Algebraic Fractions Made Easy: Simplify Like a Pro 1

Algebraic Fractions Demystified: Simplify Now with Monic Expressions by Quadratic Factorise

As an experienced mathematics tutor, I have encountered numerous students who struggle with simplifying algebraic fractions, particularly those involving quadratic terms. In this article, we will focus on non-monic expressions and learn how to simplify algebraic fractions containing quadratic terms 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 Non-Monic Quadratic Terms

Let’s learn how to simplify algebraic fractions containing non-monic quadratic terms.

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.

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

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

Let’s consider an example:

$\displaystyle \frac{x^2+6x+8}{x^2+5x+6} \times \frac{x^2+8x+15}{x^2+7x+12}$

To factorise the numerators and denominators, we can use the decomposition method:

$\displaystyle x^2+6x+8 = (x+4)(x+2)$
$\displaystyle x^2+5x+6 = (x+3)(x+2)$
$\displaystyle x^2+8x+15 = (x+5)(x+3)$
$\displaystyle x^2+7x+12 = (x+4)(x+3)$

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, $(x+2)$ is a common factor in both the numerator and denominator.
• In the second fraction, $(x+3)$ is a common factor in both the numerator and denominator.

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

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 factors $(x+2)$ in the first fraction and $(x+3)$ in the second fraction:

$\displaystyle \frac{x+4}{x+3} \times \frac{x+5}{x+4}$

Step 4: Multiply the simplified fractions

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

In our example, we get:

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

Thus, the simplified result is $\displaystyle \frac{x+5}{x+3}$.

More Examples

Let’s work through another example to reinforce the concept of simplifying algebraic fractions with non-monic quadratic terms.

Example 2

$\displaystyle \frac{2x^2+11x+15}{3x^2+14x+15} \times \frac{3x^2+19x+28}{2x^2+15x+27}$

First, let’s factorise the numerators and denominators:

• $\require{cancel} \displaystyle 2x^2+11x+15 = (2x+5)(x+3)$
• $\require{cancel} \displaystyle 3x^2+14x+15 = (3x+5)(x+3)$
• $\require{cancel} \displaystyle 3x^2+19x+28 = (3x+7)(x+4)$
• $\require{cancel} \displaystyle 2x^2+15x+27 = (2x+9)(x+3)$

We can see that:

• In the first fraction, $(x+3)$ is a common factor in both the numerator and denominator.
• In the second fraction, there are no common factors.

$\require{cancel} \displaystyle \frac{(2x+5)\cancel{(x+3)}}{(3x+5)\cancel{(x+3)}} \times \frac{(3x+7)(x+4)}{(2x+9)(x+3)}$

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

$\displaystyle \frac{2x+5}{3x+5} \times \frac{(3x+7)(x+4)}{(2x+9)(x+3)}$

Now, let’s multiply the simplified fractions:

$\displaystyle \frac{2x+5}{3x+5} \times \frac{(3x+7)(x+4)}{(2x+9)(x+3)} = \frac{(2x+5)(3x+7)(x+4)}{(3x+5)(2x+9)(x+3)}$

Thus, the simplified result is $\displaystyle \frac{(2x+5)(3x+7)(x+4)}{(3x+5)(2x+9)(x+3)}$.

Practice Problems

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

1. $\displaystyle \frac{2x^2+7x+6}{4x^2+8x+3} \times \frac{2x^2+9x+9}{x^2+6x+8}$
2. $\displaystyle \frac{3x^2+8x+4}{9x^2+12x+4} \times \frac{3x^2+13x+12}{2x^2+11x+12}$
3. $\displaystyle \frac{4x^2+4x-3}{2x^2+5x+3} \times \frac{2x^2+7x+6}{4x^2+4x-3}$
4. $\displaystyle \frac{5x^2+17x+12}{10x^2+19x+6} \times \frac{10x^2+31x+18}{15x^2+32x+12}$

Solutions:

1. $\displaystyle \frac{(2x+3)(x+2)}{(4x+3)(x+1)} \times \frac{(2x+3)(x+3)}{(x+4)(x+2)} = \frac{(2x+3)\cancel{(x+2)}}{(4x+3)(x+1)} \times \frac{\cancel{(2x+3)}(x+3)}{(x+4)\cancel{(x+2)}} = \frac{x+3}{(4x+3)(x+1)}$
2. $\displaystyle \frac{(3x+4)(x+1)}{(9x+4)(x+1)} \times \frac{(3x+4)(x+3)}{(2x+3)(x+4)} = \frac{\cancel{(3x+4)}(x+1)}{\cancel{(9x+4)}(x+1)} \times \frac{\cancel{(3x+4)}(x+3)}{\cancel{(2x+3)}(x+4)} = \frac{x+3}{x+4}$
3. $\displaystyle \frac{(4x-3)(x+1)}{(2x+3)(x+1)} \times \frac{(2x+3)(x+2)}{(4x-3)(x+1)} = \frac{\cancel{(4x-3)}(x+1)}{\cancel{(2x+3)}(x+1)} \times \frac{\cancel{(2x+3)}(x+2)}{\cancel{(4x-3)}(x+1)} = x+2$
4. $\displaystyle \frac{(5x+6)(x+2)}{(10x+6)(x+3)} \times \frac{(10x+18)(x+3)}{(15x+12)(x+2)} = \frac{\cancel{(5x+6)}(x+2)}{\cancel{(10x+6)}(x+3)} \times \frac{\cancel{(10x+18)}(x+3)}{\cancel{(15x+12)}(x+2)} = \frac{2}{3}$

Conclusion

Simplifying algebraic fractions with quadratic terms 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, whether they involve monic or non-monic quadratic terms.

• 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 the 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.

By keeping these tips in mind and practising regularly, you’ll soon be able to simplify algebraic fractions with quadratic terms like a pro, regardless of whether they are monic or non-monic expressions!

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