Question

(2x^3-x^2-3x)/(x^2+x)

a. 2x-3x
b. 2^2-3x
c. 2x^2+x
d. 2x-3

Answer 1

\[\frac{2x^3-x^2-3x}{x^2+x}=\frac{x(2x^2-x-3)}{x(x+1)}=\frac{x(2x-3)(x+1)}{x(x+1)}=\cdots \]
(eliminate the common factors in the numerator and denominator)