Answer
The domain of this function is all real numbers except $-\frac{4}{3}$ and $\frac{5}{2}$.
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 $x$:
$6x^2 - 7x - 20 = 0$
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 $-120$, 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:
$-15$ and $8$
$-20$ and $6$
$-12$ and $10$
It looks like the first combination will work. Let's split the middle term:
$6x^2 - 15x + 8x - 20 = 0$
Group the first two terms and the last two terms:
$(6x^2 - 15x) + (8x - 20) = 0$
Factor out what is common in both groups:
$3x(2x - 5) + 4(2x - 5) = 0$
Group the factors:
$(3x + 4)(2x - 5) = 0$
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:
$3x + 4 = 0$ or $2x - 5= 0$
Let's look at the first factor:
$3x + 4 = 0$
Subtract $4$ from each side of the equation:
$3x = -4$
Divide each side by $3$:
$x = -\frac{4}{3}$
Let's look at the second factor:
$2x - 5 = 0$
Add $5$ to each side:
$2x = 5$
Divide each side by $2$:
$x = \frac{5}{2}$
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 $-\frac{4}{3}$ and $\frac{5}{2}$.
