Question

Hey! I need some help. This is from the trigonometry section in my ACT pre guide.

Answer 1

We need to expand the following binomial:

`\mleft(3x^5-(1)/(9)y^3\mright)^4`

using binomial theorem. This theorem states that:

`\mleft(a+b\mright)^n=\sum ^n_(i=0)\binom{n}{i}a^(\mleft(n-i\mright))b^i`

where

`\binom{n}{i}=(n!)/(i!(n-i)!)`

In this problem, we have:

`\begin{gathered} a=3x^5 \\ \\ b=-(1)/(9)y^(3) \\ \\ n=4 \end{gathered}`

Part (a)

To express the expansion, we use the following sum in summation notation:

`\mleft(3x^5-(1)/(9)y^3\mright)^4=\sum ^4_(i=0)\binom{4}{i}(3x^5)^((4-i))\mleft(-(1)/(9)y^3\mright)^(^i)`

This can also be written as:

`\mleft(3x^5-(1)/(9)y^3\mright)^4=\sum ^4_(i=0)(4!)/(i!(4-i)!)(3x^5)^((4-i))\mleft(-(1)/(9)y^3\mright)^(^i)`

Part (b)

Simplifying the terms, we obtain:

`\begin{gathered} (4!)/(0!4!)\cdot81x^(20)+(4!)/(1!3!)\cdot27x^(15)\cdot(-(1)/(9))y^3+(4!)/(2!2!)\cdot9x^(10)\cdot(1)/(81)y^6 \\ \\ +(4!)/(3!1!)\cdot3x^5\cdot(-(1)/(729))y^9+(4!)/(4!0!)\cdot(1)/(6561)y^(12) \\ \\ =81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12) \\ \end{gathered}`

Therefore, the simplified terms of the expansion are:

`81x^(20)-12x^(15)y^3+(2)/(3)x^(10)y^6-(4)/(243)x^5y^9+(1)/(6561)y^(12)`