Let f(x)= (x^5 - x^3)/(x^2-3x + 2)

first let us factor the numerator and deniminator:

The numerator:

x^5 - x^3 = x^3(x^2 -1) = x^3 (x-1)(x+1)

The denominator:

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

Now substitute in f(x);

==> f(x) = x^3(x-1)(x+1)/(x-1)(x-2)

Reduce similar terms (x-1):

==> f(x) = x^3(x+1)/(x-2)

= (x^4+x^3)/(x-2)

**==> f(x) = (x^4+x^3)/(x-2)**

To reduce to the lowest terms, we'll have to factorize the numerator and to write the denominator as a product of linear factors.

We'll factorize the numerator by x^3:

(x^5-x^3) = x^3(x^2 - 1)

But x^2 - 1 is a difference of squares:

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

We'll compute the roots of the equation:

(x^2 - 3x + 2) = 0

x1 = 2

x2 = 1

S = 2+1 = 3

P = 2*1

The equation is written as a product of linear factors:

(x^2 - 3x + 2) = (x-x1)(x-x2)

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

We'll re-write the expression:

**x^3(x-1)(x+1)/(x-1)(x-2) = x^3(x+1)/(x-2)**

To reduce (x^5-x^3)/(x^2-3x+2) to the lowest terms.

We factor the numerator. We factor the denominator. Then if any common factor is found we cancel. Thus we arive at the simpler form of the given rational expression.

Numerator:

x^5-x^3 = x^3(x^2-1)

x^3(x^2-1) = x^3(x+1)(x-1).

Denominator:

x^2-3x+2 = x^2 -2x-x +2

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

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

Therefore the given expression (x^5-x^3)/(x^2-3x+2) = x^3(x+1)(x-1)/(x-2)(x-1).

(x^5-x^3)/(x^2-3x+2) = x^3(x+1)/(x-2) , as x-1 gets cancelled.

Therefore x^3(x+1)/(x-2) is the simple form of (x^5-x^3)/(x^2-3x+2)

We have to reduce (x^5-x^3)/(x^2 - 3x + 2)

Now : (x^5-x^3)/(x^2 - 3x + 2)

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

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

=> x^3 ( x-1) (x+1) / [x (x - 2) - 1 ( x - 2 )]

=> x^3 ( x-1) (x+1) / [ (x - 1)( x - 2 )]

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

**Therefore (x^5-x^3)/(x^2 - 3x + 2) = x^3 ( x+1) / x-2)**