Question

Can someone please help me????

Use the binomial theorem to expand (a^2 - 3)^4. Show your work.

Answer 1

Hello !

Answer:

`\large \boxed{\sf(a^2 - 3)^4 =a^8-12a^6+54a^4-108a^2+81}`

Explanation:

Let's use the binomial theorem to expand
`\sf (a^2 - 3)^4` :

Binomial theorem :

`\sf \forall n \in \mathbb N, (x+y)^n=\sum\limits_(k=0)^(n)\binom{n}{k}x^ky^(n-k)`

`\sf Where\ \binom{n}{k}=(n!)/(k!(n - k)!)`

Reminders on factorials :

• 
  `\sf \forall n \in \mathbb N,\ n!=\ \prod\limits_(i=1)^ni\ =1*2* ...* n`

• 
  `\sf 0!=1`

We have :

• 
  `\sf x=a^2`
• 
  `\sf y=-3`

Now we substitute these values into the formula :

`\sf (a^2 - 3)^4=\sum\limits^4_(k=0)\binom{4}{k}(a^2)^k(-3)^(4-k)`

`\sf =\binom{4}{0}(a^2)^0(-3)^(4)+\binom{4}{1}(a^2)^1(-3)^(3)+\binom{4}{2}(a^2)^2(-3)^(2)+\binom{4}{3}(a^2)^3(-3)^(1)+\binom{4}{4}(a^2)^4(-3)^(0)`

`\sf =\binom{4}{0}81-\binom{4}{1}27a^2+\binom{4}{2}9a^4-\binom{4}{3}3a^6+\binom{4}{4}a^8`

Let's calculate the binomial coefficients :

• 
  `\sf \binom{4}{0}=(4!)/(0!(4-0)!)=(24)/(24) =1`
• 
  `\sf \binom{4}{1}=(4!)/(1!(4-1)!)=(24)/(3!)=(24)/(6) =4`
• 
  `\sf \binom{4}{2}=(4!)/(2!(4-2)!)=(24)/(2!2!)=(24)/(4) =6`
• 
  `\sf \binom{4}{3}=(4!)/(3!(4-3)!)=(24)/(3!1!)=(24)/(6) =4`
• 
  `\sf \binom{4}{4}=(4!)/(4!(4-4)!)=(24)/(4!0!)=(24)/(24) =1`

Now we can replace the binomial coefficients with their value:

`\sf (a^2 - 3)^4 =1*81-4*27a^2+6* 9a^4-4*3a^6+1* a^8`

`\sf(a^2 - 3)^4 =81-108a^2+54a^4-12a^6+a^8`

`\boxed{\sf(a^2 - 3)^4 =a^8-12a^6+54a^4-108a^2+81}`

Have a nice day ;)