Problem 57 - Precalculus Mathematics for Calculus

Question

Perform the indicated operations and simplify. $$\left((x-1)+x^{2}\right)\left((x-1)-x^{2}\right)$$

Step-by-Step Solution

Answer

The simplified expression is \(-x^4 + x^2 - 2x + 1\).

Step 1: Use the Difference of Squares Formula

Recognize the expression as a difference of squares. The general formula is \((a+b)(a-b) = a^2 - b^2\). In our case, identify \(a = x-1\) and \(b = x^2\). Then, apply the formula: \((x-1+x^2)(x-1-x^2) = (x-1)^2 - (x^2)^2\).

Step 2: Calculate \((x-1)^2\)

Compute the square of \(x-1\). Use the formula \((a-b)^2 = a^2 - 2ab + b^2\). Here, \(a = x\) and \(b = 1\), so \((x-1)^2 = x^2 - 2x + 1\).

Step 3: Calculate \((x^2)^2\)

Compute \((x^2)^2\). This results in \(x^{4}\).

Step 4: Substitute Back to Main Expression

Substitute the results from Step 2 and Step 3 back into the expression determined in Step 1: Thus, \((x-1)^2 - (x^2)^2 = (x^2 - 2x + 1) - x^{4}\).

Step 5: Final Simplification

Simplify the expression \(x^2 - 2x + 1 - x^4\). Rearrange terms and combine like terms to obtain: \(-x^4 + x^2 - 2x + 1\).

Key Concepts

Difference of Squares

The concept of the difference of squares is a powerful and useful tool in algebra. It is particularly helpful when you have expressions that take the form \((a + b)(a - b)\). This pattern can be simplified using the formula: \(a^2 - b^2\).
In the original exercise, we recognize that the expression \((x-1) + x^2\) and \((x-1) - x^2\) are in the form of \((a + b)(a - b)\). With \(a\) being \((x-1)\) and \(b\) being \(x^2\), the difference of squares formula gives us \((x-1)^2 - (x^2)^2\).

• The application of this formula transforms the problem from a product of binomials to a subtraction of squares, simplifying the way we approach it.
• Recognizing the form early on can make solving the problem less complex and more straightforward.

Simplifying Expressions

Simplifying expressions involves breaking down complex expressions into simpler components, making them easier to understand and solve. In mathematics, simplifying helps clarify the problem and shows an equivalent expression that may be easier to work with.

• Simplification usually involves performing arithmetic operations, combining like terms, and applying algebraic rules like distributive, associative, and commutative properties.
• In our exercise, this simplification process started after identifying the expression as a difference of squares.

After using the difference of squares formula to obtain the expression \((x-1)^2 - (x^2)^2\), we took it step by step to further simplify. First, calculating \((x-1)^2\) using the formula \((a-b)^2 = a^2 - 2ab + b^2\), followed by evaluating \((x^2)^2\) as \(x^{4}\).
Once both components are computed, they are substituted back into the main expression and then terms are combined to achieve the final simplified result: \(-x^4 + x^2 - 2x + 1\). This step-by-step simplification not only offers a clearer view of what's happening in the expression but also ensures an accurate solution.

Quadratic and Higher-Degree Terms

Quadratic terms and higher-degree terms are common in algebra, each with distinct plans for solving and simplifying. Quadratic terms are those that include a variable squared, like \(x^2\), while higher-degree terms can include cubed variables like \(x^3\) or even higher, like \(x^4\).

• Quadratic expressions often form parabolas when graphed, and solving them can involve factoring or using the quadratic formula among other methods.
• Higher-degree terms expand the possibilities for polynomial behavior and are tackled by applying different algebraic strategies.

In the original problem, expressions like \(x^2 - 2x + 1\) are examples of quadratic terms, while \(x^4\) represents a fourth-degree term. Each degree in the expression introduces new dynamics in terms of solutions and shapes in graphical representations.
To solve these, it is essential to confidently perform steps such as expanding binomials, applying necessary formulas, and rearranging terms to improve our understanding and reach the simplest form possible. Knowing how to operate effectively with different degree terms is fundamental to procedural algebra and crucial for tackling more complex equations.