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Basic Properties
of Rational Expressions
A rational expression is any expression of the form
P
Q
where P and Q are polynomials and Q ≠ 0. In the following properties, no denominator is allowed to be zero.
A fraction is not defined when the denominator is zero!
Basic Property: Example:
P
Q =
R
S if and only if P • S = Q • R^
P
Q =^
P • R
Q • R
– P
Q =
–P
Q =^
P
–Q = –
–P
–Q
–^7
4 =^
P
Q = –
–P
Q = –
P
–Q =^
–P
–Q
Notice that
–P P = –1.
Since it is true that
x – a = – (–x + a)
= – (a – x),
It must also be true that
x – a a – x = –1.
W A R N I N G :
You must reduce only factors!! If the terms are not factors, they cannot be factored out.
Nonsense like "canceling out" nonfactors will not be tolerated!!
2x + 8 2 =^ x + 4
x^2 – 9 x – 3 =^ x + 3
Multiplication and Division of
Rational Expressions
To multiply two rational expression, multiply the numerators together and multiply the denominators together.
P Q •
R
S =^
P • R
Q • S.
To divide two rational expressions, invert the one immediately after the division sign, and do a rational expression multiplication.
P Q ÷
R
S =^
P
Q •^
S
R =^
P • S
Q • R
Example: Perform the indicated operation and simplify:
x^2 + 2 x – 3 x^2 – 3x – 10
÷
4x + 2 x^2 – x
2x^2 – 9 x – 5 x^2 – 2 x + 1
x^2 + 2 x – 3 x^2 – 3x – 10
x^2 – x 4x + 2 •
2x^2 – 9 x – 5 x^2 – 2 x + 1
(x + 3)(x – 1) (x – 5)(x + 2) •^
x(x – 1) 2(2x + 1) •
(2x + 1)(x – 5) (x – 1)^2
x(x + 3) 2(x + 2)
R ule for A dding or S ubtracting
F ractions with E qual D enominators
(RASFED)
To add or subtract two rational expressions whose denominators are equal, simply add or subtract the numerators. Make sure you use parentheses when appropriate!
P Q +
R
Q =^
P + R
Q
P
Q –
R
Q =^
P – R
Q
Examples: Perform the indicated operation and simplify:
a)
x x^2 – 4
x^2 – 4 b)^
x^2 x^2 – 4
4x – x^2 x^2 – 4
x + 2 x^2 – 4 =^
x^2 – (4x – x^2 ) x^2 – 4
x + 2 (x + 2)(x – 2) =^
x^2 – 4 x + x 2 x^2 – 4
=
x – 2 (^) =
2x^2 – 4x x^2 – 4
2x(x – 2) (x + 2)(x – 2)
2x (x + 2)
R ule for A dding or S ubtracting
F ractions with U nequal D enominators
(FLEAS)
1. F actor the rational expression.
2. Find the L east Common Denominator (LCD).
3. E qualize each denominators by replacing each fraction
with an equivalent one whose denominator is the LCD.
4. A dd or S ubtract using RASFED.
Example:
1 2 –
12 –^
12 +^
Examples: Perform the indicated operation and simplify.
a)
2 x^2 – 1 –^
1 x^2 + 2x + 1 b)^
x + 2 x – 2 –
x^2 + 2x x^2 – 4
=
2 (x + 1)(x – 1) –^
1 (x + 1)^2 =
x + 2 x – 2 –^
x^2 + 2x (x + 2)(x – 2)
=
2(x + 1) (x + 1)^2 (x – 1) –^
x – 1 (x + 1)^2 (x – 1) =^
(x + 2)^2 (x + 2)(x – 2) –^
x^2 + 2x (x + 2)(x – 2)
=
2(x + 1) – (x – 1) (x + 1)^2 (x – 1) =
(x^2 + 4x + 4) – (x^2 + 2x) (x + 2)(x – 2)
=
2x + 2 – x + 1 (x + 1)^2 (x – 1) =
x^2 + 4x + 4 – x^2 – 2x (x + 2)(x – 2)
=
x + 3 (x + 1)^2 (x – 1) =^
2x + 4 (x + 2)(x – 2)
=
2(x + 2) (x + 2)(x – 2)
=
2 (x – 2)
Example: Simplify the following:
a)
y^2
3 –
y
b)
x + h –
x h
x(x + h)
x + h –
x x(x + h) h =
y^2
9 –
y^2
y^2
3 –
y
=
x – (x + h) x(x + h) h
=
9y^2 – 1 3y^2 – y =
x – x – h x(x + h) h
=
(3y + 1)(3y – 1) y(3y – 1) =
–h x(x + h) h
=
3y + 1 y =
x(x + h)
Long Division of Polynomials
Monomial Denominator: When you divide a polynomial by a monomial, you use a form of the distributive property, and thus you must divide each term in the numerator by the denominator
Examples: Perform the indicated operation.
a) (x^3 – 6x^2 + 2x) ÷ 3x b) Divide 15y^3 +20y^2 –5y by 5y
Solution: a) (x^3 – 6x^2 + 2x) ÷ 3x
x^3 – 6x^2 + 2x 3x
x^3 3x –
6x^2 3x +
2x 3x
x^2 3 – 2x +
b) Divide 15y^3 +20y^2 –5y by 5y
15y^3 + 20y^2 – 5y 5y
15y^3 5y +
20y^2 5y –
5y 5y
= 3y^2 + 4y – 1
Examples: Calculate the indicated quotients by long division:
a)
x^3 – 2x^2 – 7 x + 3 x + 2
b)
x^4 – 8x^2 – 8 x^2 – x + 2
c)
6x^4 + x 3 – 9 x + 4 2x – 1
Synthetic Division of Polynomials
When you divide a polynomial by a linear polynomial with linear coefficient 1 , we can perform the division by using only the necessary coefficients.
Step 1. Write the opposite of the constant term of the divisor by itself. Write all of the coefficient of the dividend (using zero when terms are missing, of course).
Step 2. Bring down the first term of the dividend. This also becomes the current term.
Step 3. a. Multiply the current term by the divisor term. b. Put the product under the next term of the dividend. c. Add the result. The result becomes the current term.
Step 4. If there is another term of the dividend, then go to Step 3. Otherwise, go to Step 5.
Step 5. The constants of the bottom line are the coefficients of the quotient and the remainder.
Remainder Theorem
When you divide a polynomial P(x) by the factor x – c, the remainder is P(c). Thus we sometimes evaluate a polynomial P(x) when x = c by performing the appropriate synthetic division.
Examples 1: Let P(x) = 2x^3 – 4x^2 + 5.
a) By direct substitution, evaluate P(2).
b) Find the remainder when P(x) is divided by x – 2.
Examples 2: Let P(x) = 4x^6 – 25x^5 + 35x^4 + 17x^2. Find P(4)
Equations Involving Fractions
When we solve equations with fractions, we always assume that no denominator is zero. Therefore, we can find the LCD and multiply both sides by this nonzero factor. We must check to see that our solution does not cause any denominator to be zero.
To solve equations with (simple) fractions:
Step 1: Identify all fractions in the equation and find the LCD.
Step 2: Multiply the both sides of the equation by the LCD.
Step 3: Solve the resulting equation.
Step 4: Check the answer into the original problem. (At
least check to make sure no denominator can be zero.)
W A R N I N G :
You must know the difference between an expression and an equation. When you solve an equation, you may multiply both sides by the LCD and we get an equation without fractions. When you have an expression with fractions, you must perform the indicated operation and / or simplify. Nonsense like treating expressions like equations will not be tolerated!!
c)
y – 2 y – 3 = 1 –^
y^2 – 9
y – 2 y – 3 = 1 –^
(y + 3)(y – 3)
(y + 3)(y – 3)
y – 2 y – 3 = (y + 3)(y – 3)^
1 –
(y + 3)(y – 3)
(y + 3)(y – 2) = (y + 3)(y – 3) – 2
y^2 + y – 6 = y^2 – 9 – 2
y = –
d)
x^2 – 4
x^2
x^2 – 2x
2 (x + 2)(x – 2) =^
x^2
x(x – 2)
x^2 (x + 2)(x – 2)
(x + 2)(x – 2) = x
(^2) (x + 2)(x – 2)
x^2
x(x – 2)
2x^2 = (x + 2)(x – 2) + x(x + 2)
2x^2 = x^2 – 4 + x^2 + 2x
2x^2 = 2x^2 + 2x – 4
–x = –
x = 2
no solution
e) 2 +
x + 2 =^
x + 3
(x + 2)(x + 3)
2 +
x + 2 = (x + 2)(x + 3)^
x + 3
2(x + 2)(x + 3) + 10(x + 3) = 3(x + 2)
2(x^2 + 5x + 6) + 10x + 30 = 3x + 6
2x^2 + 10x + 12 + 10x + 30 = 3x + 6
2x^2 + 17x + 36 = 0
(2x + 9)(x + 4) = 0
x = –9/2, –
f)
y + 1 y + 3 +
y + 5 y – 2 = 1 +^
6y + 23 y^2 + y – 6
(y + 3)(y – 2)
y + 1 y + 3 +
y + 5 y – 2 = (y + 3)(y – 2)
1 +
6y + 23 (y+3)(y–2)
(y – 2)(y + 1) + (y + 3)(y + 5) = (y + 3)(y – 2) + (6y + 23)
y^2 – y – 2 + y^2 + 8y + 15 = y^2 + y – 6 + 6y + 23
2y^2 + 7y + 13 = y^2 + 7y + 17
y^2 = 4
y = ±
y = –
