# Algebraic Fractions: Quadratic and Linear Made Easy 3

## Simplifying Algebraic Fractions involving Quadratic and Linear Terms: Difference of Squares

As an experienced mathematics tutor, I have noticed that many students struggle with simplifying algebraic fractions, particularly those involving quadratic and linear terms. One of the most common types of quadratic expressions encountered in algebraic fractions is the difference of squares. In this article, we will explore the concept of the difference of squares and learn how to apply it to simplify algebraic fractions containing both quadratic and linear terms.

## Understanding the Difference of Squares

The difference of squares is a special type of quadratic expression that can be factored using a specific formula. It takes the form $a^2-b^2$, where $a$ and $b$ are any algebraic expressions. The factored form of the difference of squares is $(a + b)(a-b)$.

For example, let’s consider the expression $x^2-9$. We can rewrite this as the difference of squares by expressing 9 as $3^2$:

$\displaystyle x^2-9 = x^2-3^2$

Now, we can factor the expression using the difference of squares formula:

$\displaystyle x^2-3^2 = (x + 3)(x-3)$

## Simplifying Algebraic Fractions using the Difference of Squares

When simplifying algebraic fractions involving quadratic and linear terms, it’s essential to look for opportunities to apply the difference of squares formula. By factoring the numerator and denominator using this technique, we can often simplify the fraction further by cancelling common factors.

### Step 1: Identify the difference of squares

To begin, we must identify any instances of the difference of squares in the numerator and denominator of the algebraic fraction. Look for expressions in the form $a^2-b^2$.

Let’s consider an example:

$\displaystyle \frac{x^2-25}{x-5}$

In this case, the numerator is a difference of squares, as it can be rewritten as $x^2-5^2$.

### Step 2: Factor the difference of squares

Once we have identified the difference of squares, we can factor it using the formula $(a + b)(a – b)$.

In our example, we can factor the numerator as follows:

$\displaystyle x^2-25 = x^2-5^2 = (x + 5)(x-5)$

The denominator is already in its simplest form, so no further factoring is needed.

### Step 3: Cancel common factors

After factoring the numerator and denominator, we can cancel any common factors. This is similar to simplifying numerical fractions.

In our example, we can cancel the common factor $(x-5)$:

$\displaystyle \frac{(x+5)(x-5)}{x-5} = x+5$

Thus, the simplified fraction is simply $x + 5$.

## More Examples

Let’s work through a few more examples to reinforce the concept of simplifying algebraic fractions using the difference of squares.

### Example 1

$\displaystyle \frac{9x^2-16}{3x + 4}$

First, let’s identify the difference of squares in the numerator:

$\displaystyle 9x^2-16 = (3x)^2-4^2$

Now, we can factor the numerator using the difference of squares formula:

$\displaystyle (3x)^2-4^2 = (3x + 4)(3x-4)$

The denominator is already in its simplest form. However, we can now cancel the common factor $(3x + 4)$:

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

### Example 2

$\displaystyle \frac{x^2-y^2}{x-y}$

In this example, both the numerator and denominator are in the form of a difference of squares. Let’s factor them:

- Numerator: $\displaystyle x^2-y^2 = (x + y)(x-y)$
- Denominator: $\displaystyle x-y$

Now, we can cancel the common factor $(x-y)$:

$\displaystyle \frac{(x + y)(x-y)}{x-y} = x + y$

### Example 3

$\displaystyle \frac{4a^2-9b^2}{2a + 3b}$

First, let’s factor the numerator using the difference of squares formula:

$\displaystyle 4a^2-9b^2 = (2a)^2-(3b)^2 = (2a + 3b)(2a-3b)$

The denominator is already in its simplest form. However, we can now cancel the common factor $(2a + 3b)$:

$\displaystyle \frac{(2a+3b)(2a-3b)}{2a + 3b} = 2a-3b$

## Practice Problems

To reinforce your understanding of simplifying algebraic fractions using the difference of squares, try solving these practice problems:

- $\displaystyle \frac{x^2-81}{x-9}$
- $\displaystyle \frac{25x^2-36}{5x + 6}$
- $\displaystyle \frac{a^2-16b^2}{a + 4b}$
- $\displaystyle \frac{x^2-1}{x^2 + 2x + 1}$

Solutions:

- $x + 9$
- $5x-6$
- $a-4b$
- $\displaystyle \frac{x-1}{x + 1}$

## Combining the Difference of Squares with Other Techniques

In some cases, you may need to combine the difference of squares technique with other factoring methods to simplify algebraic fractions fully. For example, you might encounter a fraction where the numerator is a difference of squares, but the denominator requires factoring by grouping or the sum/difference of cubes formula.

Let’s consider an example:

$\displaystyle \frac{x^2-64}{x^3-8}$

In this case, the numerator is a difference of squares, and the denominator is a difference of cubes. We can factor them as follows:

- Numerator: $\displaystyle x^2-4 = (x + 2)(x-2)$
- Denominator: $\displaystyle x^3-8 = (x-2)(x^2 + 2x + 4)$

Now, we can simplify the fraction by dividing both the numerator and denominator by their common factor $(x-8)$:

$\displaystyle \frac{(x + 2)(x-2)}{(x-2)(x^2+2x+4)} = \frac{x + 2}{x^2+2x+4}$

The resulting fraction cannot be simplified further, so this is our final answer.

## Conclusion

Simplifying algebraic fractions involving quadratic and linear terms can be made much easier by mastering the difference of squares formula. By identifying instances of $a^2-b^2$ in the numerator and denominator, factoring them using the formula $(a + b)(a-b)$, and cancelling common factors, you can efficiently simplify even the most complex fractions.

Remember that the difference of squares technique is just one of many factoring methods you can use to simplify algebraic fractions. In some cases, you may need to combine it with other techniques, such as factoring by grouping or the sum/difference of cubes formula, to achieve the simplest possible form.

With practice and perseverance, you’ll soon be able to confidently tackle any algebraic fraction problem that comes your way.

## Transcript

It’s a square number minus the square number because you know that 16 is 4 squared, so it’s going to be, let’s get, as I said, 4 squared. So square minus square, how do we factorize that? Everyone should know it will be x plus 4, x minus 4, okay? Or the other way around, it doesn’t matter, okay? So you can see that x minus 4 is common, cross them out. So we have x plus 4 left, okay? So here, we don’t need to use the cross method because you should all remember how to factorise the difference between two squares. Okay?

Question 9. Okay! Let’s go ahead and do this. First of all, numerator, I can’t do much, but denominator, you can see that 27 and 3. 3 is a common factor but what I’m going to do is swap these around because I want my x square to come first because I always don’t know that’s the way I like to put it. I always want the x squared to come first. I always want it in order x squared x and then the constant. So that’s why I switch things around, and as I said, guys, 3 is a common factor but remember, I don’t want to have a negative in front of my x squared.

So I’m going to factorize by negative 3. So that becomes a positive x squared, which switches to a negative, so negative 9, okay? And now, guys, look at the brackets; look inside the brackets. We’ve got x squared minus 9, a difference of two squares, so 9 is 3 squared. So the difference of two squared squares minus squares. So how do we factorize that? All of you should tell me, x plus 3, x minus 3, okay? So I can cross these out, yeah? So we have that left; 1 make sure you put 1 on the numerator, 1 over negative 3x plus 3. Okay?

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