A Rational Expression Is an Expression Which Is the Ratio of Two Polynomials: Example of Rational Expressions: 2X

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A Rational Expression is an expression which is the ratio of two polynomials: Example of rational expressions: 2x − 1 1 x3 + 2x − 1 5x2 − 2x + 1 , − , , = 5x2 − 2x + 1 3x2 + 4x − 1 x2 + 1 3x + 2 1 Notice that every polynomial is a rational expression with denominator equal to 1. x2 + 5 For example, x2 + 5 = 1 Think of the relationship between rational expression and polynomial as that between rational numbers and integers.

The domain of a rational expression is all real numbers except where the denominator is 0 3x + 4 Example: Find the domain of x2 + 3x − 4 Ans: x2 + 3x − 4 = 0 ⇒ (x + 4)(x − 1) = 0 ⇒ x = −4 or x = 1 Since the denominator, x2 + 3x − 4, of this rational expression is equal to 0 when x = −4 or x = 1, the domain of this rational expression is all real numbers except −4 or 1. In general, to find the domain of a rational expression, set the denominator equal to 0 and solve the equation. The domain of the rational expression is all real numbers except the solution. To Multiply rational expressions, multiply the numerator with the numerator, and multiply denominator with denominator. Cross cancell if possible. Example: x2 + 3x + 2 x2 − 2x + 1 (x + 1)(x + 2) (x − 1)(x − 1) x − 1 · = · = x2 − 1 x2 + 5x + 6 (x − 1)(x + 1) (x + 2)(x + 3) x + 3 To Divide rational expressions, flip the divisor and turn into a multiplication. Example: x2 − 2x + 1 x2 − 1 x2 − 2x + 1 x2 − 16 (x − 1)(x − 1) (x + 4)(x − 4) ÷ = · = · x2 + 5x + 4 x2 − 16 x2 + 5x + 4 x2 − 1 (x + 4)(x + 1) (x + 1)(x − 1) (x − 1) (x − 4) x2 − 5x + 4 = · = (x + 1) (x + 1) x2 + 2x + 1 To Add or Subtract rational expressions, we need to find common denominator by finding the Least Common Multiple, or LCM of the denominators. Example: x + 1 1 + x − 1 x2 − 1 Ans: We can factor the denominator of the second fraction: x + 1 1 x + 1 1 + = + x − 1 x2 − 1 x − 1 (x − 1)(x + 1)

The LCM of the two fractions is (x − 1)(x + 1) since these are the two factors that are present in one or the other fraction. To obtain commond denominator, we need to multiply the numerator and denominator of the ﬁrst fraction by x + 1: x + 1 1 (x + 1)(x + 1) 1 + = + x − 1 (x − 1)(x + 1) (x − 1)(x + 1) (x − 1)(x + 1)

x2 + 2x + 1 1 (x2 + 2x + 1) + 1 x2 + 2x + 2 = + = = x2 − 1 x2 − 1 x2 − 1 x2 − 1

Example: 2x − 3 x + 7 2x − 3 x + 7 − = − x2 + 3x + 2 x2 − x − 6 (x + 2)(x + 1) (x + 2)(x − 3)

The L.C.M of the two fractions is (x + 2)(x + 1)(x − 3). We have:

2x − 3 x + 7 (2x − 3)(x − 3) (x + 7)(x + 1) − = − (x + 2)(x + 1) (x + 2)(x − 3) (x + 2)(x + 1)(x − 3) (x + 2)(x − 3)(x + 1)

2x2 − 9x + 9 x2 + 8x + 7 (2x2 − 9x + 9) − (x2 + 8x + 7) = − = (x + 2)(x + 1)(x − 3) (x + 2)(x − 3)(x + 1) (x + 2)(x − 3)(x + 1)

2x2 − 9x + 9 − x2 − 8x − 7 x2 − 17x + 2 = = (x + 2)(x − 3)(x + 1) (x + 2)(x − 3)(x + 1)

To perform Long Division of Polynomials, we use a process similar to long division of integers. If we were to divide two numbers, say 2137 ÷ 13, we start by writing the dividen, 2137, inside the division sign, and the divisor, 13, outside. We proceed by ﬁnding out how many times the divisor, 13, goes into the dividend/remainder: 164 13 2137 1300 837 780 57 52 5

We do long division of polynomial similarly. We put the dividend inside the division symbol and the divisor on the outside, and proceed similar to that of long division of numbers. Example: Divide the polynomial x3 + 3x2 − 3x + 4 by the polynomail x − 4. In other words, we want (x3 + 3x2 − 3x + 4) ÷ (x − 4) Ans: We ﬁrst put the dividend, x3 +3x2 −3x+4 inside of the long division symbol, and the divisor, x − 4, outside: x − 4 x3 + 3x2 − 3x + 4

We ask, what multiplied to x will give x3 ? Answer is x2, so we put x2 on the ﬁrst column on top of the long division symbol. Note that the powers of the x’s should line-up.

x2 x − 4 x3 + 3x2 − 3x + 4

We now multiply x2 to the divisor, x−4, and subtract this result from the dividend.

x2 x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2

We get a new polynomial with a lower degree. We repeat the process with this polynomial.

x2 x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x What multiplied to x will give 7x2 ? Answer is 7x, so we put 7x in the next column on top.

x2 + 7x x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x

We then repeat the process, multiply 7x to the divisor, x − 4, and subtract the result from the top:

x2 + 7x x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x − 7x2 + 28x

We get yet another expression of lower degree.

x2 + 7x x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x − 7x2 + 28x 25x + 4 We repeat the process until the degree of the remainder is strictly less than the degree of the divisor. At this stage, the degree of the remainder, 25x + 4, is 1, which is equal to the degree of the divisor, x − 4, we must still continue. We ask, what multiplied to x will give 25x? Answer is 25, so we put 25 on top:

x2 + 7x + 25 x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x − 7x2 + 28x 25x + 4

Multiply 25 to x − 4 and subtract the result from the top: x2 + 7x + 25 x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x − 7x2 + 28x 25x + 4 − 25x + 100

The remainder is now the constant 104. The degree of a constant is 0, which is strictly less than the degree of the divisor. We are done.

x2 + 7x + 25 x − 4 x3 + 3x2 − 3x + 4 − x3 + 4x2 7x2 − 3x − 7x2 + 28x 25x + 4 − 25x + 100 104

In this completed long division, the quotient is the polynomial x2 + 7x + 25, the remainder is the polynomial 104.

Example: Perform the long division and ﬁnd the quotient and remainder: (3x5 − 2x3 + 4x2 + 1) ÷ (x2 − x + 1)

Ans: Notice that the dividend, (3x5 − 2x3 + 4x2 + 1), has some missing powers of x (missing x4 and x term). When performing the long division, we should leave enough space inside the long division symbol so we can align the powers of x’s correctly. Another approach to make sure that the powers of x’s are lined-up correctly is to ﬁll in 0’s where powers of x are missing. Remember that only like powers of x’s can be subtracted or added. 3x3 + 3x2 − 2x − 1 x2 − x + 1 3x5 − 2x3 + 4x2 + 1 − 3x5 + 3x4 − 3x3 3x4 − 5x3 + 4x2 − 3x4 + 3x3 − 3x2 − 2x3 + x2 2x3 − 2x2 + 2x − x2 + 2x + 1 x2 − x + 1 x + 2

The quotient polynomial is 3x3 + 3x2 − 2x − 1. The remainder polynomial is x + 2.

Example: Perform the long division and ﬁnd the quotient and remainder:

(2x4 − x2 − 3x) ÷ (x2 − 1)

This time, both the dividend and divisor have missing powers of x. Pay close attention to spacing and alignment when trying to subtract the polynomials from each other:

2x2 + 1 x2 − 1 2x4 − x2 − 3x − 2x4 + 2x2 x2 − 3x − x2 + 1 − 3x + 1

The quotient is 2x2 + 1, remainder is −3x + 1.

Example: Perform the long division to ﬁnd quotient and remainder: (x5 − 4x3 + x2 − 3x + 2) ÷ (x − 2)

Ans: x4 + 2x3 + x − 1 x − 2 x5 − 4x3 + x2 − 3x + 2 − x5 + 2x4 2x4 − 4x3 − 2x4 + 4x3 x2 − 3x − x2 + 2x − x + 2 x − 2 0

In this example, when (x5 − 4x3 + x2 − 3x + 2) is divided by (x − 2), the remainder is 0. We say that (x − 2) is a factor of (x5 − 4x3 + x2 − 3x + 2). To solve Rational Expression Equations, multiply both sides of the equation by the LCM of the frations to eliminate the fraction. Be sure to check for Extraneous Solution. E.g. 18 2 + = 4 y + 1 5 The LCM of the fractions is 5(y + 1), so we multiply both sides of the equation by 5(y + 1). 18 2 5(y + 1) + = [5(y + 1)]4 y + 1 5 18 2 5(y + 1) + 5(y + 1) = 20(y + 1) y + 1 5 5 · 18 + 2(y + 1) = 20y + 20 90 + 2y + 2 = 20y + 20 2y + 92 = 20y + 20 18y = 72 y = 4 Any time we multiplied both sides of the equation by a variable term, we could potentially have introduced extraneous solution, which are numbers that shows up in the process of solving an equation when we perform some operations (like multiply both sides by a variable term or square both sides). An extraneous solution is not a solution to an equation. Notice the difference between an extraneous solution from a wrong solution, which is a number that simply comes from something you did incorrectly. You could have made no algebraic or other mathematical error and you could still introduce extraneous solution. Extraneous solution comes from the result of some mathematical process of solving an equation, such as multiply both sides by a variable term or square both sides. If you used such a process when solving an equation, you must check the final solution for extraneous solution, even if you believe that you made no algebraic error. If you simply made an error in solving an equation and when you check your solution the number does not wrong, it is simply you made a mistake, not an extraneous solution. Example 1 x −2 − = x + 3 x2 − 9 x − 3 The LCM is x2 − 9, we multiply this to both sides of the equation. 1 x −2 (x2 − 9) − = (x2 − 9) x + 3 x2 − 9 x − 3 1 x −2 (x2 − 9) − (x2 − 9) = (x2 − 9) x + 3 x2 − 9 x − 3 (x − 3) − (x) = (x + 3)(−2) −3 = −2x − 6 3 x = − 2 If we plug this solution back into the original equation, it solves the equation and therefore it is not an extraneous solution. E.g. 6 48 7 + = x + 3 x2 − 2x − 15 x − 5 The LCM is (x2 − 2x − 15), we multiply both sides of equation by this. 6 48 7 (x2 − 2x − 15) + = (x2 − 2x − 15) x + 3 x2 − 2x − 15 x − 5 6 48 7 (x2 − 2x − 15) + (x2 − 2x − 15) = (x2 − 2x − 15) x + 3 x2 − 2x − 15 x − 5 6(x − 5) + 48 = 7(x + 3) 6x − 30 + 48 = 7x + 21 6x + 18 = 7x + 21 x = −3 If we plug x = −3 back into the original equation, we see that this makes the denominators on the left side zero, which makes the expression undefined. So −3 is really not a solution. It is extraneous. Since the only number that we get turns out to be extraneous, our equation has no solution.