Download Example of rational expressions

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
,
− 2
,
,
= 5x2 − 2x + 1
2
3x + 4x − 1
x +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 2
x + 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 − 1
x2 − 2x + 1 x2 − 16
(x − 1)(x − 1) (x + 4)(x − 4)
x2 − 2x + 1
÷ 2
= 2
· 2
=
·
2
x + 5x + 4 x − 16
x + 5x + 4 x − 1
(x + 4)(x + 1) (x + 1)(x − 1)
(x − 1) (x − 4)
x2 − 5x + 4
=
·
= 2
(x + 1) (x + 1)
x + 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
+ 2
x−1 x −1
Ans: We can factor the denominator of the second fraction:
x+1
1
x+1
1
+ 2
=
+
x−1 x −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
=
+ 2
=
=
x2 − 1
x −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:
x+7
(2x − 3)(x − 3)
(x + 7)(x + 1)
2x − 3
−
=
−
(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.
x−4
x2
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
− x3 + 4x2
+4
We get a new polynomial with a lower degree. We repeat the process with this
polynomial.
x2
x−4
x3 + 3x2 − 3x
− x3 + 4x2
7x2 − 3x
+4
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
− x3 + 4x2
7x2 − 3x
+4
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
− x3 + 4x2
7x2 − 3x
− 7x2 + 28x
+4
We get yet another expression of lower degree.
x2 + 7x
x−4
x3 + 3x2 − 3x
− x3 + 4x2
7x2 − 3x
− 7x2 + 28x
25x
+4
+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
5
4
3
− 3x + 3x − 3x
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:
x2 − 1
2x2
+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).
2
18
+
= [5(y + 1)]4
5(y + 1)
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
x
−2
1
− 2
=
x+3 x −9 x−3
The LCM is x2 − 9, we multiply this to both sides of the equation.
x
1
−2
− 2
(x − 9)
= (x2 − 9)
x+3 x −9
x−3
1
x
−2
(x2 − 9)
− (x2 − 9) 2
= (x2 − 9)
x+3
x −9
x−3
(x − 3) − (x) = (x + 3)(−2)
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.
48
7
6
+ 2
=
x + 3 x − 2x − 15 x − 5
The LCM is (x2 − 2x − 15), we multiply both sides of equation by this.
6
48
7
(x2 − 2x − 15)
+ 2
= (x2 − 2x − 15)
x + 3 x − 2x − 15
x−5
6
48
7
(x2 − 2x − 15)
+ (x2 − 2x − 15) 2
= (x2 − 2x − 15)
x+3
x − 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.