Need help writing a binomial expansion using the binominal theorem!!

Question

Problem 10: use the binomial theorem to write the binomial expansion of (x-3y²)⁵. 

So the first step is writing out the coefficients and simplifying them. Example (5 0)=1; (a+b)⁵= 1a⁵b⁰+5a⁴b¹+10a³b²+10a²b³+5a¹b⁴+1a⁰b⁵

= 1(1x)⁵(-3y²)⁰+5(1x)⁴(-3y²)¹+10(1x)³(-3y²)²+10(1x)²(-3y²)³+5(1x)¹(-3y²)⁴+1(1x)⁰(-3y²)⁵.... I am struggling with simplifying. This is what I have but I'm pretty sure its wrong,(1)(x⁵)(1y)+(5)(x⁴)(-3y²)¹+(10)(x³)(-6y⁶)+(10)(x²)(-9y⁶)+(5)(x)¹(-12y⁸)+(1)(1x)(-15y¹⁰).. Please help!!

Answer 1

Yes Mr. Lyubov

you are 100% correct.

 

I worked the problem on paper and got the same thing.

 

The binomial expansion is of (A+B)^5 = 

(5 choose 0) A^5 B^0 + (5 choose 1) A^4 B^1 + (5 choose 2) A^3 B^2 + (5 choose 3) A^2 B^3 + (5 choose 4) A B^4 + (5 choose 5)A^0 B^5

 

(5 choose 0), (5 choose 5) and any number to the zero power are all 1.

 

(5 choose 1) = (5 choose 4)  = 5

(4 choose 2) = (5 choose 3) = 10

The expansion becomes:

 

A^5 + 5 A^4 B + 10 A^3 B^2 + 10 A^2 B^3 + 5 A B^4 + B^5

 

Substitution of A=x and B = (-3y^2) results in the answer you have given

Answer 2

= 1(1x)⁵(-3y²)⁰+5(1x)⁴(-3y²)¹+10(1x)³(-3y²)²+10(1x)²(-3y²)³+5(1x)¹(-3y²)4+1(1x)⁰(-3y²)⁵ looks correct.

 

= x⁵                    -15x⁴y²        +90x³y⁴         - ...You can finish it!

 

Note: signs alternate; powers of x go down by 1 at you move to the right, and powers of y go up by two concurrently.

 

You did well.