n. Now try the practice problems on this page. All the best!