Question

Expand the following using either the Binomial Theorem or Pascal’s Triangle. You must show your work for credit. (x - 3)^5

Answer 1

I used bionomial expansion up to and including the term
`x^(3)`


`(-3 + x)^(5)`


`-3^(5) (1 + -(1)/(3)x) ^(5)`


`-243 (1 + -(1)/(3)x) ^(5)`

n = 5
x =
`- (1)/(3)x`


Binomial expansion:


`1 + nx + (n(n-1) x^(2) )/(2!) + (n(n-1)(n-2) x^(3) )/(3!)`


Therefore:


`1 + (5*- (1)/(3)x) + (5(4) (- (1)/(3)x)^(2) )/(2) + (5(4)(3) (- (1)/(3)x)^(3) )/(6)`

=


`1 - (5)/(3) x + (10)/(9) x^(2) - (10)/(27) x^(3)`

Multiply by -243

=


`-243 + 405x - 270x^(2) + 90 x^(3)`

Answer 2

`x^5-15x^4+90x^3-270x^2+405x-243`

Explanation:

Here, the given expression is,

`(x-3)^5`

We know that by binomial theorem,

`(p+q)^n=\sum_(r=0)^(n) ^nC_r (p)^(n-r) (q)^r`

Where,

`^nC_r=(n!)/(r!(n-r)!)`

Now,

`(x-3)^5=(x+(-3))^5`

Thus, by the above theorem,

`(x+(-3))^5=\sum_(r=0)^(5) ^5C_r (x)^(5-r) (-3)^r`

`=^5C_0 (x)^(5-0) (-3)^0+^5C_1 (x)^(5-1) (-3)^1+^5C_2 (x)^(5-2) (-3)^2+^5C_3 (x)^(5-3) (-3)^3+^5C_4 (x)^(5-4) (-3)^4+^5C_5 (x)^(5-5) (-3)^5`

`=(x)^5(1)+5(x)^4(-3)+10(x)^3(-3)^2+10(x)^2(-3)^3+5(x)(-3)^4+(-3)^5`

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