Simplify (5k^2-50k+45)/(k^2-18k+81)

$\frac{5 k^{2} - 50 k + 45}{k^{2} - 18 k + 81}$

Step 1

Simplify the numerator.

Factor $5$ out of $5 k^{2} - 50 k + 45$ .

Factor $5$ out of $5 k^{2}$ .

$\frac{5 (k^{2}) - 50 k + 45}{k^{2} - 18 k + 81}$

Factor $5$ out of $- 50 k$ .

$\frac{5 (k^{2}) + 5 (- 10 k) + 45}{k^{2} - 18 k + 81}$

Factor $5$ out of $45$ .

$\frac{5 k^{2} + 5 (- 10 k) + 5 \cdot 9}{k^{2} - 18 k + 81}$

Factor $5$ out of $5 k^{2} + 5 (- 10 k)$ .

$\frac{5 (k^{2} - 10 k) + 5 \cdot 9}{k^{2} - 18 k + 81}$

Factor $5$ out of $5 (k^{2} - 10 k) + 5 \cdot 9$ .

$\frac{5 (k^{2} - 10 k + 9)}{k^{2} - 18 k + 81}$

Factor $k^{2} - 10 k + 9$ 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 $9$ and whose sum is $- 10$ .

$- 9, - 1$

Write the factored form using these integers.

$\frac{5 ((k - 9) (k - 1))}{k^{2} - 18 k + 81}$

$\frac{5 (k - 9) (k - 1)}{k^{2} - 18 k + 81}$

Step 2

Factor using the perfect square rule.

Rewrite $81$ as $9^{2}$ .

$\frac{5 (k - 9) (k - 1)}{k^{2} - 18 k + 9^{2}}$

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

$18 k = 2 \cdot k \cdot 9$

Rewrite the polynomial.

$\frac{5 (k - 9) (k - 1)}{k^{2} - 2 \cdot k \cdot 9 + 9^{2}}$

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = k$ and $b = 9$ .

$\frac{5 (k - 9) (k - 1)}{{(k - 9)}^{2}}$

Step 3

Cancel the common factor of $k - 9$ and ${(k - 9)}^{2}$ .

Factor $k - 9$ out of $5 (k - 9) (k - 1)$ .

$\frac{(k - 9) (5 (k - 1))}{{(k - 9)}^{2}}$

Cancel the common factors.

Factor $k - 9$ out of ${(k - 9)}^{2}$ .

$\frac{(k - 9) (5 (k - 1))}{(k - 9) (k - 9)}$

Cancel the common factor.

$\frac{(k - 9) (5 (k - 1))}{(k - 9) (k - 9)}$

Rewrite the expression.

$\frac{5 (k - 1)}{k - 9}$

Simplify (5k^2-50k+45)/(k^2-18k+81)

Related Questions

Support LCMGCF.com by disabling your adblocker.