Rational expressions — simplification, operations: Visual walkthrough

Before any symbol appears, let us agree on words.

Step 1 — See the expression as a picture of a fraction

WHAT. We start with the raw object $\frac{x ^{2} - 4}{x ^{2} + 5 x + 6}$ and treat it exactly like a number fraction such as $\frac{6}{8}$ .

WHY. A number fraction $\frac{6}{8}$ simplifies to $\frac{3}{4}$ because both top and bottom secretly share a factor of $2$ : $\frac{2 \cdot 3}{2 \cdot 4}$ . We want the same move here — but the shared piece is hidden inside the polynomials. So our job is to uncover it.

PICTURE. On the left, the number fraction with its hidden $2$ glowing. On the right, our polynomial fraction with a "?" where the hidden shared piece should be. Same idea, symbols swapped.

Figure — Rational expressions — simplification, operations

$top polynomial P (x) x^{2} - 4 / bottom polynomial Q (x) x^{2} + 5 x + 6$

Here $P (x)$ is just a name for the top, $Q (x)$ a name for the bottom. Nothing more mysterious than "top" and "bottom."

Step 2 — Break the top into factors

WHAT. Rewrite $x^{2} - 4$ as a product of two factors: $x^{2} - 4 = factor (x - 2) factor (x + 2)$

WHY. Canceling only works on multiplied pieces (factors). Right now $x^{2} - 4$ is a difference of two terms; we cannot grab anything to cancel. Factoring re-dresses the same quantity as a product so cancellable pieces appear. We use the difference of squares pattern $a^{2} - b^{2} = (a - b) (a + b)$ because $x^{2}$ is $x$ squared and $4$ is $2$ squared — a perfect match. (More on this in Polynomial factoring.)

PICTURE. A rectangle of area $x^{2} - 4$ is cut and rearranged into a clean $(x - 2)$ -by- $(x + 2)$ rectangle: same area, now shown as width $\times$ height, i.e. as factors.

Check by expanding: $(x - 2) (x + 2) = x^{2} + 2 x - 2 x - 4 = x^{2} - 4.$ ✓ The middle terms cancel, which is why squares leave no linear term.

Step 3 — Break the bottom into factors

WHAT. Rewrite the bottom: $x^{2} + 5 x + 6 = factor (x + 2) factor (x + 3)$

WHY. Same reason as Step 2 — we need multiplied pieces before we can cancel. To factor $x^{2} + 5 x + 6$ we hunt two numbers that multiply to $6$ (the constant) and add to $5$ (the middle coefficient). Those are $2$ and $3$ , so the factors are $(x + 2)$ and $(x + 3)$ .

PICTURE. A search grid: rows of number-pairs multiplying to $6$ , with the pair $(2, 3)$ highlighted because its sum lands on $5$ . The winning pair drops down into the two factors.

Now the whole fraction wears its factored clothes: $\frac{( x - 2 ) ( x + 2 )}{( x + 2 ) ( x + 3 )}$

Step 4 — Freeze the restrictions BEFORE touching anything

WHAT. Look at the factored bottom and write down every value of $x$ that makes it zero: $x + 2 = 0 \Rightarrow x = - 2, x + 3 = 0 \Rightarrow x = - 3.$ So $x \neq = - 2$ and $x \neq = - 3$ .

WHY. Dividing by zero is forbidden. The original bottom is $(x + 2) (x + 3)$ ; both these values kill it. We record them now, because in the next step we are about to delete the $(x + 2)$ — and if we wait, we will forget that $x = - 2$ was ever dangerous. This is exactly Domain and range thinking: the domain is decided by the original object, not the tidied one.

PICTURE. The number line with two chalk-drawn open holes punched at $x = - 2$ and $x = - 3$ — pinned in before any cancelling, like flags we plant so we don't lose the spots.

Step 5 — Cancel the shared factor

WHAT. Both top and bottom contain the identical factor $(x + 2)$ . Remove it from each: $\frac{( x - 2 ) ( x + 2 )}{( x + 2 ) ( x + 3 )} = \frac{x - 2}{x + 3} .$

WHY. The legal rule is $\frac{A \cdot C}{B \cdot C} = \frac{A}{B}$ whenever $C \neq = 0$ . Here $C = (x + 2)$ , and it is multiplied on both top and bottom, so it may go. Anywhere $x + 2 \neq = 0$ , the ratio $\frac{x + 2}{x + 2}$ equals $1$ , and multiplying by $1$ changes nothing.

PICTURE. The $(x + 2)$ tile on top and the $(x + 2)$ tile on bottom glow, then lift off together — like lifting one identical card from each hand. What remains is $\frac{x - 2}{x + 3}$ .

Step 6 — The hole that survives (the degenerate case)

WHAT. The final expression is $\frac{x - 2}{x + 3}, x \neq = - 2, x \neq = - 3.$ Notice: $\frac{x - 2}{x + 3}$ by itself is perfectly happy at $x = - 2$ (it gives $\frac{- 4}{1} = - 4$ ). But the original was undefined there. So we keep $x \neq = - 2$ as a permanent scar.

WHY. Simplifying is rewriting the same function, not replacing it with a nicer one. At $x = - 2$ the original had $\frac{0}{0}$ — genuinely undefined. Cancelling healed the algebra but not the domain. The point $x = - 2$ becomes a removable discontinuity, a single missing dot — a hole — which the study of Limits and continuity examines closely.

PICTURE. The graph of $\frac{x - 2}{x + 3}$ drawn smoothly, with one open circle at $x = - 2$ (the removable hole from the cancelled factor) and a vertical dashed asymptote at $x = - 3$ (the factor that survived in the bottom). Two different kinds of "forbidden," side by side.

Value	What happens in the ORIGINAL	Shows up as
$x = - 2$	cancelled factor $\to$ $\frac{0}{0}$	a hole (dot missing)
$x = - 3$	surviving bottom factor $\to \frac{nonzero}{0}$	a vertical asymptote

The one-picture summary

One board, top to bottom: raw fraction $\to$ factor both $\to$ plant the two hole-flags $\to$ lift the shared $(x + 2)$ $\to$ land on $\frac{x - 2}{x + 3}$ with both restrictions still attached, and the graph showing hole-at- $- 2$ , wall-at- $- 3$ .

Recall Feynman retelling — say it like a story

We had a fraction of polynomials, just a fancy $\frac{6}{8}$ . To shrink a fraction you hunt for a piece hiding in both top and bottom — but you can only pull out pieces that are multiplied, so first we re-wrote top and bottom as products (that's factoring). Before deleting anything, we walked to the bottom and wrote down every $x$ that would make it zero — those spots are forbidden forever, because you can never divide by zero, and if we don't note them now we'll forget them once we start deleting. Then we spotted the same $(x + 2)$ block on top and bottom and lifted it off both — allowed, because $\frac{x + 2}{x + 2}$ is just $1$ wherever it isn't zero. Out popped $\frac{x - 2}{x + 3}$ . But the value $x = - 2$ still can't be used, even though the new form looks fine there — because the original choked on it. That leftover forbidden point is a tiny hole in the graph, while the never-cancelled $x = - 3$ makes a full wall the curve races toward but never touches.

Recall Quick self-test

Why factor before cancelling? ::: Cancelling only works on multiplied factors; factoring turns sums into products so shareable factors appear. Why note restrictions before cancelling? ::: Once you delete a bottom factor you can no longer see the value it forbade; the original domain must be preserved. After simplifying to $\frac{x - 2}{x + 3}$ , what are the restrictions? ::: $x \neq = - 2$ (removable hole) and $x \neq = - 3$ (asymptote). What is the difference between $x = - 2$ and $x = - 3$ on the graph? ::: $x = - 2$ is a hole (cancelled factor, $\frac{0}{0}$ ); $x = - 3$ is a vertical asymptote (surviving bottom factor, nonzero $/0$ ).

Related deeper dives: Complex fractions, Rational equations, Polynomial long division, Partial fraction decomposition.