`2(x - 3)^4 + 5(x - 3)^2` Use the Binomial Theorem to expand and simplify the expression.

You need first to factorize `(x - 3)^2` , such that:

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

You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace x for x, 3for y and 2 for n, such that:

`(x-3)^2 = 2C0 (x)^2+2C1 (x)^1*(-3)^1+2C2(-3)^2`

By definition, nC0 = nCn = 1, hence `2C0 = 2C2 = 1.`

By definition `nC1 = nC(n-1) = n` , hence `2C1= 2.`

`(x-3)^2 = x^2 -...

You need first to factorize `(x - 3)^2` , such that:

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

You need to use the binomial formula, such that:

`(x+y)^n = sum_(k=0)^n ((n),(k)) x^(n-k) y^k`

You need to replace x for x, 3for y and 2 for n, such that:

`(x-3)^2 = 2C0 (x)^2+2C1 (x)^1*(-3)^1+2C2(-3)^2`

By definition, nC0 = nCn = 1, hence `2C0 = 2C2 = 1.`

By definition `nC1 = nC(n-1) = n` , hence `2C1= 2.`

`(x-3)^2 = x^2 - 6x + 9`

Replacing the binomial expansion` x^2 - 6x + 9` for `(x-3)^2`  yields:

`(x-3)^2(2(x-3)^2 + 5) = (x^2 - 6x + 9)(2(x^2 - 6x + 9) + 5)`

`(x-3)^2(2(x-3)^2 + 5) = (x^2 - 6x + 9)(2x^2 - 12x +18 + 5)`

`(x-3)^2(2(x-3)^2 + 5) = (x^2 - 6x + 9)(2x^2 - 12x +23)`

`(x-3)^2(2(x-3)^2 + 5) = (2x^4 - 12x^3 + 23x^2 - 12x^3 + 72x^2 - 138x + 18x^2 - 108x + 207)`

Grouping the like terms yields:

`2(x-3)^4 + 5(x-3)^2 = 2x^4 - 24x^3 + 113x^2 - 246x + 207`

Hence, expanding and simplifying the expression yields `2(x-3)^4 + 5(x-3)^2 = 2x^4 - 24x^3 + 113x^2 - 246x + 207.`