Problem 40 - Elementary and Intermediate Algebra

Question

Factor. If a polynomial can't be factored, write "prime." $$ x^{2}-25 $$

Step-by-Step Solution

Answer

The polynomial factors to \((x + 5)(x - 5)\).

Step 1: Identify the Type of Expression

The expression given is a quadratic polynomial: \(x^2 - 25\). Notice that it resembles the difference of squares formula.

Step 2: Recall the Difference of Squares Formula

The difference of squares formula states that \(a^2 - b^2 = (a + b)(a - b)\). Here, \(a^2 = x^2\) and \(b^2 = 25\), so \(a = x\) and \(b = 5\).

Step 3: Apply the Formula

Substitute \(a = x\) and \(b = 5\) into the difference of squares formula: \[x^2 - 25 = (x + 5)(x - 5)\].

Step 4: Verify the Result

Multiply the factored form to check: \((x + 5)(x - 5) = x^2 - 5x + 5x - 25 = x^2 - 25\). The expansion confirms the factorization is correct.

Key Concepts

Difference of Squares

The concept of the "difference of squares" is a powerful tool in algebra for simplifying certain types of polynomial expressions. A difference of squares occurs when you have an expression in the form \(a^2 - b^2\). This is a unique scenario because it allows us to apply a specific formula to factor the expression: \(a^2 - b^2 = (a + b)(a - b)\). This formula is derived from the idea that when you expand \((a+b)(a-b)\), the middle terms cancel each other out. Therefore:

• \(a^2\) represents the square of the first term in the binomials.
• \(b^2\) is the square of the second term.

The pattern is straightforward and knowing it can significantly simplify your work. For our specific example of \(x^2 - 25\), the expression is a classic difference of squares where \(x^2\) is \(a^2\) and \(25\) is \(b^2\). We apply the formula by identifying \(a = x\) and \(b = 5\), leading us to the factorization: \((x + 5)(x - 5)\). The difference of squares can also help us quickly convert unfactorable expressions into a more manageable form.

Quadratic Polynomials

Quadratic polynomials are expressions of the form \(ax^2 + bx + c\). Here, the degree of the polynomial is 2, which means the highest power of the variable (often \(x\)) is 2. Quadratic polynomials are fundamental in algebra due to their simple yet versatile structure. Many problems involve solving these types of expressions or factoring them to help simplify or find their roots. For a quadratic polynomial like \(x^2 - 25\), we encounter a specific type of quadratic known as a "perfect square" or "difference of squares," as we identified earlier.

• The leading term \(ax^2\) is simply \(x^2\) in this expression.
• \(b\) is 0 because there is no linear \(x\) term present.
• \(c\) is the constant term, which in our case is -25.

Understanding these parts can make it easier to identify which quadratics can be factored using methods such as completing the square, using the quadratic formula, or, in our case, using the difference of squares.

Polynomial Factorization

Polynomial factorization is the process of expressing a polynomial as a product of its factors. These factors can often be binomials or simpler polynomials. Factoring polynomials is an essential skill in algebra, as it helps simplify expressions, solve equations, and find roots.

• It involves looking for special patterns like the difference of squares, perfect square trinomials, and others.
• Once a pattern is identified, we can employ formulas or techniques to break down the expression into simpler terms.
• The ultimate goal is to express a polynomial as a multiplication of factors, making it easier to work with and solve.

In the case of \(x^2 - 25\), we recognized it as a difference of squares and applied the known formula to factor it into \((x + 5)(x - 5)\). This process of breakdown and reassembly not only simplifies the polynomial but also reveals its roots: the values of \(x\) that make the expression equal to zero. Thus, understanding the method of factorization enhances our ability to manipulate and solve polynomial equations efficiently.