Simplifying Algebraic Fractions with Domain Restrictions

Hi again. Last lesson established the safety rule for algebraic fractions: a denominator can never equal $0$. You found domain restrictions — הגבלות תחום / ערכים אסורים — by solving for the values that make the original denominator zero.

Now we combine that rule with factoring. By the end of this lesson, you will be able to simplify an algebraic fraction — לפשט שבר אלגברי — by factoring its numerator and denominator, cancelling only genuine common factors — גורמים משותפים — and keeping every original restriction in the final answer.

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The big idea: simplify like a numerical fraction

You already know that:

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

We “cancel” the $6$ because it is a factor multiplying the numerator and a factor multiplying the denominator. More accurately, we divide the top and bottom by the same nonzero value:

$$\frac{2\cdot6}{3\cdot6}=\frac{2\cdot1}{3\cdot1}$$

Algebraic fractions work in the same way. For example:

$$\frac{5x^2}{10x}$$

Write each part as factors:

$$\frac{5\cdot x\cdot x}{5\cdot2\cdot x}$$

Now the factors $5$ and $x$ occur in both numerator and denominator:

$$\frac{\cancel{5}\cdot\cancel{x}\cdot x} {\cancel{5}\cdot2\cdot\cancel{x}} = \frac{x}{2}$$

But there is a condition. Cancelling $x$ means dividing by $x$, which only works when $x\ne0$. Also, the original denominator $10x$ equals zero at $x=0$.

So the complete answer is:

$$\boxed{\frac{5x^2}{10x}=\frac{x}{2},\qquad x\ne0}$$

The simplified fraction looks shorter, but it must keep the original restriction.

Key vocabulary

English	Hebrew	Meaning
simplify	לפשט	Write an equivalent expression in a simpler form
factor	לפרק לגורמים	Rewrite using multiplication
factor	גורם	A quantity being multiplied
common factor	גורם משותף	A factor in both top and bottom
cancel	לצמצם	Divide matching factors from top and bottom
original expression	הביטוי המקורי	The fraction before simplifying
restriction	ערך אסור / הגבלת תחום	An $x$-value not allowed
numerator	מונה	Top of the fraction
denominator	מכנה	Bottom of the fraction

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The reliable four-step method

Use this exact order on a test:

1. Find restrictions from the original denominator.
   Set the original denominator equal to $0$, then solve.

2. Factor the numerator completely.

3. Factor the denominator completely.

4. Cancel identical factors only, then write the simplified result with the original restrictions.

Here is the structure to copy into your notebook:

$$\text{Original denominator}=0$$ $$\text{Restrictions: }x\ne\ldots$$ $$\text{Factor top and bottom}$$ $$\text{Cancel common factors}$$ $$\boxed{\text{Simplified expression, with restrictions}}$$

This order prevents the most common error: cancelling a factor and accidentally forgetting the value that made the original fraction undefined.

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Watch factoring and legal cancellation

The following video gives two useful examples: one using a greatest common factor and one using factorable trinomials. It focuses on the factoring-and-cancelling part; apply the restriction rule from the previous lesson alongside it.

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Example 1: cancelling a greatest common factor

Simplify:

$$\frac{6x^2-18x}{9x^2}$$

Step 1: find restrictions first

The original denominator is $9x^2$.

$$9x^2=0$$ $$x^2=0$$ $$x=0$$

Therefore:

$$\boxed{x\ne0}$$

Step 2: factor completely

The numerator $6x^2-18x$ has a greatest common factor of $6x$:

$$6x^2-18x=6x(x-3)$$

The denominator is:

$$9x^2=9x\cdot x$$

So the fraction becomes:

$$\frac{6x(x-3)}{9x^2}$$

Reduce the numerical factors $6$ and $9$, and cancel one factor of $x$:

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

The final answer is:

$$\boxed{\frac{2(x-3)}{3x},\qquad x\ne0}$$

Notice that one $x$ remains in the denominator. We cancelled only one factor of $x$, because $x^2=x\cdot x$.

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Factors can cancel; terms cannot

This distinction is essential.

A factor is connected by multiplication:

$$(x+2)(x-3)$$

The two grouped expressions are factors.

A term is separated by addition or subtraction:

$$x+2$$

Here, $x$ and $2$ are terms, not factors.

For example, this is not allowed:

$$\frac{x+2}{x+5}$$

You cannot “cancel the $x$” because $x+2$ is not $x\cdot2$, and $x+5$ is not $x\cdot5$.

Likewise, this is not allowed:

$$\frac{x^2+4x}{x} \ne x+4$$

at least not yet. First factor the numerator:

$$x^2+4x=x(x+4)$$

Now cancellation is legal:

$$\frac{x(x+4)}{x}=x+4$$

But the original denominator was $x$, so the answer must say:

$$\boxed{x+4,\qquad x\ne0}$$

A useful test rule is:

If there is a $+$ or $-$ inside an expression, do not cancel anything inside it. Factor first.

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Example 2: factor two trinomials, then preserve all restrictions

Simplify:

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

Step 1: restrictions from the original denominator

Set the denominator equal to zero:

$$x^2+5x+6=0$$

Factor it. We need two numbers that multiply to $6$ and add to $5$: $2$ and $3$.

$$x^2+5x+6=(x+2)(x+3)$$

Therefore:

$$(x+2)(x+3)=0$$ $$x=-2$$

or

$$x=-3$$

So write the restrictions immediately:

$$\boxed{x\ne-2,\qquad x\ne-3}$$

Step 2: factor the numerator

For $x^2-x-6$, find two numbers that multiply to $-6$ and add to $-1$. They are $-3$ and $2$:

$$x^2-x-6=(x-3)(x+2)$$

Step 3: rewrite and cancel

$$\frac{x^2-x-6}{x^2+5x+6} = \frac{(x-3)(x+2)}{(x+2)(x+3)}$$

Now $x+2$ is an identical factor on top and bottom:

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

Step 4: keep both original restrictions

$$\boxed{\frac{x-3}{x+3},\qquad x\ne-2,\quad x\ne-3}$$

Why do we still write $x\ne-2$? After simplification, it may look as if $-2$ is allowed:

$$\frac{-2-3}{-2+3}=-5$$

But in the original expression, putting $x=-2$ gives:

$$\frac{(-2)^2-(-2)-6}{(-2)^2+5(-2)+6} = \frac{0}{0}$$

That is undefined. The cancelled factor hid the problem; it did not repair it.

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See the process visually

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The guide’s main message matches the method in this lesson: first factor, then cancel full common factors, and preserve restrictions from the original denominator.

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Example 3: a cancelled factor creates a hidden restriction

Simplify:

$$\frac{x^2+3x+2}{x^2-4}$$

Start with restrictions.

$$x^2-4=0$$

Factor the difference of squares:

$$(x-2)(x+2)=0$$

So:

$$\boxed{x\ne2,\qquad x\ne-2}$$

Now factor the numerator:

$$x^2+3x+2=(x+1)(x+2)$$

Write the complete factor form:

$$\frac{(x+1)(x+2)}{(x-2)(x+2)}$$

Cancel the common factor $x+2$:

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

Final answer:

$$\boxed{\frac{x+1}{x-2},\qquad x\ne2,\quad x\ne-2}$$

The value $x=2$ is still visibly forbidden because it makes the final denominator zero. The value $x=-2$ is no longer visible in the final denominator, but it remains forbidden because it made the original denominator zero.

For this course and test preparation, the safest habit is always to list all restrictions from the original denominator.

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Common mistakes to avoid

1. Cancelling before factoring

Incorrect:

$$\frac{x^2-9}{x-3}$$

You cannot cancel “the $x$” or “the $3$.”

Correct:

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

But the original denominator gives:

$$x-3=0$$ $$x=3$$

So the complete answer is:

$$\boxed{x+3,\qquad x\ne3}$$

2. Cancelling part of a sum

Incorrect:

$$\frac{x+4}{x+7}=\frac{4}{7}$$

This is false. The $x$'s are parts of sums, not factors.

3. Forgetting a restriction after cancellation

Incorrect:

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

This is simplified correctly, but incomplete. The original denominator is $x(x+1)$, so:

$$x\ne0$$

and

$$x\ne-1$$

The complete answer is:

$$\boxed{\frac{x-2}{x},\qquad x\ne0,\quad x\ne-1}$$

4. Treating a restriction as a solution

When you solve a denominator equation, such as

$$x+2=0$$

the answer $x=-2$ is not an answer to plug in. It is an excluded value:

$$x\ne-2$$

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A final check before submitting

After simplifying, take ten seconds to check:

• Did I find restrictions using the original denominator?
• Did I factor the top and bottom completely?
• Did I cancel only whole factors?
• Did I reduce numerical factors where possible?
• Did I write every original restriction beside my final answer?

If all five answers are yes, your work is likely correct.

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Key takeaways

To simplify an algebraic fraction:

1. Find the values that make the original denominator equal to $0$.
2. Record those values as restrictions.
3. Factor numerator and denominator completely.
4. Cancel identical factors, never separate terms.
5. State the simplified expression and keep every original restriction.

For example:

$$\frac{x^2-x-6}{x^2+5x+6} = \frac{(x-3)(x+2)}{(x+2)(x+3)} = \frac{x-3}{x+3}$$

with:

$$\boxed{x\ne-2,\qquad x\ne-3}$$

Next, you will use these same factoring skills to find a least common denominator — מכנה משותף מינימלי — and rewrite algebraic fractions as equivalent fractions.