Simplify (3z^11+3z^5)/(z^12+4z^6+3)

$\frac{3 z^{11} + 3 z^{5}}{z^{12} + 4 z^{6} + 3}$

Step 1

Simplify the numerator.

Factor $3 z^{5}$ out of $3 z^{11} + 3 z^{5}$ .

Factor $3 z^{5}$ out of $3 z^{11}$ .

$\frac{3 z^{5} (z^{6}) + 3 z^{5}}{z^{12} + 4 z^{6} + 3}$

Factor $3 z^{5}$ out of $3 z^{5}$ .

$\frac{3 z^{5} (z^{6}) + 3 z^{5} (1)}{z^{12} + 4 z^{6} + 3}$

Factor $3 z^{5}$ out of $3 z^{5} (z^{6}) + 3 z^{5} (1)$ .

$\frac{3 z^{5} (z^{6} + 1)}{z^{12} + 4 z^{6} + 3}$

Rewrite $z^{6}$ as ${(z^{2})}^{3}$ .

$\frac{3 z^{5} ({(z^{2})}^{3} + 1)}{z^{12} + 4 z^{6} + 3}$

Rewrite $1$ as $1^{3}$ .

$\frac{3 z^{5} ({(z^{2})}^{3} + 1^{3})}{z^{12} + 4 z^{6} + 3}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = z^{2}$ and $b = 1$ .

$\frac{3 z^{5} ((z^{2} + 1) ({(z^{2})}^{2} - z^{2} \cdot 1 + 1^{2}))}{z^{12} + 4 z^{6} + 3}$

Simplify.

Multiply the exponents in ${(z^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3 z^{5} ((z^{2} + 1) (z^{2 \cdot 2} - z^{2} \cdot 1 + 1^{2}))}{z^{12} + 4 z^{6} + 3}$

Multiply $2$ by $2$ .

$\frac{3 z^{5} ((z^{2} + 1) (z^{4} - z^{2} \cdot 1 + 1^{2}))}{z^{12} + 4 z^{6} + 3}$

Multiply $- 1$ by $1$ .

$\frac{3 z^{5} ((z^{2} + 1) (z^{4} - z^{2} + 1^{2}))}{z^{12} + 4 z^{6} + 3}$

One to any power is one.

$\frac{3 z^{5} ((z^{2} + 1) (z^{4} - z^{2} + 1))}{z^{12} + 4 z^{6} + 3}$

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{z^{12} + 4 z^{6} + 3}$

Step 2

Simplify the denominator.

Rewrite $z^{12}$ as ${(z^{6})}^{2}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{{(z^{6})}^{2} + 4 z^{6} + 3}$

Let $u = z^{6}$ . Substitute $u$ for all occurrences of $z^{6}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{u^{2} + 4 u + 3}$

Factor $u^{2} + 4 u + 3$ 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 $3$ and whose sum is $4$ .

$1, 3$

Write the factored form using these integers.

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(u + 1) (u + 3)}$

Replace all occurrences of $u$ with $z^{6}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{6} + 1) (z^{6} + 3)}$

Rewrite $z^{6}$ as ${(z^{2})}^{3}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{({(z^{2})}^{3} + 1) (z^{6} + 3)}$

Rewrite $1$ as $1^{3}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{({(z^{2})}^{3} + 1^{3}) (z^{6} + 3)}$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = z^{2}$ and $b = 1$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) ({(z^{2})}^{2} - z^{2} \cdot 1 + 1^{2}) (z^{6} + 3)}$

Simplify.

Multiply the exponents in ${(z^{2})}^{2}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) (z^{2 \cdot 2} - z^{2} \cdot 1 + 1^{2}) (z^{6} + 3)}$

Multiply $2$ by $2$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) (z^{4} - z^{2} \cdot 1 + 1^{2}) (z^{6} + 3)}$

Multiply $- 1$ by $1$ .

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) (z^{4} - z^{2} + 1^{2}) (z^{6} + 3)}$

One to any power is one.

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) (z^{4} - z^{2} + 1) (z^{6} + 3)}$

Step 3

Reduce the expression by cancelling the common factors.

Cancel the common factor of $z^{2} + 1$ .

Cancel the common factor.

$\frac{3 z^{5} (z^{2} + 1) (z^{4} - z^{2} + 1)}{(z^{2} + 1) (z^{4} - z^{2} + 1) (z^{6} + 3)}$

Rewrite the expression.

$\frac{3 z^{5} (z^{4} - z^{2} + 1)}{(z^{4} - z^{2} + 1) (z^{6} + 3)}$

Cancel the common factor of $z^{4} - z^{2} + 1$ .

Cancel the common factor.

$\frac{3 z^{5} (z^{4} - z^{2} + 1)}{(z^{4} - z^{2} + 1) (z^{6} + 3)}$

Rewrite the expression.

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

Simplify (3z^11+3z^5)/(z^12+4z^6+3)

Related Questions

Support LCMGCF.com by disabling your adblocker.