Question 338972


{{{(2x-3y)^5}}} Start with the given expression


To expand this, we're going to use binomial expansion. So let's look at Pascal's triangle:
<center>1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;2&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;3&nbsp; &nbsp;3&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;4&nbsp; &nbsp;6&nbsp; &nbsp;4&nbsp; &nbsp;1&nbsp; &nbsp;</center>
<center>1&nbsp; &nbsp;5&nbsp; &nbsp;10&nbsp; &nbsp;10&nbsp; &nbsp;5&nbsp; &nbsp;1&nbsp; &nbsp;</center>




Looking at the row that starts with 1,5, etc, we can see that this row has the numbers:


1, 5, 10, 10, 5, and 1


These numbers will be the coefficients of our expansion. So to expand {{{(2x-3y)^5}}}, simply follow this procedure:

Write the first coefficient. Multiply that coefficient with the first binomial term {{{2x}}} and then the second binomial term {{{-3y}}}. Repeat this until all of the coefficients have been written.


Once that has been done, add up the terms like this:



{{{highlight(1)(2x)(-3y)+highlight(5)(2x)(-3y)+highlight(10)(2x)(-3y)+highlight(10)(2x)(-3y)+highlight(5)(2x)(-3y)+highlight(1)(2x)(-3y)}}} Notice how the coefficients are in front of each term.




However, we're not done yet.



{{{1(2x)^5(-3y)^0+(2x)(-3y)+5(2x)(-3y)+10(2x)(-3y)+10(2x)(-3y)+5(2x)(-3y)+1(2x)(-3y)}}} Looking at the first term {{{1(2x)(-3y)}}}, raise  {{{2x}}} to the 5th power and raise {{{-3y}}} to the 0th power.


{{{1(2x)^5(-3y)^0+(2x)^4(-3y)^1+5(2x)(-3y)+10(2x)(-3y)+10(2x)(-3y)+5(2x)(-3y)+1(2x)(-3y)}}} Looking at the  second term {{{5(2x)(-3y)}}} raise  {{{2x}}} to the 4th power and raise {{{-3y}}} to the 1st power.


Continue this until you reach the final term.



Notice how the exponents of {{{2x}}} are stepping down and the exponents of {{{-3y}}}  are stepping up.



So the fully expanded expression should now look like this:



{{{1(2x)^5(-3y)^0+5(2x)^4(-3y)^1+10(2x)^3(-3y)^2+10(2x)^2(-3y)^3+5(2x)^1(-3y)^4+1(2x)^0(-3y)^5}}}



{{{1(32x^5)(y^0)+5(16x^4)(-3y^1)+10(8x^3)(9y^2)+10(4x^2)(-27y^3)+5(2x^1)(81y^4)+1(x^0)(-243y^5)}}} Distribute the exponents



{{{1(32x^5)+5(-48x^4y)+10(72x^3y^2)+10(-108x^2y^3)+5(162xy^4)+1(-243y^5)}}} Multiply



{{{32x^5-240x^4y+720x^3y^2-1080x^2y^3+810xy^4-243y^5}}} Multiply the terms with their coefficients



So {{{(2x-3y)^5}}} expands and simplifies to {{{32x^5-240x^4y+720x^3y^2-1080x^2y^3+810xy^4-243y^5}}}.



In other words, {{{(2x-3y)^5=32x^5-240x^4y+720x^3y^2-1080x^2y^3+810xy^4-243y^5}}}



From the last equation, we can see that the 2nd term is {{{-240x^4y}}} and the 5th term is {{{810xy^4}}}



Shortcut: You can find the kth term of the expansion {{{(a+b)^n}}} by using the formula {{{(nCk)*a^(n-k)*b^k}}} where {{{nCk}}} is the coefficient (either found with a calculator or by using Pascal's triangle).



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Jim