The problem asks to simplify the product of three rational algebraic fractions.

Factor Each Expression

To simplify the product, we first factor each numerator and denominator:

• First fraction: $\frac{x+3}{x^2-2x}$
  • Numerator: $x+3$ (already factored)
  • Denominator: $x^2-2x = x(x-2)$
• Second fraction: $\frac{2x-1}{x^2+2x+4}$
  • Numerator: $2x-1$ (already factored)
  • Denominator: $x^2+2x+4$ (irreducible over real numbers, related to difference of cubes)
• Third fraction: $\frac{x^4-8x}{2x^2+5x-3}$
  • Numerator: $x^4-8x = x(x^3-8)$. Using the difference of cubes formula $a^3-b^3 = (a-b)(a^2+ab+b^2)$, we get $x^3-8 = x^3-2^3 = (x-2)(x^2+2x+4)$. So, the numerator is $x(x-2)(x^2+2x+4)$.
  • Denominator: $2x^2+5x-3$. We look for two numbers that multiply to $2 \times -3 = -6$ and add to $5$. These are $6$ and $-1$. So, $2x^2+6x-x-3 = 2x(x+3)-1(x+3) = (2x-1)(x+3)$.

Rewrite and Simplify Product

Now, substitute the factored forms back into the expression:

Cancel out the common factors in the numerators and denominators:

• The term $(x+3)$ cancels out.
• The term $x$ cancels out.
• The term $(x-2)$ cancels out.
• The term $(2x-1)$ cancels out.
• The term $(x^2+2x+4)$ cancels out.

After canceling all common factors, the expression simplifies to:

Therefore, the value of the given expression is 1.