Divide ((6x^2-18)/(12x))÷((x^2-3x+18)/(x^2-7x+6))

$\frac{6 x^{2} - 18}{12 x} \div \frac{x^{2} - 3 x + 18}{x^{2} - 7 x + 6}$

Step 1

To divide by a fraction, multiply by its reciprocal.

$\frac{6 x^{2} - 18}{12 x} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Step 2

Cancel the common factor of $6 x^{2} - 18$ and $12$ .

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

$\frac{6 (x^{2}) - 18}{12 x} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Factor $6$ out of $- 18$ .

$\frac{6 x^{2} + 6 \cdot - 3}{12 x} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Factor $6$ out of $6 x^{2} + 6 \cdot - 3$ .

$\frac{6 (x^{2} - 3)}{12 x} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Cancel the common factors.

Factor $6$ out of $12 x$ .

$\frac{6 (x^{2} - 3)}{6 (2 x)} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Cancel the common factor.

$\frac{6 (x^{2} - 3)}{6 (2 x)} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Rewrite the expression.

$\frac{x^{2} - 3}{2 x} \cdot \frac{x^{2} - 7 x + 6}{x^{2} - 3 x + 18}$

Step 3

Factor $x^{2} - 7 x + 6$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $6$ and whose sum is $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

$\frac{x^{2} - 3}{2 x} \cdot \frac{(x - 6) (x - 1)}{x^{2} - 3 x + 18}$

Step 4

Multiply $\frac{x^{2} - 3}{2 x}$ by $\frac{(x - 6) (x - 1)}{x^{2} - 3 x + 18}$ .

$\frac{(x^{2} - 3) (x - 6) (x - 1)}{2 x (x^{2} - 3 x + 18)}$

Divide ((6x^2-18)/(12x))÷((x^2-3x+18)/(x^2-7x+6))

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