Question

Simplify the expression, if possible. $$ \frac{32 x^4-50}{4 x^3-12 x^2-5 x+15} $$

Short answer

The simplified expression is \(\frac{2(4x^2-5)(4x^2+5)}{(4x^2-5)(x-3)}\)

Step-by-step solution

Factor the Numerator

The numerator \(32 x^4 - 50\) can be factored by removing the common factor of 2 and then factoring the resulting difference of squares:\n\n\(32 x^4-50 = 2(16 x^4 - 25) = 2((4x^2)^2 - 5^2) = 2(4x^2 - 5)(4x^2 + 5)\)

Factor the Denominator

The denominator \(4 x^3 - 12 x^2 - 5 x + 15\) is a polynomial of degree 3, so it's a little bit more complex. Using factor by grouping, we get:\n\n\(4 x^3 - 12 x^2 - 5 x + 15 = 4x^2(x - 3) - 5(x - 3) = (4x^2 - 5)(x - 3)\)

Simplify the Expression

Now that both the numerator and the denominator are factored, we can see if there are any common factors that we can cancel out. However, there are no common factors. So the simplified form of the expression is \(\frac{2(4x^2 - 5)(4x^2 + 5)}{(4x^2 - 5)(x - 3)}\)

Key concepts

Factoring Polynomials

Factoring polynomials is a fundamental skill in simplifying algebraic expressions. It involves finding the expressions that multiply to give the original polynomial. This process is essential when you want to simplify, solve, or analyze polynomials.

To begin with, always look for a greatest common factor (GCF) that can be factored out. In the exercise, the polynomial in the numerator, \(32x^4 - 50\), has a GCF of 2. Once the GCF is factored out, you may look for further factoring possibilities.

After factoring the GCF, the expression becomes \(2(16x^4 - 25)\). We then check if the remaining polynomial is factorizable by other methods. Sometimes, you may need to experiment with different factoring techniques, like grouping or recognizing patterns such as trinomials or special products.

Note that factoring might not always be straightforward. Practice identifying polynomial types and applying appropriate strategies to gain confidence in this area. With time, recognizing factorable expressions will become more intuitive.

Differences of Squares

Differences of squares is a specific factoring technique used when you have two perfect squares separated by a subtraction sign. The general formula for factoring a difference of squares is: \(a^2 - b^2 = (a - b)(a + b)\).

In the given exercise, after factoring out a 2 from the numerator, we have \(16x^4 - 25\). Notice that \(16x^4\) is \((4x^2)^2\) and 25 is \(5^2\). This fits perfectly with the difference of squares pattern.

Using the formula, \((4x^2)^2 - 5^2\) becomes \((4x^2 - 5)(4x^2 + 5)\). This technique is extremely handy in simplifying algebraic fractions or any problems involving polynomials where you'll encounter differences of squares.

Always keep an eye out for squared terms separated by a minus sign as a potential difference of squares, and remember that this only applies when both terms are perfect squares.

Polynomial Long Division

Polynomial long division is analogous to numeric long division. It helps simplify complex rational expressions by dividing polynomials, particularly when other simplifying methods like factoring aren't immediately evident.

Although the exercise focused on factoring, there are times when polynomial long division plays a crucial role, typically with more complex polynomial expressions in both the numerator and the denominator.

Polynomial long division involves:

• Dividing the term with the highest exponent in the numerator by the highest degree term in the divisor.
• Multiplying the entire divisor by that quotient and subtracting the result from the original polynomial.
• Repeating the process with the remainder until the remainder's degree is less than the divisor's.

This step-by-step process systematically reduces the complexity of a polynomial fraction.

Even if the exercise didn't require it fully, understanding polynomial long division can enhance comprehension of algebraic fractions, helping to tackle expressions that resist simpler factoring approaches.