Answer
The domain of this function is all real numbers except $-\frac{3}{2}$ and $5$.
Work Step by Step
To find the domain of this function, we need to find which values are excluded for $x$. In a rational function, the denominator cannot equal $0$ because the function would be undefined. Therefore, we need to set the denominator equal to $0$ and solve for $x$:
$2x^2 - 7x - 15 = 0$
We can factor this equation to solve for $x$.
We see that we have a quadratic equation, which is given by the formula:
$ax^2 + bx + c$, where $a$, $b$, and $c$ are all real numbers.
To factor this equation, we want to find which factors when multiplied will give us the product of the $a$ and $c$ terms, which is $-30$, but when added together will give us the $b$ term, which is $-7$. This means that one factor should be negative and one factor should be positive, but the negative factor should have the greater absolute value.
Let's look at possible factors:
$-10$ and $3$
$-15$ and $2$
It looks like the first combination will work. Let's split the middle term:
$2x^2 - 10x + 3x - 15 = 0$
Group the first two terms and the last two terms:
$(2x^2 - 10x) + (3x - 15) = 0$
Factor out what is common in both groups:
$2x(x - 5) + 3(x - 5) = 0$
Group the factors:
$(2x + 3)(x - 5) = 0$
According to the zero product property, if the product of two factors equals $0$, then either factor can be $0$; therefore, we can set each of these factors equal to $0$ and solve:
$2x + 3 = 0$ or $x - 5 = 0$
Let's look at the first factor:
$2x + 3 = 0$
Subtract $3$ from each side of the equation:
$2x = -3$
Divide each side by $2$:
$x = -\frac{3}{2}$
Let's look at the second factor:
$x - 5 = 0$
Add $5$ to each side:
$x = 5$
By solving for $x$ in the denominator, we find what numbers $x$ cannot be. Therefore, the domain of this function is all real numbers except $-\frac{3}{2}$ and $5$.
