## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$\text{improper}$; $1+\dfrac{-2x+12}{x^{2}+x-12}$
A rational expression $\dfrac{A(x)}{B(x)}$ is said to be proper if the degree of polynomial in numerator is less than the degree of its denominator.If it does not happen , then it is said to be improper rational polynomial. Here, the degree of the numerator $A(x)= x(x-1)=x^2-x$ is $2$ and the degree of the denominator ; $B(x)=(x+4)(x-3)=x^2+x-12$ is $2$. We see that the given rational expression is improper, because the degree of the polynomial in the numerator is equal to the degree of the polynomial in the denominator. To make the rational expression proper, we will solve as: \begin{align*} \dfrac{x(x-1)}{(x+4)(x-3)}&=\dfrac{x^2-x}{x(x-3)+4(x-3)} \\ \\&=\dfrac{x^2-x}{x^2-3x+4x-12} \\ \\&=\dfrac{x^2-x}{x^2+x-12} \\ \\&=\dfrac{(x^2+x-12)-2x+12}{x^{2}+x-12} \\ \\&=\dfrac{x^2+x-12}{x^{2}+x-12}+\dfrac{-2x+12}{x^{2}+x-12} \\ \\&=1+\dfrac{-2x+12}{x^{2}+x-12} \end{align*}