Simplify ((16x^2-8x+1)/(x^3-7x^2+12x))÷((20x^2-5x)/(15x^3))

$\frac{16 x^{2} - 8 x + 1}{x^{3} - 7 x^{2} + 12 x} \div \frac{20 x^{2} - 5 x}{15 x^{3}}$

To divide by a fraction, multiply by its reciprocal.

$\frac{16 x^{2} - 8 x + 1}{x^{3} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor using the perfect square rule.

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

$\frac{{(4 x)}^{2} - 8 x + 1}{x^{3} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Rewrite $1$ as $1^{2}$ .

$\frac{{(4 x)}^{2} - 8 x + 1^{2}}{x^{3} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$8 x = 2 \cdot (4 x) \cdot 1$

Rewrite the polynomial.

$\frac{{(4 x)}^{2} - 2 \cdot (4 x) \cdot 1 + 1^{2}}{x^{3} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = 4 x$ and $b = 1$ .

$\frac{{(4 x - 1)}^{2}}{x^{3} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Simplify the denominator.

Factor $x$ out of $x^{3} - 7 x^{2} + 12 x$ .

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

$\frac{{(4 x - 1)}^{2}}{x \cdot x^{2} - 7 x^{2} + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x$ out of $- 7 x^{2}$ .

$\frac{{(4 x - 1)}^{2}}{x \cdot x^{2} + x (- 7 x) + 12 x} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x$ out of $12 x$ .

$\frac{{(4 x - 1)}^{2}}{x \cdot x^{2} + x (- 7 x) + x \cdot 12} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x$ out of $x \cdot x^{2} + x (- 7 x)$ .

$\frac{{(4 x - 1)}^{2}}{x (x^{2} - 7 x) + x \cdot 12} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x$ out of $x (x^{2} - 7 x) + x \cdot 12$ .

$\frac{{(4 x - 1)}^{2}}{x (x^{2} - 7 x + 12)} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x^{2} - 7 x + 12$ 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 $12$ and whose sum is $- 7$ .

$- 4, - 3$

Write the factored form using these integers.

$\frac{{(4 x - 1)}^{2}}{x ((x - 4) (x - 3))} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

$\frac{{(4 x - 1)}^{2}}{x (x - 4) (x - 3)} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Simplify terms.

Cancel the common factor of $x$ .

Factor $x$ out of $x (x - 4) (x - 3)$ .

$\frac{{(4 x - 1)}^{2}}{x ((x - 4) (x - 3))} \cdot \frac{15 x^{3}}{20 x^{2} - 5 x}$

Factor $x$ out of $15 x^{3}$ .

$\frac{{(4 x - 1)}^{2}}{x ((x - 4) (x - 3))} \cdot \frac{x (15 x^{2})}{20 x^{2} - 5 x}$

Cancel the common factor.

$\frac{{(4 x - 1)}^{2}}{x ((x - 4) (x - 3))} \cdot \frac{x (15 x^{2})}{20 x^{2} - 5 x}$

Rewrite the expression.

$\frac{{(4 x - 1)}^{2}}{(x - 4) (x - 3)} \cdot \frac{15 x^{2}}{20 x^{2} - 5 x}$

Multiply $\frac{{(4 x - 1)}^{2}}{(x - 4) (x - 3)}$ by $\frac{15 x^{2}}{20 x^{2} - 5 x}$ .

$\frac{{(4 x - 1)}^{2} (15 x^{2})}{(x - 4) (x - 3) (20 x^{2} - 5 x)}$

Factor $5 x$ out of $20 x^{2} - 5 x$ .

Factor $5 x$ out of $20 x^{2}$ .

$\frac{{(4 x - 1)}^{2} \cdot 15 x^{2}}{(x - 4) (x - 3) (5 x (4 x) - 5 x)}$

Factor $5 x$ out of $- 5 x$ .

$\frac{{(4 x - 1)}^{2} \cdot 15 x^{2}}{(x - 4) (x - 3) (5 x (4 x) + 5 x (- 1))}$

Factor $5 x$ out of $5 x (4 x) + 5 x (- 1)$ .

$\frac{{(4 x - 1)}^{2} \cdot 15 x^{2}}{(x - 4) (x - 3) (5 x (4 x - 1))}$

$\frac{{(4 x - 1)}^{2} \cdot 15 x^{2}}{(x - 4) (x - 3) \cdot 5 x (4 x - 1)}$

Cancel the common factor of ${(4 x - 1)}^{2}$ and $4 x - 1$ .

Factor $4 x - 1$ out of ${(4 x - 1)}^{2} \cdot 15 x^{2}$ .

$\frac{(4 x - 1) ((4 x - 1) \cdot 15 x^{2})}{(x - 4) (x - 3) \cdot 5 x (4 x - 1)}$

Cancel the common factors.

Factor $4 x - 1$ out of $(x - 4) (x - 3) \cdot 5 x (4 x - 1)$ .

$\frac{(4 x - 1) ((4 x - 1) \cdot 15 x^{2})}{(4 x - 1) ((x - 4) (x - 3) \cdot 5 x)}$

Cancel the common factor.

$\frac{(4 x - 1) ((4 x - 1) \cdot 15 x^{2})}{(4 x - 1) ((x - 4) (x - 3) \cdot 5 x)}$

Rewrite the expression.

$\frac{(4 x - 1) \cdot 15 x^{2}}{(x - 4) (x - 3) \cdot 5 x}$

Cancel the common factor of $15$ and $5$ .

Factor $5$ out of $(4 x - 1) \cdot 15 x^{2}$ .

$\frac{5 ((4 x - 1) \cdot 3 x^{2})}{(x - 4) (x - 3) \cdot 5 x}$

Cancel the common factors.

Factor $5$ out of $(x - 4) (x - 3) \cdot 5 x$ .

$\frac{5 ((4 x - 1) \cdot 3 x^{2})}{5 ((x - 4) (x - 3) x)}$

Cancel the common factor.

$\frac{5 ((4 x - 1) \cdot 3 x^{2})}{5 ((x - 4) (x - 3) x)}$

Rewrite the expression.

$\frac{(4 x - 1) \cdot 3 x^{2}}{(x - 4) (x - 3) x}$

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

Factor $x$ out of $(4 x - 1) \cdot 3 x^{2}$ .

$\frac{x ((4 x - 1) \cdot 3 x)}{(x - 4) (x - 3) x}$

Cancel the common factors.

Factor $x$ out of $(x - 4) (x - 3) x$ .

$\frac{x ((4 x - 1) \cdot 3 x)}{x ((x - 4) (x - 3))}$

Cancel the common factor.

$\frac{x ((4 x - 1) \cdot 3 x)}{x ((x - 4) (x - 3))}$

Rewrite the expression.

$\frac{(4 x - 1) \cdot 3 x}{(x - 4) (x - 3)}$

Simplify the expression.

Move $3$ to the left of $4 x - 1$ .

$\frac{3 \cdot (4 x - 1) x}{(x - 4) (x - 3)}$

Reorder factors in $\frac{3 (4 x - 1) x}{(x - 4) (x - 3)}$ .

$\frac{3 x (4 x - 1)}{(x - 4) (x - 3)}$

Simplify ((16x^2-8x+1)/(x^3-7x^2+12x))÷((20x^2-5x)/(15x^3))

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