Examine the expression (5x + 15)/5x. Long division would have students divide the fraction: 5x + 15 ÷ 5x. In this example, 5x goes into (5x + 15) one time with a remainder of 15. Because of the variable, you cannot divide 15 by 5x. However, with short division, students visually cancel out the like terms.
Cross out the common factors in both the numerator and denominator, which in this case is 5x.
Write the remaining term, which is 15.
Divide x^2 - 49 by x + 7. Write it as a fractional division problem: (x^2 - 49)/(x + 7).
Factor the numerator (x^2 - 49) as a product of two binomials: (x + 7)(x - 7).
Write the binomials over the denominator: (x + 7)(x - 7)/(x + 7). Notice that the numerator and denominator have a common factor, the term (x + 7).
Cross out the like factors and write the quotient 1 in their places: 1(x - 7)/1.
Multiply the numerator out to simplify: 1 x (x - 7) = x - 7/1, which simplifies further to x - 7.
Divide 15x^3 - 20x^2 by 5x or write it as a fractional division problem (15x^3 - 20x^2)/5x.
Write each part of the numerator over the denominator: 15x^3/5x + -20x^2/5x.
Notice that 5 is common factor of both the numerators and denominators.
Cancel each numerator and denominator by crossing out the common factors and then write the quotients in their places: 3x^3/1x + -4x^2/1x. There are still the common factors of x to deal with.
Divide the variables in the numerators by the variables in the denominator. According to the variable-exponent division rule, simply subtract the denominator's exponent from the numerator's exponent. Tthere is always an understood 1 for every term; in this case the denominators are actually x^1. Therefore, x^3 - x^1 = x^2 and x^2 - x^1 = x^1.
Write the simplified solution: 3x^2 - 4x. (Even though the exponent 1 has been omitted from the solution, it is understood as present.)