Question

Binomial Expansion Without Pascal Triangle

Answer 1

Suppose you wanted to expand \((4x - 5y)^4\) without depending on Pascal's Triangle but do not know exactly how to proceed? The following steps will hopefully reveal a way for you to expand binomials in a manner that is more intuitive than applying Pascal's Triangle.

1st, what you want to do is write out each term. Begin by writing only the \(x\) variable} expressions (in descending order) as demonstrated below:
\begin{array}{|c|c|c|c|c|c|c|c|}
\hline
(4x)^4&⁣⁣⁣⁣⁣\phantom{-}&(4x)^3&⁣⁣\phantom{-}&(4x)^2&⁣⁣\phantom{-}&(4x)^1&⁣⁣\phantom{-}&(4x)^0
\\\hline
\end{array}
Notice, the spaces between the expressions are reserved for placing plus or minus symbols later on.

2nd, Next write out the \(y\) variable expressions to the immediate right of the \(x\) variable expressions in ascending order like so:
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline(4x)^4(5y)^0 &\phantom{-}&(4x)^3(5y)^1 &\phantom{-}& (4x)^2(5y)^2 &\phantom{-}&(4x)^1(5y)^3 &\phantom{-}&(4x)^0(5y)^4\\\hline\end{array}

3rd, In a toggle-like manner, add plus or minus symbols between each term beginning with a + for the 1st term and a minus between the 1st and 2nd terms:
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline(4x)^4(5y)^0 &- &(4x)^3(5y)^1 & + & (4x)^2(5y)^2 &- &(4x)^1(5y)^3 &+ &(4x)^0(5y)^4\\\hline\end{array}

4th, Write the term number directly beneath each corresponding term:
\begin{array}{|c|c|c|c|c|c|c|c|c|}\hline(4x)^4(5y)^0 &- &(4x)^3(5y)^1 & + & (4x)^2(5y)^2 &- &(4x)^1(5y)^3 &+ &(4x)^0(5y)^4\\\hline1& &2& &3& &4& &5 \\\hline\end{array}

5th, Apply the Coefficient Generating Algorithm:}\)

\(\text{Coefficient of the \(n\)th term = }\)
\(\dfrac{\text{The coefficient of the previous term \(\times x\) exponent of previous term}}{\text{Number of the previous term:}}\)

As demonstrated here:
\begin{array}
{|c|c|c|}
\hline
\text{Term Number}&\text{Coefficient Algorithm}&\text{Resultant Term}
\\\hline
1&\text{No Previous Term}&(4x)^4(5y)^0
\\\hline
2&\dfrac{1 \times 4}{1}=4&-4(4x)^3(5y)^1
\\\hline
3&\dfrac{4 \times 3}{2}=6&6(4x)^2(5y)^2
\\\hline
4&\dfrac{6 \times 2}{3}=4&-4(4x)^1(5y)^3
\\\hline
5&\dfrac{4 \times 1}{4}=1&1(4x)^0(5y)^4
\\\hline
\end{array}

6th, At this point, your binomial expansion should look like this:
\((4x)^4(5y)^0-4(4x)^3(5y)^1+6(4x)^2(5y)^2 - 4(4x)^1(5y)^3+(4x)^0(5y)^4\)


7th, You can simplify it further by removing zero exponents.
\((4x)^4-4(4x)^3(5y)+6(4x)^2(5y)^2-4(4x)(5y)^3+(5y)^4\)

8th, If you expand completely, you end up with:
\(256x^4 - 1280x^3y + 2400x^2y^2 - 2000xy^3+625y^4\)

Admittedly this a bit involved, however, once you are comfortable enough, you may find this approach to binomial expansion far easier and intuitive to use than its predecessor.

\(\textbf{Pascal who?}\)