Simplify (k^10+9k^5+18)/(k^10+6k^5+9)

$\frac{k^{10} + 9 k^{5} + 18}{k^{10} + 6 k^{5} + 9}$

Simplify the numerator.

Rewrite $k^{10}$ as ${(k^{5})}^{2}$ .

$\frac{{(k^{5})}^{2} + 9 k^{5} + 18}{k^{10} + 6 k^{5} + 9}$

Let $u = k^{5}$ . Substitute $u$ for all occurrences of $k^{5}$ .

$\frac{u^{2} + 9 u + 18}{k^{10} + 6 k^{5} + 9}$

Factor $u^{2} + 9 u + 18$ 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 $18$ and whose sum is $9$ .

$3, 6$

Write the factored form using these integers.

$\frac{(u + 3) (u + 6)}{k^{10} + 6 k^{5} + 9}$

Replace all occurrences of $u$ with $k^{5}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{k^{10} + 6 k^{5} + 9}$

Simplify the denominator.

Rewrite $k^{10}$ as ${(k^{5})}^{2}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{{(k^{5})}^{2} + 6 k^{5} + 9}$

Let $u = k^{5}$ . Substitute $u$ for all occurrences of $k^{5}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{u^{2} + 6 u + 9}$

Factor using the perfect square rule.

Rewrite $9$ as $3^{2}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{u^{2} + 6 u + 3^{2}}$

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

$6 u = 2 \cdot u \cdot 3$

Rewrite the polynomial.

$\frac{(k^{5} + 3) (k^{5} + 6)}{u^{2} + 2 \cdot u \cdot 3 + 3^{2}}$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = u$ and $b = 3$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{{(u + 3)}^{2}}$

Replace all occurrences of $u$ with $k^{5}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{{(k^{5} + 3)}^{2}}$

Cancel the common factors.

Factor $k^{5} + 3$ out of ${(k^{5} + 3)}^{2}$ .

$\frac{(k^{5} + 3) (k^{5} + 6)}{(k^{5} + 3) (k^{5} + 3)}$

Cancel the common factor.

$\frac{(k^{5} + 3) (k^{5} + 6)}{(k^{5} + 3) (k^{5} + 3)}$

Rewrite the expression.

$\frac{k^{5} + 6}{k^{5} + 3}$

Simplify (k^10+9k^5+18)/(k^10+6k^5+9)

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