The Basics (Junior Cert Mathematics): Revision Notes

Dividing Expressions

What Does It Mean to Divide Expressions?

When you divide expressions in algebra, you're taking a complex expression and breaking it down into simpler parts.

We can divide expressions in two ways:

1. Dividing Expressions by Direct Division
2. Dividing Expressions by Factorising

Key Concepts to Remember

• Coefficients: These are the numbers in front of the variables, like the "3" in $3x$. You divide these numbers just like you would in regular arithmetic.
• Variables: The letters like $x$ or $y$ represent unknowns. When you divide variables that are the same, you subtract the powers (exponents).

The Rule for Dividing Variables

If you have a term like $\frac{x^m}{x^n}$, you subtract the exponent in the denominator from the exponent in the numerator: $\frac{x^m}{x^n} = x^{m-n}$ For example, $\frac{x^5}{x^2} = x^{5-2} = x^3$.

Dividing Expressions by Direct Division

Let's start with the most straightforward way to divide expressions: by directly dividing the coefficients and then the variables.

Example 1: Simplifying a Basic Expression

Example 2: Dividing Expressions with Multiple Variables

Example 3: Dividing a Polynomial by a Monomial

Dividing Expressions by Factorising

Sometimes, instead of directly dividing, it's easier to factorise the expression first. Instead of directly dividing terms, you can factorise the numerator (and sometimes the denominator), and then cancel out common factors. This approach can make the division easier and help avoid mistakes.

For the Junior Cycle Maths exam, we will learn how to use these methods of factorisation:

3. Taking out the HCF (Highest common factor)
4. Grouping
5. Quadratic
6. Difference of two squares These will be explained in greater detail in the 'Factorising' chapter.

Steps for Dividing by Factorising

1. Factorise the numerator: Break down the numerator (the expression on top) into its factors.
2. Factorise the denominator: If possible, factorise the denominator (the expression on the bottom) as well.
3. Cancel common factors: Simplify the expression by cancelling out any common factors in the numerator and denominator.
4. Simplify the remaining expression: After cancelling, simplify what's left.

What does factorising mean? Factorising a quadratic expression like $x^2 + 5x + 6$ means breaking it down into two simpler expressions that, when multiplied together, give you back the original quadratic.

This is similar to how you would break down the number 12 into $3 \times 4$ because $3 \times 4 = 12$.

How do we factorise $x^2 + 5x + 6$ ?

1. Identify the coefficients: In $x^2 + 5x + 6$, the coefficient of $x^2$ is 1, the coefficient of $x$ is 5, and the constant term (the number without an $x$) is 6.
2. Look for two numbers that multiply to give the constant term 6 and add up to give the coefficient of $(x)$$ (5).

• We need two numbers that multiply to 6 and add up to 5.
• Let's list the pairs of numbers that multiply to 6:
• 1 $times$ 6 $= 6$
• 2 $times$ 3 $= 6$
• Now, check which pair adds up to 5:
• $1 + 6 = 7$ (This doesn't work)
• $2 + 3 = 5$ (This works!)

3. Write the quadratic as a product of two binomials: Now that we have found the numbers 2 and 3, we can factorise the quadratic expression.

• The factorised form of$x^2 + 5x + 6$ is $(x + 2)(x + 3)$. Why does this work?

• If you expand$(x + 2)(x + 3)$, you should get back the original quadratic expression: $(x + 2)(x + 3) = x(x + 3) + 2(x + 3) = x^2 + 3x + 2x + 6 = x^2 + 5x + 6$

• As you can see, multiplying$(x + 2)(x + 3)$ gives us $x^2 + 5x + 6$, so our factorisation is correct.

---

Step 2: Factorise the Denominator

The denominator is $x + 2$, which is already in its simplest form. There's no need to factorise it further.

---

Step 3: Cancel Out Common Factors

Now that we have factorised the numerator, our expression looks like this:

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

Notice that both the numerator and the denominator have a common factor of x + 2.

• We can cancel out the $x + 2$ term from both the numerator and the denominator: $\frac{(x + 2)(x + 3)}{x + 2} = x + 3$ Why can we cancel?

• You can cancel out common factors in a fraction because dividing something by itself equals 1. In this case, $frac{x + 2}{x + 2} = 1$, so it effectively disappears from the expression.

Final Answer

After cancelling out the common factor, the simplified expression is: $x + 3$

Summary: In this example, we started by Factorising the quadratic expression $x^2 + 5x + 6$ into $(x + 2)(x + 3)$. Then, we cancelled out the common factor of $x + 2$ from both the numerator and the denominator to simplify the expression to $x + 3$.

This method of factorising before dividing helps to simplify expressions more easily, especially in cases like this where the denominator is a factor of the numerator.

:::

Practice Problems: Dividing Expressions

---

Solutions

Solution to Problem 1

Step 1: Divide the coefficients (numbers)

• The coefficients are 15 and 5.

• Divide them: $\frac{15}{5} = 3$. Step 2: Divide the variables (letters)

• You have $x^4$ divided by $x^2$.

• Subtract the exponents: $\frac{x^4}{x^2} = x^{4-2} = x^2$ Final Answer: $\frac{15x^4}{5x^2} = 3x^2$

Explanation: We divided the coefficient 15 by 5 to get 3, and then subtracted the exponents for the $x$ terms, resulting in $x^2$.

---

Solution to Problem 2

Step 1: Divide the coefficients

• The coefficients are 24 and 6.

• Divide them:$\frac{24}{6} = 4 .$ Step 2: Divide the $y$ variables

• You have$y^5$ divided by $y^2$.

• Subtract the exponents: $\frac{y^5}{y^2} = y^{5-2} = y^3$ Step 3: Divide the (z) variables

• You have $z^3$ divided by $z$.

• Subtract the exponents: $\frac{z^3}{z} = z^{3-1} = z^2$ Final Answer: $\frac{24y^5z^3}{6y^2z} = 4y^3z^2$

Explanation: We divided the coefficients, then handled each variable separately by subtracting the exponents.

---

Solution to Problem 3

Step 1: Divide each term in the polynomial by the monomial

We divide each term separately by $5x$:

• First term: $\frac{10x^3}{5x}$
  • Divide the coefficients: $\frac{10}{5} = 2$
  • Divide the variables: $\frac{x^3}{x} = x^{3-1} = x^2$
  • So,$\frac{10x^3}{5x} = 2x^2$
• Second term:$\frac{-20x^2}{5x}$
  • Divide the coefficients: $\frac{-20}{5} = -4$
  • Divide the variables: $\frac{x^2}{x} = x^{2-1} = x$
  • So, $\frac{-20x^2}{5x} = -4x$
• Third term: $\frac{30x}{5x}$
  • Divide the coefficients: $\frac{30}{5} = 6$
  • Divide the variables: $\frac{x}{x} = x^{1-1} = 1$
  • So, $\frac{30x}{5x} = 6$ Final Answer: $\frac{10x^3 - 20x^2 + 30x}{5x} = 2x^2 - 4x + 6$

Explanation: Each term in the polynomial was divided separately by $5x$, leading to the simplified expression.

---

Solution to Problem 4

Step 1: Divide the coefficients

• The coefficients are -18 and 6.
• Divide them: $\frac{-18}{6} = -3$.

Step 2: Divide the $x$ variables

• You have $x^3$ divided by $x$.

• Subtract the exponents: $\frac{x^3}{x} = x^{3-1} = x^2$ Step 3: Divide the $y$ variables

• You have $y^2$ divided by $y$.

• Subtract the exponents: $\frac{y^2}{y} = y^{2-1} = y$ Final Answer: $\frac{-18x^3y^2}{6xy} = -3x^2y$

Explanation: We divided the coefficients and then subtracted the exponents for both $x$ and $(y)$, leading to the final simplified expression.

---

Solution to Problem 5

Step 1: Factorise the numerator

The numerator $x^2 + 7x + 12$ is a quadratic expression. We can factorise it by finding two numbers that multiply to 12 (the constant term) and add to 7 (the coefficient of $x$.

• The numbers 3 and 4 work because:
  • $3 \times 4 = 12$ (they multiply to 12)
  • $3 + 4 = 7$ (they add to 7) So, we factorise the numerator: $x^2 + 7x + 12 = (x + 3)(x + 4)$

Step 2: Simplify by cancelling common factors

Now, substitute the factorised form into the expression: $\frac{(x + 3)(x + 4)}{x + 3}$ We can cancel out the common factor of x + 3 from the numerator and denominator: $= x + 4$

Final Answer: $\frac{x^2 + 7x + 12}{x + 3} = x + 4$

Explanation: We factorised the quadratic expression and then cancelled out the common factor, simplifying the expression to $x + 4$.

---

These practice problems should help you get more comfortable with dividing expressions in algebra. Remember, each step is important, and understanding how to break down and simplify these expressions will help you a lot in your Junior Cycle Maths exams. Keep practising, and don't hesitate to revisit these examples if you need to!

---