Lesson: February 5, 2025. Quiz: February 13, 2025.

  • Understand that rational expressions form a system analogous to the system of rational numbers and use that understanding to add and subtract rational expressions.

Assignment

The LCM of Polynomials

The least common multiple of two numbers involved each numbers prime factors. In fact, the LCM is the product of multiplying the highest power of each prime number together. To find the LCM of 12 and 18 we need prime factorization of each.

\[\begin{align} 12 &=2^2 \cdot 3^1\\ 18 &= 2^1 \cdot 3^2 \\[1em] \text{lcm}(12,18) &= 2^2 \cdot 3^2 \\ &= 36 \end{align}\]

The LCM of polynomials is similar, except your prime factors are just the result of the factoring you have been doing since the beginning. Let’s find the LCM of $x^2+4x+4$ and $x^2+5x+6$.

\[\begin{align} x^2+4x+4 &= (x+2)^2\\ x^2+5x+6 &= (x+2)(x+3) \\[1em] \text{LCM: }& (x+2)^2(x+3) \end{align}\]

Now we can go about adding rational expressions with unlike denominators.

Adding Rational Expressions With Unlike Denominators

The LCM will help us identify what is missing from each denominator. Once we determine what additional factors are needed to hit the LCM, we’ll multiply each fraction by what it needs.

\[\begin{align} \frac{x+3}{x^2-1} + \frac{2}{x^2-3x+2} &= \frac{x+3}{(x-1)(x+1)} + \frac{2}{(x-2)(x-1)} \\[1em] &= \frac{x+3}{(x-1)(x+1)} \cdot {\color{red}\frac{x-2}{x-2}} + \frac{2}{(x-2)(x-1)} \cdot {\color{red}\frac{x+1}{x+1}} \\[1em] \end{align}\]

From here, I recommend leaving your denominator factored and only simplify your numerator. Then you can refactor and divide out what you can.

\[\begin{align} \frac{x+3}{(x-1)(x+1)} \cdot {\frac{x-2}{x-2}} + \frac{2}{(x-2)(x-1)} \cdot {\frac{x+1}{x+1}} &= \frac{(x^2+x-6)+(2x+2))}{(x-1)(x+1)(x-2)} \\[1em] &= \frac{x^2+3x-4}{(x-1)(x+1)(x-2)} \\[1em] &= \frac{(x+4)(x-1)}{(x-1)(x+1)(x-2)} \\[1em] &= \frac{x+4}{(x+1)(x-2)} \end{align}\]

Keep in mind your original domain from the start of the problem. Despite appearances, $x\neq\{2,\pm1\}$.

Simplifying Compound Fractions

Remember that division is just multiplication of the inverse.

\[\begin{align} \frac{\frac{1}{x}+\frac{2}{x+1}}{\frac{1}{y}} &= \left(\frac{1}{x}+\frac{2}{x+1}\right)\left(\frac{y}{1}\right) \\[1em] \end{align}\]

From here, I recommend working with the two rational expressions in the group on the left, then distributing the $y$ after the fact.

\[\begin{align} \left(\frac{1}{x}+\frac{2}{x+1}\right)\left(\frac{y}{1}\right) &= \left(\frac{(x+1)+2x}{x(x+1)}\right)(y) \\[1em] &= \frac{y(3x+1)}{x(x+1)} \end{align}\]

For $(15)$, adding those two fractions can be done quickly if you remember that the quickest way to a common denominator is to multiply them by each other.

\[\begin{align} \frac{a}{b}+\frac{c}{d} = \frac{ad+bc}{ad} \end{align}\]

Multiplying both numerator and denominator by the other denominator will get adding quickly, but sometimes leading to bigger headaches. If we tried this with what we did in $(8)$, we would end up with a cubic and make our factoring and simplifying more difficult.