Question

simplify the rational expression.\\(\frac{k^{2}+18k + 81}{k^{2}+6k - 27}\\)\
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simplify the rational expression.\\(\frac{c^{2}-17c + 72}{c^{2}-3c - 54}\\)\
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Explanation:

First Rational Expression: $\boldsymbol{\frac{k^2 + 18k + 81}{k^2 + 6k - 27}}$

Step1: Factor numerator

The numerator $k^2 + 18k + 81$ is a perfect square trinomial. We know that $(a + b)^2 = a^2 + 2ab + b^2$. Here, $a = k$, $2ab = 18k$ (so $b = 9$), and $b^2 = 81$. Thus, $k^2 + 18k + 81 = (k + 9)^2$.

Step2: Factor denominator

For the denominator $k^2 + 6k - 27$, we need two numbers that multiply to $-27$ and add to $6$. The numbers are $9$ and $-3$ (since $9 \times (-3) = -27$ and $9 + (-3) = 6$). So, $k^2 + 6k - 27 = (k + 9)(k - 3)$.

Step3: Simplify the expression

Now we have $\frac{(k + 9)^2}{(k + 9)(k - 3)}$. We can cancel out the common factor $(k + 9)$ (assuming $k
eq -9$ to avoid division by zero). This gives $\frac{k + 9}{k - 3}$.

Step1: Factor numerator

For the numerator $c^2 - 17c + 72$, we need two numbers that multiply to $72$ and add to $-17$. The numbers are $-8$ and $-9$ (since $(-8) \times (-9) = 72$ and $(-8) + (-9) = -17$). So, $c^2 - 17c + 72 = (c - 8)(c - 9)$.

Step2: Factor denominator

For the denominator $c^2 - 3c - 54$, we need two numbers that multiply to $-54$ and add to $-3$. The numbers are $-9$ and $6$ (since $(-9) \times 6 = -54$ and $(-9) + 6 = -3$). So, $c^2 - 3c - 54 = (c - 9)(c + 6)$.

Step3: Simplify the expression

Now we have $\frac{(c - 8)(c - 9)}{(c - 9)(c + 6)}$. We can cancel out the common factor $(c - 9)$ (assuming $c
eq 9$ to avoid division by zero). This gives $\frac{c - 8}{c + 6}$.

Answer:

$\frac{k + 9}{k - 3}$

Second Rational Expression: $\boldsymbol{\frac{c^2 - 17c + 72}{c^2 - 3c - 54}}$