Question

How do you use the Binomial Theorem to expand `(x-3)^5`?

Answer 1

`(x-3)^5=x^5-15x^4+90x^3-270x^2+405x-243.`

Explanation:

The Binomial Theorem Expansion `: (x+y)^n=x^n+""nC_1x^(n-1)y+""nC_2x^(n-2)y^2+""nC_3x^(n-3)y^3+.........+""nC_rx^(n-r)y^r+... ...+y^n.`

Letting `y=-3, n=5`, we get,
`(x-3)^5=x^5+""5C_1x^4(-3)^1+""5C_2x^3(-3)^2+""5C_3x^2(-3)^3+""5C_4x^1(-3)^4+(-3)^5.`

Here, `""5C_1=""5C_(5-1)=""5C_4=5, ""5C_2=""5C_3=(5*4)/(1*2)=10`

So, `(x-3)^5=x^5+5x^4(-3)+10x^3(9)+10x^2(-27)+5x(81)+(-243).`
`=x^5-15x^4+90x^3-270x^2+405x-243.`