Question 1141476
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The binomial theorem looks scary to most students when they first see it.  But with a little experience it is fairly easy to use -- especially if you understand WHY it works.<br>
So let me start you on this with a demonstration; then you can finish.<br>
The expression means you have five identical factors of (x-3y).  When you expand the expression, you are going to get terms of x^5, x^4y, x^3y^2, x^2y^3, xy^4, and y^5.<br>
Think about how many ways you are going to get each of those kinds of terms.<br>
(1) x^5...<br>
To get an x^5 term, you clearly need to choose the "x" term from all 5 factors (and the "-3y" term in 0 of them).  The number of ways you can choose the "x" term in all 5 of the 5 factors is "5 choose 5" = C(5,5) = 1.<br>
So the x^5 term will be {{{C(5,5)((x)^5)((-3y)^0) = (1)(x^5)(1) = x^5}}}.<br>
(2) x^4y...<br>
To get an x^4y term, you need to choose the "x" term from 4 of the 5 factors and the "-3y" term in 1 of them.  The number of ways you can choose the x term in 4 of the 5 factors is "5 choose 4" = C(5,4) = 5.<br>
So the x^4y term will be {{{C(5,4)((x)^4)((-3y)^1) = (5)(x^4)(-3y) = -15x^4y}}}.<br>
(3) x^3y^2...<br>
To get an x^3y^2 term, you need to choose the "x" term from 3 of the 5 factors and the "-3y" term in 2 of them.  The number of ways you can choose the x term in 3 of the 5 factors is "5 choose 3" = C(5,3) = 10.<br>
So the x^3y^2 term will be {{{C(5,3)((x)^3)((-3y)^2) = (10)(x^3)(9y^2) = 90x^3y^2}}}.<br>
I'll let you fill in the calculations to finish the expansion.  The remaining terms are<br>
{{{C(5,2)((x)^2)((-3y)^3)}}}<br>
{{{C(5,1)((x)^1)((-3y)^4)}}}<br>
{{{C(5,0)((x)^0)((-3y)^5)}}}<br>