Rational Equations and Partial Fraction Decomposition 4.6 Rational Equations and Partial Fraction Decomposition

are rational expressions. For example, A rational expression is a fraction with polynomials for the numerator and denominator. are rational expressions. For example, If x is replaced by a number making the denominator of a rational expression zero, the value of the rational expression is undefined. Example: Evaluate for x = –3, 0, and 1. x  3 undefined undefined 9 undefined 1 Rational Expression

A rational equation is an equation between rational expressions. For example, and are rational equations. To solve a rational equation: 1. Find the LCM of the denominators. 2. Clear denominators by multiplying both sides of the equation by the LCM. 3. Solve the resulting polynomial equation. 4. Check the solutions. Rational Equation

Examples: 1. Solve: . LCM = x – 3. 1 = x + 1 x = 0 (0) LCM = x(x – 1). Find the LCM. 1 = x + 1 Multiply by LCM = (x – 3). x = 0 Solve for x. (0) Check. Substitute 0. Simplify. True. 2. Solve: . LCM = x(x – 1). Find the LCM. Multiply by LCM. x – 1 = 2x Simplify. x = –1 Solve. Examples: Solve

In this case, the value is not a solution of the rational equation. After clearing denominators, a solution of the polynomial equation may make a denominator of the rational equation zero. In this case, the value is not a solution of the rational equation. It is critical to check all solutions. Example: Solve: . Since x2 – 1 = (x – 1)(x + 1), LCM = (x – 1)(x + 1). 3x + 1 = x – 1 2x = – 2  x = – 1 Check. Since – 1 makes both denominators zero, the rational equation has no solutions. Example: Solve

Example: Solve: . x2 – 8x + 15 = (x – 3)(x – 5) x(x – 5) = – 6 Factor. The LCM is (x – 3)(x – 5). Original Equation. x(x – 5) = – 6 Polynomial Equation. x2 – 5x + 6 = 0 Simplify. (x – 2)(x – 3) = 0 Factor. Check. x = 2 is a solution. x = 2 or x = 3 Check. x = 3 is not a solution since both sides would be undefined. Example: Solve

Sometimes we need more tools to help with rational expressions… We will learn to perform a process known as partial fraction decomposition…

To find partial fractions for an expression, we need to reverse the process of adding fractions. We will also develop a method for reducing a fraction to 3 partial fractions.

We’ll start by adding 2 fractions. e.g. The partial fractions for are

To find the partial fractions, we start with The expressions are equal for all values of x so we have an identity. The identity will be important for finding the values of A and B.

To find the partial fractions, we start with Multiply by the denominator of the l.h.s. So, If we understand the cancelling, we can in future go straight to this line from the 1st line.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. Why should I choose x = 2 ? ANS: x = 2 means the coefficient of B is zero, so B disappears and we can solve for A.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. What value would you substitute next ? ANS: Any value would do but x = - 1 is good.

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

This is where the identity is important. The expressions are equal for all values of x, so I can choose to let x = 2. If we chose x = 1 instead, we get 4 = 2A – B, giving the same result. So,

e.g. 2 Express the following as 2 partial fractions. Solution: Let Multiply by : It’s very important to write this each time

So, We never leave fractions piled up like this, so The “halves” are written in the denominators ( as 2s ) and the minus sign is moved to the front of the 2nd fraction. Finally, we need to check the answer. A thorough check would be to reverse the process and put the fractions together over a common denominator.

Another check is to use the “cover-up” method: To check A, find the value of x that makes the factor under A equal to zero ( x = 3 ) Cover-up on the l.h.s. and substitute x = 3 into the l.h.s. only We get To check B, substitute x = 1 in the l.h.s. but cover-up

The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors,

The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, ( we may have to factorise to find them )

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree than the denominator The degree of a polynomial is given by the highest power of x.

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree than the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree then the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree 2

The degree of a polynomial is given by the highest power of x. The method we’ve used finds partial fractions for expressions I’ll call Type 1 e.g. where, the denominator has 2 linear factors, and the numerator is a polynomial of lower degree then the denominator The degree of a polynomial is given by the highest power of x. Here the numerator is of degree 1 and the denominator of degree 2

SUMMARY To find partial fractions for expressions like Let Multiply by the denominator of the l.h.s. Substitute a value of x that makes the coefficient of B equal to zero and solve for A. Substitute a value of x that makes the coefficient of A equal to zero and solve for B. Check the result by reversing the method or using the “cover-up” method.

Exercises Express each of the following in partial fractions. 1. 2. 3. 4.