You need to find the roots of polynomial `x^3-4x^2+5x-2 = 0` , hence you should find the divisors of the constant terms such that:

`D_2:{+-1;+-2}`

You need to substitute each of these divisors in polynomial to check if one of them cancels it.

Substituting -1 for x yields:

`(-1)^3 - 4*(-1)^2 + 5*(-1) - 2 != 0`

`-1 - 4 - 5 - 2 != 0`

Substituting 1 for x yields:

`(1)^3 - 4*(1)^2 + 5*(1) - 2 = 0`

`1 - 4 + 5 - 2 = -3 + 3 = 0`

Notice that x = 1 is a root of polynomial, hence you may use reminder theorem such that:

`x^3-4x^2+5x-2 = (x-1)(ax^2+bx+c)`

Opening the brackets yields:

`x^3-4x^2+5x-2 = ax^3 + bx^2 + cx - ax^2 - bx - c`

Equating coefficients of like terms yields:

`a=1`

`b-a=-4 =gt b-1=-4 =gt b=-3`

`c-b=5 =gt c+3=5 =gt c=2`

Hence, if `x^3-4x^2+5x-2 =(x-1)(x^2-3x+2)`  and `x^3-4x^2+5x-2 = 0` , then `(x^2-3x+2) = 0` .

You need to use quadratic formula to find the roots of equation `x^2-3x+2 = 0`  such that:

`x_(1,2) = (3+-sqrt(9-8))/2 =gt x_(1,2) = (3+-1)/2`

`x_1 = 2; x_2 = 1`

Hence, you need to write the factored form of polynomial `x^3-4x^2+5x-2`  such that: `x^3 - 4x^2 + 5x - 2 = (x - 1)^2(x - 2).`

You need to write the factored form of difference `27a^3-64b^3` , hence you may use the following formula of difference of cubes such that:

`x^3 - y^3 = (x-y)(x^2+xy+y^2)`

`27a^3-64b^3 = (3a - 4b)(9a^2 + 12ab + 16b^2)`

Hence, evaluating the factorized form of `27a^3-64b^3`  yields `27a^3-64b^3 = (3a - 4b)(9a^2 + 12ab + 16b^2).`

The expression `x^3-4x^2+5x-2` has to be factored.

`x^3-4x^2+5x-2`

=> `x^3 - x^2 - 3x^2 + 3x + 2x - 2`

=> `x^2(x - 1) - 3x(x - 1) + 2(x - 1)`

=> `(x^2 - 3x + 2)(x - 1)`

=> `(x^2 - 2x - x + 2)(x - 1)`

=> `(x(x - 2) - 1(x - 2))(x - 1)`

=> `(x - 1)(x - 2)(x - 1)`

=> `(x - 2)(x - 1)^2`

`27a^3 - 64b^3`

=> `(3a)^3 - (4b)^3`

=> `(3a - 4b)(9a^2 + 16b^2 + 12ab)`

The required factorized form of `x^3-4x^2+5x-2 = (x - 2)(x - 1)^2` and `27a^3 - 64b^3 = (3a - 4b)(9a^2 + 16b^2 + 12ab)`