Division of Polynomial Fractions

Polynomial fraction is in the form of the ratio of two polynomials like $\frac{P(x)}{Q(x)}$ where divisible of zero is not allowed,like $Q(x)\neq 0$ . Various operations can be performed same as we do in simple arithmetic such as add, divide, multiply and subtract.Polynomial fraction is an expression of a polynomial divided by another polynomial. Let P(x) and Q(x), where Q(x) cannot be zero.

$F(x)=\frac{P(x)}{Q(x)}$ = $\frac{15y-6-25}{y^{2}-y-5}$ $\leftarrow Numerator\\\leftarrow Denominator$

The principle which we apply while dividing two fraction i.e. $\frac{a}{b}\div \frac{c}{d} = \frac{a}{b}*\frac{d}{c} = \frac{a*d}{b*c}$ where $b\neq 0$ , $d\neq 0$ and $c\neq 0$ , the same principle is being applied while dividing two polynomial fractions containing variables and coeficient in it. There are two ways to divide rational expression i.e. reduce the fraction or simplify it and the another one is to do the long polynomial division. Remember that division of fractions of rational expressions is the same as multiplication by the reciprocal of the divisor. To divide polynomial, first step is to reciprocal the fraction which appears after the division sign and change into multiplication sign.

Example 1: Divide $\frac{2}{6} \div \frac{4}{6}$

Solution: Given expression $\frac{2}{6} \div \frac{4}{6}$ = $\frac{2}{6} * \frac{6}{4}$ = $\frac{2*1}{1*4}$ = $\frac{2}{4}$ = $\frac{1}{2}$

Example 2: Divide $\frac{3}{2} \div \frac{7}{4}$

Solution: Given expression $\frac{3}{2} \div \frac{7}{4}$ = $\frac{3}{2} * \frac{4}{7}$ = $\frac{3*2}{1*7}$ = $\frac{6}{7}$

Steps for dividing polynomial fractions

Factor each the numerators and denominators of all fractions completely.
Reciprocal the fraction which appears after the division sign and changes it into multiplication sign. Remember it is essential to change the fraction and multiply thereafter to cancel it further.
Cancel or reduce the fractions. keep in mind that to reduce fractions; you’ll be able to cancel something within the numerator with one thing within the denominator, however, so as to cancel something within the numerator and denominator the 2 factors should be precisely the same.
Rewrite the remaining factor. Notice that you simply don’t need to really to multiply something within numerator or denominator.

Example 1: Divide $\frac{x(x+2)}{x^2-1} \div \frac{x}{x-1}$

Solution: 1. By factoring completely the numerator and denominator,if possible we get $\frac{x(x+2)}{x^2-1} \div \frac{x}{x-1}$ = $\frac{x(x+2)}{(x-1)(x+1)} \div \frac{x}{x-1}$

2. By flipping the fraction after the division sign and change it into multipilcation sign $\frac{x(x+2)}{(x-1)(x+1)} \div \frac{x}{x-1}$ = $\frac{x(x+2)}{(x-1)(x+1)} * \frac{x-1}{x}$

3. Reducing the fraction by cancelling out common factors that appears in both the numerators and dinominator and then rewriting the fractions: $\frac{x(x+2)}{(x-1)(x+1)} * \frac{x-1}{x}$ = $\frac{x+2}{x+1}$

Example 2: Divide $\frac{2x+18}{x^2-x} \div \frac{5x+45}{x+1}$

Solution: 1. By factoring completely the numerator and denominator,if possible we get $\frac{2x+18}{x^2-x} \div \frac{5x+45}{x+1}$ = $\frac{2(x+9)}{x(x+1)} \div \frac{5(x+9)}{x+1}$

2. By flipping the fraction after the division sign and change it into multipilcation sign $\frac{2(x+9)}{x(x+1)} \div \frac{5(x+9)}{x+1}$ = $\frac{2(x+9)}{x(x+1)} * \frac{x+1}{5(x+9)}$

3. Reducing the fraction by cancelling out common factors that appears in both the numerators and dinominator and then rewriting the fractions: $\frac{2(x+9)}{x(x+1)} * \frac{x+1}{5(x+9)}$ = $\frac{2}{5x}$

Example 3: Divide $\frac{6x^2+11x+3}{2x^2-7x-15} \div \frac{9x^2-1}{2x^2-9x-5}$

Solution: 1. By factoring completely the numerator and denominator,if possible we get $\frac{6x^2+11x+3}{2x^2-7x-15} \div \frac{9x^2-1}{2x^2-9x-5}$ = $\frac{(3x+1)(2x+3)}{(2x+3)(x-5)} \div \frac{(3x+1)(3x-1)}{(x-5)(2x+1)}$

2. By flipping the fraction after the division sign and change it into multipilcation sign $\frac{(3x+1)(2x+3)}{(2x+3)(x-5)} \div \frac{(3x+1)(3x-1)}{(x-5)(2x+1)}$ = $\frac{(3x+1)(2x+3)}{(2x+3)(x-5)} * \frac{(x-5)(2x+1)}{(3x+1)(3x-1)}$

3. Reducing the fraction by cancelling out common factors that appears in both the numerators and dinominator and then rewriting the fractions: $\frac{(3x+1)(2x+3)}{(2x+3)(x-5)} * \frac{(x-5)(2x+1)}{(3x+1)(3x-1)}$ = $\frac{2x+1}{3x-1}$