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

$\frac{6 x^{3}}{x^{2} - 4} \cdot \frac{x^{2} - 3 x + 1}{2 x^{2}}$

Step 1

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$\frac{6 x^{3}}{x^{2} - 2^{2}} \cdot \frac{x^{2} - 3 x + 1}{2 x^{2}}$

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 2$ .

$\frac{6 x^{3}}{(x + 2) (x - 2)} \cdot \frac{x^{2} - 3 x + 1}{2 x^{2}}$

Step 2

Simplify terms.

Cancel the common factor of $2 x^{2}$ .

Factor $2 x^{2}$ out of $6 x^{3}$ .

$\frac{2 x^{2} (3 x)}{(x + 2) (x - 2)} \cdot \frac{x^{2} - 3 x + 1}{2 x^{2}}$

Cancel the common factor.

$\frac{2 x^{2} (3 x)}{(x + 2) (x - 2)} \cdot \frac{x^{2} - 3 x + 1}{2 x^{2}}$

Rewrite the expression.

$\frac{3 x}{(x + 2) (x - 2)} \cdot (x^{2} - 3 x + 1)$

Multiply $\frac{3 x}{(x + 2) (x - 2)}$ by $x^{2} - 3 x + 1$ .

$\frac{3 x (x^{2} - 3 x + 1)}{(x + 2) (x - 2)}$