Questions: Find all holes of the following function. If a hole exists, write your answer in simplest form. f(x) = (5x^2+33x-56) / (3x^2+3x)

Transcript text: Find all holes of the following function. If a hole exists, write your answer in simplest form. \[ f(x)=\frac{5 x^{2}+33 x-56}{3 x^{2}+3 x} \]

Solution

Solution Steps

To find the holes of the given function \( f(x) = \frac{5x^2 + 33x - 56}{3x^2 + 3x} \), we need to identify any common factors in the numerator and the denominator that can be canceled out. If such factors exist, the values of \( x \) that make these factors zero are the locations of the holes.

Solution Approach

Factor both the numerator and the denominator.
Identify and cancel out any common factors.
The values of \( x \) that make the canceled factors zero are the holes.

Step 1: Factor the Numerator and Denominator

The given function is: \[ f(x) = \frac{5x^2 + 33x - 56}{3x^2 + 3x} \]

First, we factor the numerator and the denominator: \[ \text{Numerator: } 5x^2 + 33x - 56 = (x + 8)(5x - 7) \] \[ \text{Denominator: } 3x^2 + 3x = 3x(x + 1) \]

Step 2: Simplify the Function

Next, we simplify the function by canceling out any common factors. In this case, there are no common factors between the numerator and the denominator: \[ f(x) = \frac{(x + 8)(5x - 7)}{3x(x + 1)} \]

Step 3: Identify the Holes

Since there are no common factors to cancel, there are no holes in the function. The common factors are: \[ \text{Common Factors: } 1 \] Thus, the values of \( x \) that would create holes are: \[ \text{Holes: } \text{None} \]