Question 183017



{{{(3r-2)^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 {{{(3r-2)^5}}}, simply follow this procedure:

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


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



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




However, we're not done yet.



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


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


Continue this until you reach the final term.



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



So the fully expanded expression should now look like this:



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



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



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



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



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



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