Problem 37 - Algebra and Trigonometry

Question

In Exercises 15–58, find each product. $$ \left(4 x^{2}+5 x\right)\left(4 x^{2}-5 x\right) $$

Step-by-Step Solution

Answer

The product is \( 16x^4 - 25x^2 \).

Step 1: Identify the form

The given expression is in the form of \( (a+b)(a-b) \), with \( a = 4x^2 \) and \( b = 5x \).

Step 2: Apply the difference of squares formula

The difference of squares formula states that \( (a+b)(a-b) = a^2 - b^2 \). By substituting \( a = 4x^2 \) and \( b = 5x \) into this formula, the solution becomes \( (4x^2)^2 - (5x)^2 \).

Step 3: Calculate the powers and simplify

Calculate \( (4x^2)^2 \) and \( (5x)^2 \) to obtain \( 16x^4 - 25x^2 \). This is the simplified form of the given expression.

Key Concepts

Polynomial Multiplication

Polynomial multiplication is an essential concept in algebra that involves combining two or more polynomials to produce a new polynomial. It is similar to multiplying numbers, but instead, it involves algebraic terms. To multiply polynomials, we distribute each term in the first polynomial by each term in the second polynomial.

For example, consider the expression \( (4 x^{2}+5 x)(4 x^{2}-5 x) \). To multiply these two polynomials, we use the distributive property, also known as the FOIL method (First, Outer, Inner, Last), which stipulates that we multiply each term in the first polynomial by each term in the second.

In this case:

• First: \( 4x^2 \times 4x^2 = 16x^4 \)
• Outer: \( 4x^2 \times -5x = -20x^3 \)
• Inner: \( 5x \times 4x^2 = 20x^3 \)
• Last: \( 5x \times -5x = -25x^2 \)

Here, the middle terms (outer and inner) cancel each other out because they are equal in magnitude but opposite in sign, leaving us with \( 16x^4 - 25x^2 \).
However, recognizing the special pattern of a difference of squares can make this process much quicker, as was applied in the provided solution.

Algebraic Expressions

Algebraic expressions are composed of variables, numbers, and arithmetic operations. They can represent virtually any quantity in mathematics and are integral in illustrating relationships between variables. Understanding how to manipulate these expressions is fundamental in algebra.

An important skill in working with algebraic expressions is recognizing patterns, such as the difference of squares, which is a specific type of algebraic identity used to simplify expressions. When faced with an expression like \( (a+b)(a-b) \) where \( a \) and \( b \) are any algebraic expressions, it can be rapidly simplified to \( a^2 - b^2 \) without the need for extensive polynomial multiplication.

Ingeterpreting our original example of \( (4 x^{2}+5 x)(4 x^{2}-5 x) \), by identifying \( a \) as \( 4x^2 \) and \( b \) as \( 5x \) and employing the difference of squares identity, we immediately arrive at the simplified expression \( 16x^4 - 25x^2 \) without the need for long multiplication.

Factoring Polynomials

Factoring polynomials is the process of breaking down a complex polynomial into simpler polynomials that, when multiplied together, give you the original polynomial. It is essentially the reverse of polynomial multiplication. This approach is particularly useful in solving equations, simplifying expressions, and finding zeroes of polynomials.

One of the most recognized techniques in factoring is the use of the difference of squares. When a polynomial is expressed as \( a^2 - b^2 \) where \( a \) and \( b \) are algebraic expressions, it can be factored back into \( (a+b)(a-b) \) readily. This special case arises quite often in mathematics, so identifying it quickly can greatly simplify many problems.

In the context of our original problem, the expression \( 16x^4 - 25x^2 \) is already in its simplest factored form, showing the result of the multiplication \( (4 x^{2}+5 x)(4 x^{2}-5 x) \) but can also be seen as the conclusive step if we were factoring the polynomial \( 16x^4 - 25x^2 \) by recognizing it as a difference of squares.