Chapter 1 · Algebraic Proofs · Concept 1
1-2 Multinomial Theorem
Say you want to expand something with three terms in it: (x + y + z)⁴. Without doing the multiplication, how can you find the coefficient of the term with x²yz in it? And how do you write it down without opening the brackets?
Guiding question
How do you find one coefficient of a three‑term expansion without multiplying it all out?
Learning outcome. Applies the multinomial theorem to find the expansion of algebraic expressions and determine the coefficients of certain terms.
In lesson 1-1 the bracket had two parts, like (2x − 3). Now it has three, like (a + b + c). That is the only change.
Lesson 1-1
Two parts. You picked one number, r, and the other count was whatever was left.
how many ways = n! ⁄ (r! (n−r)!)
Lesson 1-2
Three parts. You pick two counts, p and q, and the third is whatever is left.
how many ways = n! ⁄ (p! q! r!)
Look at the two lines. They are the same rule. In 1-1 there were two counts on the bottom; here there are three. If you ever had four parts, there would be four.
The rule
When you expand (a + b + c)n, the term with ap bq cr in it is
n! ⁄ (p! q! r!) ap bq cr
The three powers p, q and r always add up to n. So if the question gives you two of them, the third one is free.
(a + b + c)8 means eight brackets standing in a row. Each bracket hands you one letter. A term like a4b3c is one way of doing that. The coefficient is just how many ways there are.
Here are 6 brackets in a row. Each one hands over one letter. You asked for 3 a's, 2 b's — so the rest, 1, must be c's.
Every shuffle is a different way to get the same term a3b2c.
How many different ways are there in total?
6! ⁄ (3! × 2! × 1!) = 60
And that number is the coefficient. Nothing else is going on.
Change any number and watch the four lines change with it. The third power is filled in for you, because it is never a choice.
- The third power. 8 − 4 − 3 = 1
- How many ways. 8! ⁄ (4! × 3! × 1!) = 280
- What each part brings. (1)4 × (-2)3 × (3)1 = -24
- Multiply. -6720 a4b3c
The book's own two questions. Find the coefficient of a4b3c in each expansion. Try them first, then open the steps.
(a) (a + b + c)8
- 1
Read off the three powers.
a has power 4, b has power 3, c has power 1Where no power is written, it is 1. So c means c¹.
(b) (a − 2b + 3c)8
- 1
The powers are the same as before, so the counting part is the same.
8! ⁄ (4! × 3! × 1!) = 280The numbers in front of the letters change nothing about how many ways there are.
Work with a partner. Answer question 2 with your group, then talk about the answer. You can send any question below to your notebook. Tap it, and it is waiting there for you.
1. Find the coefficient of these terms in the expansion of (a + b + c)6.
2. Find the coefficient of these terms in the expansion of (a + 3b − 2c)6.
1. Find the coefficient of these terms in the expansion of (x + y + z)7.
2. Find the coefficient of these terms in the expansion of (x + y + z)9.
3. Find the coefficient of these terms in the expansion of (a + 2b − 3c)7.
4. Find the coefficient of these terms in the expansion of (2a − b + 3c)8.
Find the greatest coefficient in the expansion of (1 + x + x2)4.
The three powers must add to n.
In (a + b + c)⁶ there is no term a³b³c². Those powers add to 8, not 6. Check this first — it takes two seconds and saves the whole question.
A missing power is 1, not 0.
In a⁴b³c the c has power 1, so the bottom of the fraction is 4! × 3! × 1!. Reading it as 0 gives the wrong count.
The counting part is not the whole answer.
For (a − 2b + 3c)⁸ the answer is 280 × (−2)³ × 3, not 280. Every number in front of a letter has to be raised to that letter’s power.
Take the minus with the number.
In (a + 2b − 3c)⁷ the third part is −3c. So c⁵ brings (−3)⁵ = −243, not 243. Odd powers stay negative; even powers turn positive.
Lesson 1-3 uses the binomial theorem to prove things.
Keep reading — this lesson marks itself done automatically.