Question

given the rational function $f(x) = \frac{x^2 + 13x + 36}{x^2 + 5x + 4}$ find any hole(s) of the function.\
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$\circ x = -4, x = -1$\
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$\circ x = -9$\
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$\circ x = -1$\
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$\circ x = -4$\
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$\circ x = -4, x = -9$\
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$\circ$ there are no holes

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 + 13x + 36\): \(x^2 + 13x + 36=(x + 4)(x + 9)\)
Factor \(x^2 + 5x + 4\): \(x^2 + 5x + 4=(x + 4)(x + 1)\)

Step2: Simplify the function

\(f(x)=\frac{(x + 4)(x + 9)}{(x + 4)(x + 1)}=\frac{x + 9}{x + 1}\) (for \(x
eq - 4\))

Step3: Identify holes

A hole occurs at the root of the common factor \((x + 4)\). Set \(x+4 = 0\), so \(x=-4\).

Answer:

\(x = - 4\)