Problem 47 - Algebra A Combined Function

Question

Multiply. $$ \left(2 y^{2}-1\right)\left(2 y^{2}+1\right) $$

Step-by-Step Solution

Answer

The product is \(4y^4 - 1\).

Step 1: Identify the Expression

The expression given is \((2y^2 - 1)(2y^2 + 1)\). Each part inside the parentheses is a binomial.

Step 2: Recognize the Pattern

Notice that the expression is of the form \((a-b)(a+b)\), which is the difference of squares formula: \((a-b)(a+b) = a^2 - b^2\).

Step 3: Define the Variables

Assign \(a = 2y^2\) and \(b = 1\) as per the pattern recognized in the previous step.

Step 4: Apply the Difference of Squares Formula

Substitute \(a\) and \(b\) into the formula: \[a^2 - b^2 = (2y^2)^2 - 1^2\]

Step 5: Simplify the Expression

Calculate \((2y^2)^2\) and \(1^2\):\[(2y^2)^2 = 4y^4\]\[1^2 = 1\]So the expression becomes:\[4y^4 - 1\]

Step 6: Final Result

The product of the original expression is the simplified form derived from using the difference of squares formula.\[(2y^2 - 1)(2y^2 + 1) = 4y^4 - 1\]

Key Concepts

Difference of Squares

The difference of squares is a unique and important algebraic expression pattern. The formula for the difference of squares is \((a - b)(a + b) = a^2 - b^2\). This pattern allows us to simplify expressions quickly by recognizing that two binomials differ only by the sign in between terms, making computation easier. When you see an expression like \((2y^2 - 1)(2y^2 + 1)\), it matches this pattern.

• First term: \((2y^2)\) in both binomials
• Second term: \((1)\) in both binomials, but with opposite signs

By applying the difference of squares, you avoid multiplying each term directly and instead use the shortcut to find the result as \(4y^4 - 1\). Remember, if you spot two terms that only differ by a plus and minus sign, it's likely you can apply this concise formula.

Multiplying Binomials

Multiplying binomials often involves ensuring each term in one binomial multiplies with every term in the other. While this can seem complex, especially when expressions grow in size, there's a systematic way to approach it.For our given expression \((2y^2 - 1)(2y^2 + 1)\), you initially think to use the "FOIL" method (First, Outer, Inner, Last) for multiplying, which is a common mnemonic.

• First: Multiply the first terms \((2y^2)(2y^2) = 4y^4\)
• Outer: Multiply the outer terms \((2y^2)(1) = 2y^2\)
• Inner: Multiply the inner terms \((-1)(2y^2) = -2y^2\)
• Last: Multiply the last terms \((-1)(1) = -1\)

Adding up these helps in deriving \(4y^4 + 2y^2 - 2y^2 - 1\).Notice that \(+2y^2\) and \(-2y^2\) cancel each other out, simplifying your expression further. In this particular case, recognizing the difference of squares can be more efficient.

Simplifying Expressions

Simplifying expressions is critical in algebra to obtain a concise and understandable form. Simplification involves reducing expressions to their simplest form using arithmetic operations and algebraic rules.In the exercise given, once we recognize the expression as a difference of squares, we'll have:\((2y^2)^2 - 1^2\).Calculating gives:

• \((2y^2)^2 = 4y^4\)
• \(1^2 = 1\)

This results in\(4y^4 - 1\), a neatly simplified polynomial. Notice how using the difference of squares pattern gets to the simplest form directly, avoiding redundant calculations.Simplifying expressions not only makes them easier to work with but also helps in solving equations and comparing expressions more efficiently.