Simplify (x^2-16x+64)/(4x^2-36x+32)

$\frac{x^{2} - 16 x + 64}{4 x^{2} - 36 x + 32}$

Factor using the perfect square rule.

Rewrite $64$ as $8^{2}$ .

$\frac{x^{2} - 16 x + 8^{2}}{4 x^{2} - 36 x + 32}$

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

$16 x = 2 \cdot x \cdot 8$

Rewrite the polynomial.

$\frac{x^{2} - 2 \cdot x \cdot 8 + 8^{2}}{4 x^{2} - 36 x + 32}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = x$ and $b = 8$ .

$\frac{{(x - 8)}^{2}}{4 x^{2} - 36 x + 32}$

Simplify the denominator.

Factor $4$ out of $4 x^{2} - 36 x + 32$ .

Factor $4$ out of $4 x^{2}$ .

$\frac{{(x - 8)}^{2}}{4 (x^{2}) - 36 x + 32}$

Factor $4$ out of $- 36 x$ .

$\frac{{(x - 8)}^{2}}{4 (x^{2}) + 4 (- 9 x) + 32}$

Factor $4$ out of $32$ .

$\frac{{(x - 8)}^{2}}{4 x^{2} + 4 (- 9 x) + 4 \cdot 8}$

Factor $4$ out of $4 x^{2} + 4 (- 9 x)$ .

$\frac{{(x - 8)}^{2}}{4 (x^{2} - 9 x) + 4 \cdot 8}$

Factor $4$ out of $4 (x^{2} - 9 x) + 4 \cdot 8$ .

$\frac{{(x - 8)}^{2}}{4 (x^{2} - 9 x + 8)}$

Factor $x^{2} - 9 x + 8$ 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 $8$ and whose sum is $- 9$ .

$- 8, - 1$

Write the factored form using these integers.

$\frac{{(x - 8)}^{2}}{4 ((x - 8) (x - 1))}$

$\frac{{(x - 8)}^{2}}{4 (x - 8) (x - 1)}$

Cancel the common factor of ${(x - 8)}^{2}$ and $x - 8$ .

Factor $x - 8$ out of ${(x - 8)}^{2}$ .

$\frac{(x - 8) (x - 8)}{4 (x - 8) (x - 1)}$

Cancel the common factors.

Factor $x - 8$ out of $4 (x - 8) (x - 1)$ .

$\frac{(x - 8) (x - 8)}{(x - 8) (4 (x - 1))}$

Cancel the common factor.

$\frac{(x - 8) (x - 8)}{(x - 8) (4 (x - 1))}$

Rewrite the expression.

$\frac{x - 8}{4 (x - 1)}$

Simplify (x^2-16x+64)/(4x^2-36x+32)

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