Multiply Rational Expressions

Here are incredibly detailed notes on multiplying rational expressions and simplifying them, based on "Multiplying Rational Expressions" from "The Organic Chemistry Tutor" YouTube channel [1].

Multiplying Rational Expressions and Simplifying

The general approach involves factoring completely all numerators and denominators and then canceling common terms that appear in both the numerator and the denominator [1]. If a part of the expression is not a fraction, it should be written over one to begin the process, making it a rational expression [1].

Example 1: Multiplying (7x + 14) / (2x^2 - 8) by (x^2 + 3x - 10)

1. Rewrite the second part as a fraction: If an expression is not explicitly a fraction, place it over $1$ to make it a rational expression [1].

   • $(x^2 + 3x - 10)$ becomes $(x^2 + 3x - 10) / 1$ [1].

2. Factor each component completely: This is the most crucial step. Every polynomial in the numerator and denominator of both fractions must be factored into its simplest irreducible components [1].

   • Numerator of the first fraction 7x + 14: This is a binomial.

     • Identify the Greatest Common Factor (GCF). Both terms, $7x$ and $14$, are divisible by $7$ [1].

     • Factor out the GCF: $7(x + 2)$ [1].

   • Denominator of the first fraction 2x^2 - 8: This is a binomial.

     • Identify the GCF. Both terms, $2x^2$ and $-8$, are divisible by $2$ [1].

     • Factor out the GCF: $2(x^2 - 4)$ [1].

     • Further factor $(x^2 - 4)$ using the difference of perfect squares technique. This technique applies to expressions in the form $(a^2 - b^2)$ which factors into $(a + b)(a - b)$ [1].

       • Identify $a$ and $b$: The square root of $x^2$ is $x$ (so $a=x$), and the square root of $4$ is $2$ (so $b=2$) [1].

       • Apply the formula: $(x^2 - 4)$ becomes $(x + 2)(x - 2)$ [1].

       • The fully factored denominator is $2(x + 2)(x - 2)$ [1].

   • Numerator of the second fraction x^2 + 3x - 10: This is a trinomial with a leading coefficient of one (the coefficient of $x^2$ is $1$) [1].

     • Find two numbers that multiply to the constant term ($-10$) and add to the middle coefficient ($3$) [1].

     • Consider pairs of factors for $-10$: $(1, -10), (-1, 10), (2, -5), (-2, 5)$ [1].

     • The pair $+5$ and $-2$ satisfies both conditions ($5 imes -2 = -10$ and $5 + (-2) = 3$) [1].

     • So, the factored form is $(x + 5)(x - 2)$ [1].

3. Assemble the completely factored expression: Rewrite the entire multiplication problem with all components factored [1].

   • $[7(x + 2) / (2(x + 2)(x - 2))] imes [(x + 5)(x - 2) / 1]$ [1].

4. Cancel common factors in the numerator and denominator: Any factor that appears identically in a numerator and a denominator (across both fractions) can be canceled [1].

   • Cancel $(x + 2)$ from the numerator of the first fraction and the denominator of the first fraction [1].

   • Cancel $(x - 2)$ from the denominator of the first fraction and the numerator of the second fraction [1].

5. Write the final simplified answer: Multiply the remaining terms in the numerator and the remaining terms in the denominator [1].

   • The remaining terms in the numerator are $7$ and $(x + 5)$ [1].

   • The remaining terms in the denominator are $2$ and $1$ [1].

   • Final answer: $7(x + 5) / 2$ [1].

Example 2: Multiplying (5x^2 - 15x) / (2x^2 + 11x + 12) by (3x^2 - 48) / (10x^3 - 70x^2 + 120x)

1. Factor each component completely: Apply appropriate factoring techniques to each polynomial [1].

   • Numerator of the first fraction 5x^2 - 15x: Binomial.

     • GCF is $5x$ [1].

     • Factor out the GCF: $5x(x - 3)$ [1].

   • Numerator of the second fraction 3x^2 - 48: Binomial.

     • GCF is $3$ [1].

     • Factor out the GCF: $3(x^2 - 16)$ [1].

     • Factor $(x^2 - 16)$ using the difference of squares technique: $(x + 4)(x - 4)$ [1].

     • The fully factored numerator is $3(x + 4)(x - 4)$ [1].

   • Denominator of the second fraction 10x^3 - 70x^2 + 120x: Trinomial.

     • GCF is $10x$ [1].

     • Factor out the GCF: $10x(x^2 - 7x + 12)$ [1].

     • Factor the trinomial $(x^2 - 7x + 12)$ (leading coefficient of one) [1].

       • Find two numbers that multiply to $12$ and add to $-7$ [1].

       • These numbers are $-3$ and $-4$ ($-3 imes -4 = 12$ and $-3 + -4 = -7$) [1].

       • So, $(x^2 - 7x + 12)$ becomes $(x - 3)(x - 4)$ [1].

     • The fully factored denominator is $10x(x - 3)(x - 4)$ [1].

   • Denominator of the first fraction 2x^2 + 11x + 12: Trinomial with a leading coefficient ($2$) not equal to one [1].

     • Method: Factor by Grouping [1]:

       1. Multiply the leading coefficient ($2$) by the constant term ($12$): $2 imes 12 = 24$ [1].

       2. Find two numbers that multiply to $24$ and add to the middle coefficient ($11$) [1].

          • Consider factors of $24$: $(1,24), (2,12), (3,8), (4,6)$ [1].

          • The pair $8$ and $3$ satisfy both conditions ($8 imes 3 = 24$ and $8 + 3 = 11$) [1].

       3. Replace the middle term 11x) with these two numbers as coefficients of x: Rewriting $(+11x)$ as $(+8x + 3x)$ gives: $2x^2 + 8x + 3x + 12$ [1].

       4. Factor by grouping: Group the first two terms and the last two terms, then find the GCF of each group [1].

          • From the first two terms $(2x^2 + 8x)$, the GCF is $2x$, leaving $2x(x + 4)$ [1].

          • From the last two terms $(3x + 12)$, the GCF is $3$, leaving $3(x + 4)$ [1].

          • The expression now is $2x(x + 4) + 3(x + 4)$ [1].

       5. Factor out the common binomial: Notice that $(x + 4)$ is a common factor to both terms; factor it out [1].

          • The factored form is $(x + 4)(2x + 3)$ [1].

     • The fully factored denominator is $(x + 4)(2x + 3)$ [1].

2. Assemble the completely factored expression: Write out the full multiplication problem with all factored components [1].

   • $[5x(x - 3) / ((x + 4)(2x + 3))] imes [3(x + 4)(x - 4) / (10x(x - 3)(x - 4))]$ [1].

3. Cancel common factors in the numerator and denominator: Identify and cancel identical factors [1].

   • Cancel $(x - 3)$ [1].

   • Cancel $(x - 4)$ [1].

   • Cancel $(x + 4)$ [1].

   • Cancel $x$ [1].

   • Reduce the constant terms $5$ and $10$: $5/10$ reduces to $1/2$. This means the $5$ in the numerator cancels, and the $10$ in the denominator becomes $2$ [1]. (Alternatively, think of $10 / 5 = 2$ remaining in the denominator).

4. Write the final simplified answer: Multiply remaining terms [1].

   • The remaining term in the numerator is $3$ [1].

   • The remaining terms in the denominator are $2$ (from the reduction of $10$) and $(2x + 3)$ [1].

   • Final answer: $3 / (2(2x + 3))$ [1].

5. Identify Excluded Values (Points of Discontinuity): These are values of $x$ for which the original expression would be undefined because they would make any denominator zero [1]. It's crucial to consider all factors in all denominators before cancellation [1].

   • From $(x + 4)$: Set $x + 4 = 0$ $<br>ightarrow$ $x = -4$ [1].

   • From $(2x + 3)$: Set $2x + 3 = 0$ $<br>ightarrow$ $2x = -3$ $<br>ightarrow$ $x = -3/2$ [1].

   • From $x$ (in $10x$): Set $x = 0$ [1].

   • From $(x - 3)$: Set $x - 3 = 0$ $<br>ightarrow$ $x = 3$ [1].

   • From $(x - 4)$: Set $x - 4 = 0$ $<br>ightarrow$ $x = 4$ [1].

   • Excluded Values: $x$ cannot equal $-4$, $4$, $0$, $3$, or $-3/2$. These are the points of discontinuity for the original expression [1].

Example 3: Multiplying (3x^3 - 24) / (2x^2 - 14x + 20) by (4x^3 - 20x^2 + 3x - 15) / (3x^2 + 6x + 12)

1. Factor each component completely: This example uses more advanced factoring techniques [1].

   • Numerator of the first fraction 3x^3 - 24: Binomial.

     • GCF is $3$ [1].

     • Factor out the GCF: $3(x^3 - 8)$ [1].

     • Factor $(x^3 - 8)$ using the difference of perfect cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$ [1].

       • Identify $a$ and $b$: If $a^3 = x^3$, then $a = x$. If $b^3 = 8$, then $b = 2$ (the cube root of $8$) [1].

       • Substitute $a=x$ and $b=2$ into the formula:

         • First binomial part: $(x - 2)$ [1].

         • Second trinomial part: $(x^2 + (x imes 2) + 2^2)$ which simplifies to $(x^2 + 2x + 4)$ [1].

     • The fully factored numerator is $3(x - 2)(x^2 + 2x + 4)$ [1].

   • Denominator of the first fraction 2x^2 - 14x + 20: Trinomial.

     • GCF is $2$ (all coefficients are even) [1].

     • Factor out the GCF: $2(x^2 - 7x + 10)$ [1].

     • Factor the trinomial $(x^2 - 7x + 10)$ (leading coefficient of one) [1].

       • Find two numbers that multiply to $10$ and add to $-7$ [1].

       • These numbers are $-2$ and $-5$ ($-2 imes -5 = 10$ and $-2 + -5 = -7$) [1].

       • So, $(x^2 - 7x + 10)$ becomes $(x - 5)(x - 2)$ [1].

     • The fully factored denominator is $2(x - 5)(x - 2)$ [1].

   • Numerator of the second fraction 4x^3 - 20x^2 + 3x - 15: This is a four-term polynomial, which strongly suggests factoring by grouping [1].

     • Check for grouping suitability: A quick check involves seeing if the ratio of the first two coefficients is the same as the ratio of the last two coefficients. Here, $-20 / 4 = -5$ and $-15 / 3 = -5$. Since they match, factoring by grouping is likely to work [1].

     • Group the first two terms and the last two terms: $(4x^3 - 20x^2) + (3x - 15)$ [1].

     • From the first group $(4x^3 - 20x^2)$, GCF is $4x^2$, leaving $4x^2(x - 5)$ [1].

     • From the second group $(3x - 15)$, GCF is $3$, leaving $3(x - 5)$ [1].

     • The expression becomes $4x^2(x - 5) + 3(x - 5)$ [1].

     • Factor out the common binomial $(x - 5)$: $(x - 5)(4x^2 + 3)$ [1].

     • The fully factored numerator is $(x - 5)(4x^2 + 3)$ [1].

   • Denominator of the second fraction 3x^2 + 6x + 12: Trinomial.

     • GCF is $3$ (all coefficients are divisible by $3$) [1].

     • Factor out the GCF: $3(x^2 + 2x + 4)$ [1].

     • The trinomial $(x^2 + 2x + 4)$ cannot be factored further over real numbers. This type of trinomial often appears as part of the difference or sum of cubes formula and is considered an irreducible quadratic factor [1].

2. Assemble the completely factored expression: Create the multiplication problem with all factored polynomials [1].

   • $[3(x - 2)(x^2 + 2x + 4) / (2(x - 5)(x - 2))] imes [(x - 5)(4x^2 + 3) / (3(x^2 + 2x + 4))]$ [1].

3. Cancel common factors in the numerator and denominator: Carefully identify and cancel [1].

   • Cancel $(x - 2)$ [1].

   • Cancel $(x^2 + 2x + 4)$ [1].

   • Cancel $(x - 5)$ [1].

   • Cancel the constant factor $3$ [1].

4. Write the final simplified answer: Combine the remaining terms [1].

   • The remaining term in the numerator is $(4x^2 + 3)$ [1].

   • The remaining term in the denominator is $2$ [1].

   • Final answer: $(4x^2 + 3) / 2$ [1].

Key Factoring Techniques Used

• Greatest Common Factor (GCF): Always the first step in factoring any polynomial. This involves finding the largest monomial that divides each term in the polynomial and factoring it out [1].

• Difference of Perfect Squares: Used for binomials in the form $a^2 - b^2$, which factors into $(a + b)(a - b)$ [1]. To apply, ensure both terms are perfect squares and are separated by a subtraction sign.

• Trinomials (leading coefficient of one): For trinomials like $x^2 + bx + c$, find two numbers that multiply to the constant term ($c$) and add to the middle coefficient ($b$) [1]. The factors will be $(x + ext{number }1)(x + ext{number }2)$.

• Trinomials (leading coefficient not equal to one): For trinomials like $ax^2 + bx + c$ where $a e 1$, use the factor by grouping method [1]. This involves:

  1. Multiplying $a$ by $c$.

  2. Finding two numbers that multiply to this product $(ac)$ and add to the middle term's coefficient ($b$).

  3. Replacing the middle term ($bx$) with two new terms using these numbers as coefficients.

  4. Grouping the resulting four terms and factoring out common factors to arrive at the final binomial factors.

• Difference of Perfect Cubes: Used for binomials in the form $a^3 - b^3$, which factors into $(a - b)(a^2 + ab + b^2)$ [1]. Recognize perfect cubes like $x^3, 8, 27, 64$, etc.

• Factoring by Grouping (four-term polynomials): For polynomials with four terms, group the first two terms and the last two terms. Find the GCF of each group. If a common binomial factor appears, factor it out [1]. This method is often indicated when the ratio of coefficients of the first two terms matches the ratio of coefficients of the last two terms.

• Irreducible Trinomials: Some trinomials, especially those of the form $x^2 + 2x + 4$ (and similar forms derived from sum/difference of cubes), may not factor further over real numbers [1]. However, they can still be canceled if they appear identically in both the numerator and denominator of a rational expression.

Identifying Excluded Values (Points of Discontinuity)

• Rule: The denominator of a rational expression cannot be zero because division by zero is undefined [1].

• To find excluded values, you must set every unique factor in all denominators of the original expression (before any cancellation) equal to zero and solve for $x$ [1