The Binomial Expander
Section 13.8 -- Binomial expansion
Step 1 of 8
Let's start simple. What is \((a + b)^1\)?
It is just \(a + b\). The coefficients are: 1, 1.
Not much to see yet. One row of coefficients: 1, 1.
Step 2 of 8
Now FOIL \((a + b)^2 = (a+b)(a+b)\). Which of these is the result?
Step 3 of 8
Expand \((a + b)^3\) by multiplying \((a^2 + 2ab + b^2)\) by \((a + b)\). What are the coefficients?
Step 4 of 8
Now \((a + b)^4\). The coefficients for \((a+b)^3\) were 1, 3, 3, 1. Multiply that result by \((a + b)\) one more time. What coefficients do you get?
Step 5 of 8
Look at the triangle you just built. The 6 in row 4 sits below two numbers. What are they?
Step 6 of 8
Expand \((a + b)^7\) by reading row 7 off the triangle, instead of multiplying seven times. The exponent pattern: \(a\) counts down from \(n\) to 0, \(b\) counts up from 0 to \(n\), and every term's exponents sum to \(n\).
First, predict: what is the third term of \((a+b)^7\)?
Step 7 of 8
Where do these numbers come from? The 6 in row 4 is \(\binom{4}{2} = \frac{4!}{2!\,2!} = 6\). Why?
When you expand \((a+b)(a+b)(a+b)(a+b)\), each term in the result picks either \(a\) or \(b\) from each of the four factors. The \(a^2b^2\) terms are exactly the ways to choose which 2 of the 4 factors give you a \(b\). There are \(\binom{4}{2} = 6\) such choices.
The coefficient is a count. It counts arrangements.
What is \(\binom{7}{3}\)?
Step 8 of 8
Coin flips. Five flips, exactly three heads. How many ways can that happen?
Explore the Binomial Expansion
Pascal's Triangle
Expansion
Custom Binomial
Set \(a\) and \(b\) terms with coefficients and signs, then see the expansion.
Select a power \(n\) to see the corresponding row of Pascal's Triangle highlighted and the full binomial expansion displayed below. Click any coefficient in the triangle to see its factorial computation.
Try This
Expand \((2x - 3)^4\) using the binomial theorem, not by multiplying. Watch the signs: \(b = -3\), so odd powers of \(b\) are negative.
The coefficients from row 4 of Pascal's Triangle are: 1, 4, 6, 4, 1.
Which of these is the correct expansion?
Find the coefficient of \(x^3\) in \((x + 2)^7\) without expanding the whole thing. Which single term do you need, and how do you find it directly?
The term containing \(x^3\) in \((x+2)^7\) has the form \(\binom{7}{k} x^{7-k} \cdot 2^k\). You need \(7 - k = 3\), so \(k = 4\).
What is the coefficient of \(x^3\)?
Two carriers of a recessive trait have 5 children. Each child independently has a 0.25 chance of expressing it. What is the probability that exactly 2 of the 5 express it?
Set it up as \(\binom{5}{2}(0.25)^2(0.75)^3\) and compute.
What is the probability (to four decimal places, within 0.001)?
Tolerance: within 0.001
Then explain: what is the \(\binom{5}{2}\) counting, and why do you need it at all?
- Expand \((a+b)^n\) using the Binomial Theorem
- Construct Pascal's Triangle and use it to find binomial coefficients
- Compute \(\binom{n}{k}\) using the factorial formula
- Identify the exponent pattern in a binomial expansion
- Connect binomial coefficients to counting problems and probability
- Handle negative terms in binomial expansions
Quick Check
What is the coefficient of the \(a^3b^4\) term in the expansion of \((a+b)^7\)?
Instructor Notes
Teaching Notes
The Discover sequence is designed to make students feel the tedium of repeated multiplication before revealing Pascal's Triangle as the shortcut. Steps 1 through 4 are deliberately mechanical. The payoff arrives in Step 5 when the pattern emerges, and Step 6 when they can expand a 7th-power binomial without any multiplication at all.
Step 7 connects the coefficients to combinatorics. The key insight is that each coefficient counts something -- the number of ways to arrange a's and b's -- rather than being an arbitrary number that happens to appear.
Step 8 bridges algebra and probability. Students who built the triangle for an algebra reason discover it answers a probability question they already care about.
Common Student Errors
- Dropping signs when b is negative. In \((2x - 3)^4\), students often forget that \((-3)^3 = -27\), not \(+27\). The Starter problem in Apply mode targets this directly.
- Thinking coefficients are arbitrary. Step 7 addresses this by showing that each coefficient counts the number of arrangements. Students who skip this step treat the triangle as a lookup table without understanding why the numbers work.
- Confusing \(\binom{n}{k}\) with \(n \cdot k\). Some students compute \(\binom{7}{3}\) as \(7 \times 3 = 21\) and get the right answer for the wrong reason. Probe with \(\binom{7}{4}\) to catch this.
Discussion Questions
- Why does row \(n\) of Pascal's Triangle always start and end with 1?
- What is the sum of all entries in row \(n\)? Can you see why from the expansion of \((1+1)^n\)?
- If you expand \((1-1)^n\), what do you get? What does that tell you about alternating sums of Pascal's Triangle rows?
- Why is \(\binom{n}{k} = \binom{n}{n-k}\)? Can you explain this both algebraically and in terms of counting?
Exam Connection
Binomial expansion problems on the exam typically ask for a specific term (like "find the coefficient of \(x^5\) in \((2x-1)^8\)") rather than the full expansion. Students who can identify the correct \(k\) value and apply \(\binom{n}{k} a^{n-k} b^k\) directly will save significant time. The Stretch problem in Apply mode practices exactly this skill.