Answer
$\dfrac{g(x)}{f(x)} = x - 1$
$\text{The domain is all real numbers except }2$.
Work Step by Step
In this problem, we are asked to divide one function from another. Let us go ahead and set up the division problem:
$\dfrac{g(x)}{f(x)} = \dfrac{x^2 - 3x + 2}{x - 2}$
Let's factor the numerator to see if we can cancel one of the factors with the denominator.
To factor a quadratic polynomial in the form $ax^2 + bx + c$, we look at factors of $c$ that when added together equals $b$.
For the expression $x^2 - 3x + 2$, $c=2$, so look for factors of $2$ that when added together will equal $b$ or $-3$. We need two negative numbers because two negative numbers multiplied will yield a positive number, but when added together will yield a negative number.
We came up with the following possibility:
$2=(-2)(-1)$
$(-2)+(-1) = -3$
Thus, the factored form of the trinomial is $(x-2)(x-1)$.
Therefore the function above can be rewritten as:
$\dfrac{g(x)}{f(x)} = \dfrac{x - 2)(x - 1)}{x - 2}$
Simplify the expression by cancelling the common factor $x - 2$ to obtain:
$\dfrac{g(x)}{f(x)} = x - 1$
For the domain, we look to the denominator to see what types of restrictions we need for values of $x$. We cannot have a situation in which the denominator equals $0$ because the fraction would be undefined. Therefore, we set the denominator equal to $0$ to find which values of $x$ we cannot use.
$x - 2 = 0$
$x = 2$
This means that $x$ cannot equal $2$ because that would make the denominator $0$:
The domain is all real numbers except $2$.
