Simplifying Rational Expressions

➗ Simplifying Rational Expressions

Part 1 of 5 — What Is a Rational Expression?

Topics in This Part

🔑 Key Concept: A rational expression is just a fraction whose top and bottom are polynomials. Simplifying one works exactly like reducing $\dfrac{12}{18}$ to $\dfrac{2}{3}$ — you cancel what the top and bottom have in common. The whole skill comes down to one word: factor.

Rational Expressions Are Polynomial Fractions

A rational expression is a quotient $\dfrac{P}{Q}$ where $P$ and $Q$ are polynomials and $Q \ne 0$.

Reducing a numerical fraction relies on this fact:

$\frac{12}{18} = \frac{6\cdot 2}{6\cdot 3} = \frac{\cancel{6}\cdot 2}{\cancel{6}\cdot 3} = \frac{2}{3}$

You cancelled the common factor $6$. Rational expressions work the same way — but first you have to factor the top and bottom so the common pieces are visible.

🔑 Key Idea: To simplify $\dfrac{P}{Q}$, factor $P$ and $Q$ completely, then divide out any factor that appears in both.

Concept Check 🎯

The Golden Rule: Cancel Factors, Never Terms

You may only cancel something that is multiplied across the entire numerator and the entire denominator — a common factor.

$\frac{5\cdot(x+1)}{8\cdot(x+1)} = \frac{5}{8} \qquad \color{green}{\checkmark \;\text{factor cancelled}}$

You may never cancel a term — something connected by $+$ or $-$:

$\frac{x+1}{x+5} \ne \frac{1}{5} \qquad \color{red}{✗ \;\text{you cannot cancel the } x}$

The $x$ in $x+1$ is added to $1$, not multiplied by the whole top. There is no common factor here, so nothing cancels.

⚠️ The #1 mistake in this entire topic: cancelling across a $+$ or $-$ sign. If a piece isn't multiplying the whole numerator and the whole denominator, it stays put.

Legal or Illegal? 🎯

Decide whether each cancellation is allowed.

Where We're Headed

Almost every rational expression starts out looking like nothing cancels:

$\frac{x^2 - 9}{x^2 + 5x + 6}$

But once both polynomials are factored, common factors appear:

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

So the real skill is factoring. In Part 2 we sharpen the four factoring tools you'll use over and over, and in Part 3 we put them to work cancelling.

Warm-Up: Spot the Common Factor 🧮

Reduce each numerical fraction to lowest terms by cancelling the greatest common factor. Enter your answer as a fraction like 2/3.

1) $\dfrac{14}{21} = \,?$ 2) $\dfrac{15(x+1)}{25(x+1)} = \,?$ (cancel the $(x+1)$, then reduce the numbers)

➗ Simplifying Rational Expressions

Part 2 of 5 — The Factoring Toolkit

🔑 The Idea: You can't cancel a common factor until you can see one. These four factoring moves reveal the hidden factors in almost every Algebra 2 rational expression.

Four Factoring Moves

For the trinomial $x^2 + bx + c$, find two numbers that multiply to $c$ and add to $b$.

For $x^2 + 5x + 6$: which two numbers multiply to $6$ and add to $5$? → $2$ and $3$. So $(x+2)(x+3)$.

💡 Always check for a GCF first. Factoring $2x^2 - 8$ as $2(x-2)(x+2)$ exposes three possible cancelling factors instead of leaving it tangled.

Concept Check 🎯

Pick the Right Factoring 🔽

Match each expression to its factored form.

Factor the Trinomial 🧮

For each $x^2 + bx + c$, find the two numbers that multiply to $c$ and add to $b$. Enter them smaller first, separated by your two boxes.

1) $x^2 + 8x + 15$: the two numbers are $\,?$ and $\,?$ 2) $x^2 - 7x + 10$: the two numbers are $\,?$ and $\,?$

You Now Have the Tools

With GCF, difference of squares, and trinomial factoring in hand, you can break almost any Algebra 2 numerator or denominator into a product of simple factors.

In Part 3 we combine these moves with the Golden Rule: factor both parts, then cancel the common factors.

➗ Simplifying Rational Expressions

Part 3 of 5 — Factor, Then Cancel

🔑 The Method: (1) Factor the numerator completely. (2) Factor the denominator completely. (3) Divide out every factor common to both. (4) Write what's left.

Worked Example: $\dfrac{x^2 - 9}{x^2 + 5x + 6}$

Step 1 — Factor the top. Difference of squares: $x^2 - 9 = (x-3)(x+3)$

Step 2 — Factor the bottom. Multiply to $6$, add to $5$ → $2$ and $3$: $x^2 + 5x + 6 = (x+2)(x+3)$

Step 3 — Cancel the common factor $(x+3)$: $\frac{(x-3)\cancel{(x+3)}}{(x+2)\cancel{(x+3)}} = \frac{x-3}{x+2}$

Step 4 — Final answer: $\frac{x^2 - 9}{x^2 + 5x + 6} = \frac{x-3}{x+2}$

✅ Check with a number: let $x = 1$. Original: $\dfrac{1-9}{1+5+6} = \dfrac{-8}{12} = -\dfrac{2}{3}$. Simplified: $\dfrac{1-3}{1+2} = \dfrac{-2}{3}$. Same value ✓

Worked Example: $\dfrac{2x^2 - 8}{x^2 + 4x + 4}$

Top: GCF first, then difference of squares: $2x^2 - 8 = 2(x^2 - 4) = 2(x-2)(x+2)$

Bottom: perfect-square trinomial (multiply to $4$, add to $4$ → $2$ and $2$): $x^2 + 4x + 4 = (x+2)(x+2)$

Cancel one $(x+2)$: $\frac{2(x-2)\cancel{(x+2)}}{(x+2)\cancel{(x+2)}} = \frac{2(x-2)}{x+2}$

⚠️ Only one $(x+2)$ cancels — the top has a single factor of $(x+2)$, so only one pair can divide out. The leftover $(x+2)$ stays on the bottom.

Concept Check 🎯

Order the Steps 🔽

You're simplifying $\dfrac{x^2 + 2x}{x^2 - 4}$. Choose what happens at each stage.

Simplify and Evaluate 🧮

Simplify each expression, then evaluate the simplified form at the given value.

1) $\dfrac{x^2 - 1}{x + 1}$ simplifies to $x - 1$. Its value at $x = 5$ is $\,?$ 2) $\dfrac{x^2 + 6x + 8}{x + 2}$ simplifies to $x + 4$. Its value at $x = 3$ is $\,?$

➗ Simplifying Rational Expressions

Part 4 of 5 — Domain Restrictions & Opposite Factors

🔑 Two things that trip everyone up: (1) the excluded values that make the denominator zero — they survive even after you cancel, and (2) factors like $(2-x)$ and $(x-2)$ that look uncancellable but are actually opposites.

Excluded Values (Domain Restrictions)

A rational expression is undefined wherever its original denominator equals zero. Those $x$-values are excluded from the domain.

Example: $\dfrac{x^2 - 9}{x^2 + 5x + 6}$

Factor the denominator: $(x+2)(x+3)$. It is zero when $x = -2$ or $x = -3$.

So the expression is undefined at $x = -2$ and $x = -3$, even though the simplified form $\dfrac{x-3}{x+2}$ no longer shows the $(x+3)$.

⚠️ Find restrictions from the ORIGINAL denominator, before cancelling. A cancelled factor like $(x+3)$ still forbids its value — that's where a graph gets a hole.

Concept Check 🎯

The Opposite-Factor Trick

Factors like $(2 - x)$ and $(x - 2)$ aren't identical, so they don't cancel directly. But they are opposites:

$2 - x = -(x - 2)$

Factoring out $-1$ flips the order. That lets you create a matching factor and cancel — leaving a $-1$ behind.

Example: $\dfrac{x - 5}{5 - x}$

Rewrite the bottom: $5 - x = -(x - 5)$.

$\frac{x-5}{5-x} = \frac{x-5}{-(x-5)} = \frac{\cancel{(x-5)}}{-\,\cancel{(x-5)}} = -1$

💡 Shortcut: whenever a factor and its reverse appear (like $a-b$ over $b-a$), they cancel to $-1$. Don't expect a clean $1$ — the flip always costs a negative sign.

Opposites & Restrictions 🔽

List the Excluded Values 🧮

Find every value of $x$ that makes the denominator zero. Enter them smaller first.

1) $\dfrac{x+1}{(x-3)(x+6)}$: excluded values are $\,?$ and $\,?$ 2) $\dfrac{x^2}{x^2 - 49}$: excluded values are $\,?$ and $\,?$

➗ Simplifying Rational Expressions

Part 5 of 5 — Mixed Practice & Mastery Check

You can now (1) recognize rational expressions, (2) factor numerator and denominator, (3) cancel common factors, (4) handle opposite factors, and (5) state domain restrictions. Let's put it all together.

Quick Reference

⚠️ Remember: always pull a GCF first, find restrictions before cancelling, and never cancel a term across a $+$ or $-$ sign.

Mixed Practice 🎯

One More Simplification 🧮

Simplify $\dfrac{4x + 12}{x^2 - 9}$ to the form $\dfrac{A}{x - 3}$.

1) First factor the top: $4x+12 = 4(x + \,?)$ 2) The simplified expression is $\dfrac{A}{x-3}$. Enter $A = \,?$

Exit Quiz ✅

Answer all three to finish the lesson.