# Algebraic Fractions Made Easy: Simplify Like a Pro 2

## Algebraic Fractions Demystified: Simplify by Difference of Squares and Quadratic Factorise

As an experienced mathematics tutor, I have encountered numerous students who struggle with simplifying algebraic fractions, particularly those involving quadratic terms and the difference of squares. In this article, we will focus on these concepts and learn how to simplify algebraic fractions containing quadratic terms and the difference of squares step by step. By the end of this guide, you will have the tools and knowledge needed to tackle these problems confidently.

## Simplifying Algebraic Fractions with Quadratic Terms and Difference of Squares

Let’s learn how to simplify algebraic fractions containing quadratic terms and the difference of squares.

### Step 1: Factorise the numerator and denominator of each fraction

The first step in simplifying algebraic fractions is to factorise the numerator and denominator of each fraction. To do this, we need to find the factors of the quadratic terms and look for the difference in squares.

For quadratic expressions, we can use various factorisation techniques such as:

• Trial and error
• Grouping
• Decomposition
• Sum or difference of cubes

For the difference of squares, we use the formula: $\displaystyle a^2-b^2 = (a+b)(a-b)$

Let’s consider an example:

$\displaystyle \frac{x^3+3x^2}{x^2+9x+20} \times \frac{x^2-16}{x^2-x-12}$

To factorise the numerators and denominators, we can use the decomposition method for the quadratic terms and the difference of squares formula:

$\displaystyle x^3+3x^2 = x^2(x+3)$
$\displaystyle x^2+9x+20 = (x+4)(x+5)$
$\displaystyle x^2-16 = (x+4)(x-4)$ (difference of squares)
$\displaystyle x^2-x-12 = (x+3)(x-4)$

### Step 2: Identify common factors in each fraction

After factorising the numerators and denominators, we can identify any common factors within each fraction. These common factors can be numbers, variables, or even more complex expressions.

In our example, we can see that:

• In the first fraction, there are no common factors in the numerator and denominator.
• In the second fraction, $(x-4)$ is a common factor in both the numerator and denominator.

$\require{cancel} \displaystyle \frac{x^2(x+3)}{(x+4)(x+5)} \times \frac{(x+4)\cancel{(x-4)}}{(x+3)\cancel{(x-4)}}$

### Step 3: Divide out common factors in each fraction

To simplify each algebraic fraction, we divide both the numerator and denominator by their common factors.

In our example, we can divide out the common factor $(x-4)$ in the second fraction:

$\displaystyle \frac{x^2(x+3)}{(x+4)(x+5)} \times \frac{x+4}{x+3}$

### Step 4: Multiply the simplified fractions

After simplifying each fraction, we can multiply the resulting fractions together.

In our example, we get:

$\require{cancel} \displaystyle \frac{x^2(x+3)}{(x+4)(x+5)} \times \frac{x+4}{x+3} = \frac{x^2\cancel{(x+3)}}{\cancel{(x+4)}(x+5)} \times \frac{\cancel{(x+4)}}{\cancel{(x+3)}} = \frac{x^2}{x+5}$

Thus, the simplified result is $\displaystyle \frac{x^2}{x+5}$.

## More Examples

Let’s work through another example to reinforce the concept of simplifying algebraic fractions with quadratic terms and the difference of squares.

### Example 2

$\displaystyle \frac{x^2-9}{x^2+5x+4} \times \frac{x^2-1}{x^2+6x+9}$

First, let’s factorise the numerators and denominators:

• $\displaystyle x^2-9 = (x+3)(x-3)$ (difference of squares)
• $\displaystyle x^2+5x+4 = (x+1)(x+4)$
• $\displaystyle x^2-1 = (x+1)(x-1)$ (difference of squares)
• $\displaystyle x^2+6x+9 = (x+3)^2$

We can see that:

• In the first fraction, $(x+1)$ is a common factor in the denominator.
• In the second fraction, $(x+3)$ is a common factor in the denominator.

$\displaystyle \frac{(x+3)(x-3)}{(x+1)(x+4)} \times \frac{(x+1)(x-1)}{(x+3)(x+3)}$

Dividing out the common factor $(x+1)$ in the first fraction and $(x+3)$ in the second fraction, we get:

$\displaystyle \frac{\cancel{(x+3)}(x-3)}{\cancel{(x+1)}(x+4)} \times \frac{\cancel{(x+1)}(x-1)}{\cancel{(x+3)}(x+3)}$

Thus, the simplified result is $\displaystyle \frac{(x-3)(x-1)}{(x+4)(x+3)}$.

## Practice Problems

To reinforce your understanding of simplifying algebraic fractions with quadratic terms and the difference of squares, try solving these practice problems:

1. $\displaystyle \frac{x^2-4}{x^2+3x+2} \times \frac{x^2+7x+12}{x^2-36}$
2. $\displaystyle \frac{x^2-8x+16}{x^2+6x+9} \times \frac{x^2-9}{x^2-16}$
3. $\displaystyle \frac{x^2-1}{x^2+4x+3} \times \frac{x^2-9}{x^2+x-6}$
4. $\displaystyle \frac{x^2-64}{x^2+10x+25} \times \frac{x^2-4x+4}{x^2-81}$

Solutions:

1. $\displaystyle \frac{(x-2)(x+2)}{(x+1)(x+2)} \times \frac{(x+3)(x+4)}{(x-6)(x+6)} = \frac{(x-2)\cancel{(x+2)}}{(x+1)\cancel{(x+2)}} \times \frac{(x+3)(x+4)}{(x-6)(x+6)} = \frac{(x-2)(x+3)(x+4)}{(x+1)(x-6)(x+6)}$
2. $\require{cancel} \displaystyle \frac{(x-4)^2}{(x+3)^2} \times \frac{(x-3)(x+3)}{(x-4)(x+4)} = \frac{\cancel{(x-4)}^2}{\cancel{(x+3)}^2} \times \frac{(x-3)\cancel{(x+3)}}{\cancel{(x-4)}(x+4)} = \frac{(x-4)(x-3)}{(x+3)(x+4)}$
3. $\displaystyle \frac{(x-1)(x+1)}{(x+1)(x+3)} \times \frac{(x-3)(x+3)}{(x-2)(x+3)} = \frac{(x-1)\cancel{(x+1)}}{\cancel{(x+1)}(x+3)} \times \frac{(x-3)\cancel{(x+3)}}{(x-2)\cancel{(x+3)}} = \frac{(x-1)(x-3)}{(x+3)(x-2)}$
4. $\displaystyle \frac{(x-8)(x+8)}{(x+5)^2} \times \frac{(x-2)^2}{(x-9)(x+9)} = \frac{(x-8)(x+8)(x-2)^2}{(x+5)^2(x-9)(x+9)}$

## Conclusion

Simplifying algebraic fractions with quadratic terms and the difference of squares may seem daunting at first, but with practice and the right strategies, it becomes much easier. Remember to factorise the numerators and denominators, identify common factors, and divide them out to simplify each fraction. Then, multiply the simplified fractions together to obtain the final result. By mastering these techniques, you’ll be able to confidently tackle a wide range of algebraic fraction problems involving quadratic terms and the difference of squares.

• When factorising quadratic expressions, always look for common factors first. This can save you time and simplify the process.
• If you’re having trouble factorising a quadratic expression, try using the quadratic formula: $\displaystyle x = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a}$, where $a$, $b$, and $c$ are the coefficients of the quadratic expression in standard form $(ax^2 + bx + c)$.
• Remember that if the numerator and denominator have no common factors, the fraction is already in its simplest form.
• When simplifying algebraic fractions, be careful not to divide by zero. If you end up with a denominator that equals zero for a specific value of the variable, note that the fraction is undefined for that value.
• Always keep an eye out for the difference of squares, as it can often simplify the factorisation process and lead to more efficient simplification of the algebraic fractions.

By remembering these tips and practising regularly, you’ll soon be able to simplify algebraic fractions with quadratic terms and the difference of squares like a pro!

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