Problem 5 - Precalculus: Mathematics for Calculus

Question

The Special Product Formula for the "product of the sum and difference of terms" is \((A+B)(A-B)=\) ____. \(\operatorname{So}(5+x)(5-x)=\) _____.

Step-by-Step Solution

Answer

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

Step 1: Identify the Special Product Formula

The special product formula for the product of the sum and difference of the same two terms is given by \[(A + B)(A - B) = A^2 - B^2\]. This is known as the difference of squares formula. We need to apply this formula to find the result.

Step 2: Assign Values for A and B

In the given expression \((5 + x)(5 - x)\), identify \(A\) and \(B\). Here, \(A = 5\) and \(B = x\). These will be used in our special product formula.

Step 3: Apply the Formula

Apply the difference of squares formula using the values for \(A\) and \(B\) we identified: \[(A + B)(A - B) = A^2 - B^2\] becomes \[(5 + x)(5 - x) = 5^2 - x^2\].

Step 4: Calculate the Squares

Calculate \(5^2\) and \(x^2\):\[5^2 = 25\] and \[x^2 = x^2\]. So the expression simplifies to \[25 - x^2\].

Step 5: Combine the Results

Combine the results from the calculation:\[(5 + x)(5 - x) = 25 - x^2\]. This is the simplified expression.

Key Concepts

Special Product Formula

The special product formula, known as the "product of the sum and difference" formula, is a powerful tool in algebra. It simplifies expressions where you multiply two binomials that are the sum and difference of the same terms. This formula is given by

• \((A + B)(A - B) = A^2 - B^2\)

This is called the difference of squares formula. To use this, identify what your \(A\) and \(B\) are from your expression. Then, you simply square these terms and subtract one square from the other.
For example, with \((5 + x)(5 - x)\), once you know \(A = 5\) and \(B = x\), it becomes easy to apply this formula.
The calculation results in \(25 - x^2\). This approach cuts down on lengthy distribution calculations by providing a shortcut to the result.

Factoring Polynomials

Factoring polynomials is about breaking them down into simpler components or products that, when multiplied together, give back the original polynomial.
In the context of the special product formula, you’re working backwards from \(A^2 - B^2\) back to \((A + B)(A - B)\).

• Recognize that the squared terms in the difference of squares result from original, simpler expressions.
• If you're given \(A^2 - B^2\), you can factor it back into \((A + B)(A - B)\).

This is especially useful because factoring makes solving equations or simplifying expressions easier.
Knowing the difference of squares helps in identifying patterns in polynomials quickly, aiding in faster solutions.

Simplifying Expressions

Simplifying expressions involves rewriting them in a simpler or more convenient form. It often requires identifying patterns or using algebraic formulas like the difference of squares.
When you apply the special product formula, you're simplifying a product of two binomials into a single expression.

• In \((5 + x)(5 - x)\), the product simplifies directly to \(25 - x^2\).

This reduces complexity, making further calculations more straightforward.
Normally, simplifying expressions might require expanding using distribution or combining like terms, but formulas like this can make the process faster and less error-prone.
Remember, the goal is to make equations or expressions as uncomplicated as possible, facilitating easier manipulation or solving.