The Binomial Theorem (College Board AP® Precalculus): Revision Note

Binomial theorem

What is a binomial?

• A binomial is an expression consisting of two terms added together

  • E.g. $(a+b)$, $(x+3)$

• Expanding a binomial raised to a power means writing it as a polynomial in standard form

  • E.g. $(x+3{)}^2=x^2+6x+9$

What is Pascal's Triangle?

• Pascal's Triangle is a triangular arrangement of numbers where each entry is the sum of the two entries directly above it

• The rows of Pascal's Triangle are numbered starting from row $0$:

  • Row 0: $1$

  • Row 1: $11$

  • Row 2: $121$

  • Row 3: $1331$

  • Row 4: $14641$

  • Row 5: $15101051$

• The table can be extended to new rows by

  • adding the two numbers above each position

  • and putting 1s on the outside

• The entries in row $n$ are the coefficients needed to expand $(a+b{)}^n$

How do I use Pascal's Triangle to expand (a+b)ⁿ?

• To expand $(a+b{)}^n$

  • Look up row $n$ of Pascal's Triangle for the coefficients

  • Write out terms where for each term

    • the power of $a$ decreases from $n$ to $0$

    • and the power of $b$ increases from $0$ to $n$

      • There will be $n+1$ terms in total

  • Multiply each term by the corresponding coefficient from Pascal's Triangle

    • and then add all the terms together

  • Simplify

• E.g. to expand $(a+b{)}^3$:

  • Row 3 of Pascal's Triangle gives coefficients: $1, 3, 3, 1$

  • The terms are $a^3{b}^0,a^2{b}^1,a^1{b}^2,a^0{b}^3$

  • Multiplying by the coefficients and adding together gives

    • $(a+b{)}^3=1·a^3{b}^0+3·a^2{b}^1+3·a^1{b}^2+1·a^0{b}^3$

  • Simplify, including using $a^0=b^0=1$, $a^1=a$ and $b^1=b$

    • $(a+b{)}^3=a^3+3a^{2}b+3a{b}^2+b^3$

• Following the same procedure for $(a+b{)}^4$ gives

  • $(a+b{)}^4=a^4+4a^{3}b+6a^2{b}^2+4a{b}^3+b^4$

How do I expand polynomial functions of the form (x+c)ⁿ?

• The binomial theorem applies directly to expressions of the form $(x+c{)}^n$

  • where $c$ is a constant

• This is just the general $(a+b{)}^n$ expansion with $a=x$ and $b=c$

  • After expanding, simplify by evaluating the powers of $c$ and collecting terms

• E.g. to expand $(x+2{)}^3$:

  • Row 3 of Pascal's Triangle gives coefficients: $1, 3, 3, 1$

  • The terms are $x^3·2^0,x^2·2^1,x^1·2^2,x^0·2^3$

  • Multiplying by the coefficients and adding together gives

    • $(x+2{)}^3=1·x^3·2^0+3·x^2·2^1+3·x^1·2^2+1·x^0·2^3$

  • Simplify, including using $x^0=2^0=1$ and $x^1=x$

    • $\begin{array}{rcl}(x+2{)}^3& =& 1·x^3·1+3·x^2·2+3·x·4+1·1·8\\ & =& x^3+6x^2+12x+8\end{array}$

• Or to expand $(x-4{)}^4$:

  • Here $b=-4$

    • (note the negative sign, i.e. treat the subtraction as addition of a negative)

  • Row 4 of Pascal's Triangle gives coefficients: $1, 4,6, 4, 1$

  • The terms are $x^4·{\left(-4\right)}^0,x^3·{\left(-4\right)}^1,x^2·{\left(-4\right)}^2,x^1·{\left(-4\right)}^3,x^0·{\left(-4\right)}^4$

  • Multiplying by the coefficients and adding together gives

    • ${\left(x-4\right)}^4=1·x^4·{\left(-4\right)}^0+4·x^3·{\left(-4\right)}^1+6·x^2·{\left(-4\right)}^2+4·x^1·{\left(-4\right)}^3+1·x^0·{\left(-4\right)}^4$

    • Simplify, including using $x^0={\left(-4\right)}^0=1$ and $x^1=x$

      • $\begin{array}{rcl}{\left(x-4\right)}^4& =& 1·x^4·1+4·x^3·\left(-4\right)+6·x^2·16+4·x·\left(-64\right)+1·1·256\\ & =& x^4-16x^3+96x^2-256x+256\end{array}$