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# Rational expressions ☰

Rational expressions are expressions that are composed of one or more polynomial terms divided by another polynomial term. A polynomial is a mathematical expression that consists of one or more variables and coefficients, and is built up of powers of these variables. For example, \( 3x^2 –2x+1 \) is a polynomial expression in the variable \(x\).

The general form of a rational expression is: \(\frac{p(x)}{q(x)}\)

where \( p(x) \) and \( q(x) \) are polynomials and \( q(x) \) is not equal to zero.

Some examples of rational expressions are:

- \(\frac{2x^2+ 5x - 1}{x^2- 4}\)
- \( \frac{4x^3- 6x^2+ 2x}{2x^2- 6x} \)
- \( \frac{x^2- 4}{x + 2} \)

Any polygon can be represented in the form of a fraction with a numerator of 1. This means that polygons are also rational expressions.

The sum, difference, product, and quotient of rational expressions are also rational expressions. Rational expressions that are in the form of a fraction with a variable in both the numerator and the denominator are called **rational algebraic expressions**.

The possible values of variables that give meaning to the expression are called the *Domain of Meaningful Values (DMV)*. In the case of fractional expressions with variables, the expression may not have meaning for certain values of the variable.

For example:

The expression \( \frac{2x^2+ 5x - 1}{x^2- 4} \) is meaningless when \(x=2\).

This is because when \(x=2\), the fraction becomes \(0\), and division by zero is undefined, making the expression meaningless.

# Simplifying rational expressions ☰

Simplifying rational expressions involves reducing an expression to its simplest form by factoring the numerator and denominator and canceling out any common factors. The goal is to express the expression in a form that is easier to work with or that better represents the original expression.

To simplify a rational expression, the first step is to factor both the numerator and the denominator. Factoring involves breaking the expression down into its constituent parts, which can be further simplified if necessary. This process can be done by finding common factors or using methods such as factoring by grouping, the difference of squares, or the sum or difference of cubes.

Once the numerator and denominator are factored, the next step is to cancel out any common factors that appear in both the numerator and denominator. This is equivalent to dividing both the numerator and denominator by the same factor.

When two expressions are equal for all possible values of the variables, they are called identical or equivalent expressions.

If we multiply or divide both the numerator and the denominator of a fraction by the same non-zero number, the value of the fraction remains unchanged.

That is, when \(b\neq0\) and \(c\neq0\), the equality \( \frac{a}{b}=\frac{ac}{bc} \) holds true. This property is also valid for similar rational expressions.

For example:

\( \frac{3}{x} \) and \( \frac{3x+3}{x^2+x} \) are equivalent expressions. This is because if we multiply both the numerator and the denominator of \( \frac{3}{x} \) by \((x+1)\), we obtain the expression \( \frac{3x+3}{x^2+x} \).

- Multiplying by \((x+1)\):

\( \frac{3}{x}=\frac{3(x+1)}{x(x+1)}= \frac{3x+3}{x^2+x} \) - Dividing by \((x+1)\):

\( \frac{3x+3}{x^2+x}=\frac{3(x+1)}{x(x+1)} =\frac{3}{x} \)

Here is an example of simplifying a rational expression:

Simplify the expression \( \frac{6x^2 + 9x}{3x} \)

Step 1: Factor both the numerator and denominator.

\( \frac{6x^2+ 9x}{3x}= \frac{3x(2x+3)}{3x} \)

Step 2: Cancel out any common factors.

\( \frac{3x(2x+3)}{3x}=2x+3 \)

The simplified expression is: \(2x+3\).

Here's another example:

Simplify the expression \( \frac{2x^2– 8x}{6x^2– 18x} \)

Step 1: Factor both the numerator and denominator.

\( \frac{2x^2– 8x}{6x^2– 18x}= \frac{2x(x-4)}{6x(x-3)} \)

Step 2: Cancel out any common factors.

\( \frac{2x(x-4)}{6x(x-3)} = \frac{x-4}{3(x-3)} \)

The simplified expression is \( \frac{x-4}{3(x-3)} \).

In some cases, after factoring the numerator and denominator, it may not be possible to cancel out any common factors. In these cases, the expression is already in its simplest form and cannot be simplified further.

Simplifying rational expressions is an important skill in algebra and is used in many areas of mathematics and science, such as calculus, physics, and engineering. a

# Addition, subtraction, multiplication, division, and exponentiation of rational expressions. ☰

**Addition and subtraction of rational expressions:**

To add or subtract rational expressions, the first step is to find a common denominator. This is the same as finding a common multiple of the denominators. Once a common denominator is found, the numerators can be added or subtracted.

For example, to add the rational expressions \( \frac{2}{x}+\frac{3}{x+1} \) , we need to find a common denominator, which in this case is \( x(x+1) \). We then rewrite the expressions with the common denominator:

\( \frac{2(x+1)+ 3x}{x(x+1)} =\frac{5x+2}{x(x+1)} \)

Similarly, to subtract the rational expressions \( \frac{2}{x}-\frac{3}{x+1} \) , we find the common denominator \( x(x+1) \) and rewrite the expressions:

\( \frac{2(x+1)-3x}{x(x+1)} = -\frac{x-2}{x(x+1)}\)

**Multiplication of rational expressions:**

To multiply rational expressions, we multiply the numerators together and the denominators together. We should also simplify the resulting expression, if possible, by canceling out any common factors in the numerator and denominator.

For example, to multiply the rational expressions \( \frac{2}{x}\cdot \frac{3}{x+1}\) ,we multiply the numerators together and the denominators together:

\( \frac{2\cdot3}{x\cdot (x+1)} =\frac{6}{x^2+x} \)

**Division of rational expressions:**

To divide rational expressions, we invert the second expression (the denominator), then multiply the first expression (the numerator) by the inverted second expression. We should also simplify the resulting expression, if possible, by canceling out any common factors in the numerator and denominator.

For example, to divide the rational expressions \( \frac{(\frac{2}{x})}{(\frac{3}{x+1})} \) , we invert the second expression and multiply:

\( \frac{2}{x}\cdot \frac{x+1}{3}=\frac{2(x+1)}{3x} \)

**Exponentiation of rational expressions:**

To raise a rational expression to an exponent, we raise the numerator and denominator to the power of the exponent separately. We should also simplify the resulting expression, if possible, by canceling out any common factors in the numerator and denominator.

The general rule: \( (\frac{A}{B})^n=\frac{A^n}{B^n} \)

For example, to raise the rational expression \(\frac{2}{x}\) to the power of 3, we raise the numerator and denominator to the power of 3 separately:

\(\frac{2^3}{x^3} =\frac{8}{x^3}\)

Example:

1. Performing the addition operation for \( \frac{3}{x}+\frac{5}{2x}\)

Step 1: Find the common denominator, which is the smallest number that both denominators can evenly divide. In this case, the common denominator is \(2x\).

Step 2: Convert both fractions to have the common denominator.

Step 3: Add the fractions:

\( \frac{3}{x}+\frac{5}{2x}=\frac{3\cdot 2}{2x}+\frac{5}{2x}=\frac{11}{2x}\)

2. Subtract the rational expressions \( \frac{7}{x}–\frac{2}{x+1}\). Simplify your answer as much as possible.

Answer: Find the common denominator. In this case, the common denominator is \(x(x+1)\), because it's the least common multiple of \(x\) and \(x+1\)

\( \small \frac{7}{x} - \frac{2}{x + 1} = \frac{7 \cdot (x + 1)}{x(x + 1)} - \frac{2x}{x(x + 1)} = \frac{7x + 7 - 2x}{x(x + 1)} = \frac{5x + 7}{x(x + 1)}\)

3. Perform the multiplication.

\(\frac{2x}{x^2 + 2x} \cdot \frac{3x^2 + 6x}{x^2 + 4x}\)

Solution:

\(\frac{2x}{x^2 + 2x} \cdot \frac{3x^2 + 6x}{x^2 + 4x}\) First, we simplify both rational expressions separately.

Simplifying the first rational expression, we get:

\( \frac{2x}{x^2 + 2x} = \frac{2x}{x(x+2)} = \frac{2}{x+2} \)

Simplifying the second rational expression, we get:

\( \frac{3x^2 + 6x}{x^2 + 4x} = \frac{3x(x+2)}{x(x+4)} = \frac{3(x+2)}{(x+4)} \)

Now, we multiply the simplified rational expressions:

\( \frac{2}{x+2} \cdot \frac{3(x+2)}{(x+4)} = \frac{6(x+2)}{(x+2)(x+4)} = \frac{6}{x+4} \)

# Reciprocal Function and its graph ☰

The function \(y=\frac{k}{x}\) , where \(k\) is a constant, is an example of a rational function, also known as a reciprocal function. It is a mathematical relationship that describes how the output value of \(y\) changes as the input value of \(x\) changes.

In this function, \(k\) is a constant value that does not change as \(x\) changes. As \(x\) increases or decreases, the value of \(y\) changes in a way that depends on \(k\).

When \(x\) is positive, the value of \(y\) is positive, and as \(x\) increases, the value of \(y\) decreases. This is because as \(x\) gets larger, the denominator of the fraction, which is \(x\), gets larger, causing the value of the fraction to become smaller.

Conversely, as \(x\) gets smaller, the denominator gets smaller, causing the value of the fraction to become larger. Therefore, as \(x\) approaches zero, \(y\) approaches infinity, and as \(x\) approaches infinity, \(y\) approaches zero.

The function has a vertical asymptote at \(x=0\), which means that the value of \(y\) becomes infinitely large or small as \(x\) approaches 0 from either direction. Additionally, the function has a horizontal asymptote at \(y=0\), which means that as \(x\) approaches infinity or negative infinity, the value of \(y\) approaches 0.

In summary, the function \(y=\frac{k}{x}\) is a reciprocal function that describes how the output value of \(y\) changes as the input value of \(x\) changes. Its behavior is determined by the constant \(k\), and it has a vertical asymptote at \(x=0\) and a horizontal asymptote at \(y=0\).

**The graph of the function \(y=\frac{k}{x}\) is a hyperbola.** It consists of two branches, one in the first quadrant and one in the third quadrant, with the \(y\)-axis as the vertical asymptote and the \(x\)-axis as the horizontal asymptote.

The shape of the hyperbola is determined by the value of \(k\). If \(k\) is positive, the hyperbola will open up and to the right, and if \(k\) is negative, the hyperbola will open down and to the left.

As \(x\) approaches 0 from either side, the function approaches positive or negative infinity, depending on the sign of \(k\). As \(x\) approaches infinity or negative infinity, the function approaches zero.

Here is an example graph of \(y=\frac{1}{x}\):

As you can see, the graph has two branches, one in the first quadrant and one in the third quadrant, with the \(y\)-axis and \(x\)-axis as the vertical and horizontal asymptotes, respectively. As \(x\) approaches zero, the function approaches positive or negative infinity, and as \(x\) approaches infinity or negative infinity, the function approaches zero.