Dividing Expressions (Junior Cert Mathematics): Revision Notes

Dividing Expressions

What Does It Mean to Divide Expressions?

When you divide expressions in algebra, you are simplifying a more complicated expression into a simpler one.
We can divide expressions in two main ways:

1. Direct division
2. Division by factorising first

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Key Concepts to Remember

• Coefficients: The numbers in front of variables (e.g., in $3x$, the coefficient is $3$). Divide these as you would ordinary numbers.
• Variables: Letters such as $x$ or $y$ represent unknowns. When dividing the same variable, subtract the powers (exponents).

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The Rule for Dividing Variables

If you have a term like $\dfrac{x^m}{x^n}$ (with the same base $x$), subtract the exponent in the denominator from the exponent in the numerator:

$$\frac{x^m}{x^n} = x^{m-n}$$

Example: $\dfrac{x^5}{x^2} = x^{5-2} = x^3$.

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Dividing Expressions by Direct Division

Start by dividing the coefficients, then divide the variables (subtracting exponents when the base is the same).

Example 1 — Simplifying a basic expression
Problem: Simplify $\dfrac{12x^3}{4x}$.

Step 1 (coefficients): $\dfrac{12}{4} = 3$
Step 2 (variables): $\dfrac{x^3}{x} = x^{3-1} = x^2$
Final answer: $3x^2$

Explanation: Divide the numbers, then subtract the exponents of $x$.

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Example 2 — Dividing with multiple variables
Problem: Simplify $\dfrac{18y^4z^2}{6y^2z}$.

Step 1 (coefficients): $\dfrac{18}{6} = 3$
Step 2 ($y$ variables): $\dfrac{y^4}{y^2} = y^{4-2} = y^2$
Step 3 ($z$ variables): $\dfrac{z^2}{z} = z^{2-1} = z$
Final answer: $3y^2z$

Explanation: Divide the coefficients, then simplify each variable by subtracting exponents.

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Example 3 — Dividing a polynomial by a monomial
Problem: Simplify $\dfrac{15x^3 + 10x^2 - 5x}{5x}$.

Divide each term by $5x$:

• $\dfrac{15x^3}{5x} = 3x^2$
• $\dfrac{10x^2}{5x} = 2x$
• $\dfrac{-5x}{5x} = -1$ (here we use $x/x = 1$)

Final answer: $3x^2 + 2x - 1$

Why this matters: In exam questions, you must divide every term in the numerator by the monomial in the denominator.

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Dividing Expressions by Factorising

Sometimes it’s easier to factorise first and then cancel common factors.

In Junior Cycle Ordinary Level, useful factorising methods include:

• Taking out the HCF (highest common factor)
• Grouping
• Quadratic factorisation (where possible)
• Difference of two squares

These are covered in detail in the Factorising chapter.

Steps for Dividing by Factorising

1. Factorise the numerator.
2. Factorise the denominator (if possible).
3. Cancel common factors from top and bottom.
4. Simplify what remains.

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Example 1 — Factorise and divide
Problem: Simplify $\dfrac{6x^2 - 18x}{3x}$.

Step 1 (factorise numerator):
$6x^2 - 18x = 6x(x - 3)$

Step 2 (denominator): $3x$ is already simplest.

Step 3 (cancel common factors):
$\dfrac{6x(x - 3)}{3x} = 2(x - 3)$

Final answer: $2(x - 3)$

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Example 2 — Quadratic over a binomial
Problem: Simplify $\dfrac{x^2 + 5x + 6}{x + 2}$.

Step 1 (factorise numerator):
$x^2 + 5x + 6 = (x + 2)(x + 3)$

Step 2 (cancel common factor):
$\dfrac{(x + 2)(x + 3)}{x + 2} = x + 3$

Final answer: $x + 3$

Why you can cancel: A factor divided by the same factor equals $1$, so it disappears from the fraction.

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Practice Problems: Dividing Expressions

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Solutions

Solution to Problem 1
$\dfrac{15x^4}{5x^2} = 3x^{4-2} = 3x^2$

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Solution to Problem 2
$\dfrac{24y^5z^3}{6y^2z} = 4y^{5-2}z^{3-1} = 4y^3z^2$

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Solution to Problem 3
$\dfrac{10x^3 - 20x^2 + 30x}{5x} = \dfrac{10x^3}{5x} - \dfrac{20x^2}{5x} + \dfrac{30x}{5x} = 2x^2 - 4x + 6$

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Solution to Problem 4
$\dfrac{-18x^3y^2}{6xy} = -3x^{3-1}y^{2-1} = -3x^2y$

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Solution to Problem 5
$\dfrac{x^2 + 7x + 12}{x + 3} = \dfrac{(x + 3)(x + 4)}{x + 3} = x + 4$

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