Problem 14
Question
Factor. \(25-t^{2}\)
Step-by-Step Solution
Verified Answer
The expression \(25-t^{2}\) factors to \((5-t)(5+t)\).
1Step 1: Recognize the Difference of Squares Formula
We are asked to factor the expression \(25 - t^2\). Notice that the expression is in the form \(a^2 - b^2\) where \(a = 5\) and \(b = t\). This resembles the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\). First, identify each square: \(5^2 = 25\) and \(t^2\) is already in the correct form.
2Step 2: Apply the Difference of Squares Formula
Using the formula \(a^2 - b^2 = (a - b)(a + b)\), plug in \(a = 5\) and \(b = t\) to get: \((5 - t)(5 + t)\). This is the factored form of \(25 - t^2\).
Key Concepts
Difference of SquaresAlgebraPolynomials
Difference of Squares
The difference of squares is a specific type of mathematical expression that can be factored easily. The formula looks like this:
The expression \(25 - t^2\) is a great example of this. Here, 25 is the square of 5, and \(t^2\) is already a perfect square. Thus, we identify \(a = 5\) and \(b = t\), fitting into the difference of squares formula.
By recognizing the pattern, we can quickly factor the equation to \((5-t)(5+t)\), keeping the entire process efficient and straightforward.
- \(a^2 - b^2 = (a - b)(a + b)\)
The expression \(25 - t^2\) is a great example of this. Here, 25 is the square of 5, and \(t^2\) is already a perfect square. Thus, we identify \(a = 5\) and \(b = t\), fitting into the difference of squares formula.
By recognizing the pattern, we can quickly factor the equation to \((5-t)(5+t)\), keeping the entire process efficient and straightforward.
Algebra
Algebra is a branch of mathematics that involves variables and numerical coefficients. It provides tools and concepts that allow us to model real-world scenarios and solve equations.
One of the basic operations in algebra is factoring, which helps simplify equations and expressions to make them easier to work with. Recognizing patterns like the difference of squares is one of the critical skills.
Factoring this way makes solving equations more manageable. For instance, the expression \(25 - t^2\) becomes simpler, as it breaks down into \((5-t)(5+t)\). This makes it easier to see the roots or solutions when set equal to zero, demonstrating algebra's core principle of simplification through transformation.
One of the basic operations in algebra is factoring, which helps simplify equations and expressions to make them easier to work with. Recognizing patterns like the difference of squares is one of the critical skills.
Factoring this way makes solving equations more manageable. For instance, the expression \(25 - t^2\) becomes simpler, as it breaks down into \((5-t)(5+t)\). This makes it easier to see the roots or solutions when set equal to zero, demonstrating algebra's core principle of simplification through transformation.
Polynomials
Polynomials are expressions composed of variables raised to non-negative integer powers and coefficients. An example is \(ax^n + bx^{n-1} + \ldots + c\).
In the case of the exercise, \(25 - t^2\) is a polynomial with a degree of 2, as the highest term is \(t^2\). Such polynomials are called 'quadratic' since they involve terms up to the power of two.
In this context, you reduce the quadratic \(25 - t^2\) into its linear factors: \((5-t)(5+t)\), simplifying the polynomial and making it easier to solve.
In the case of the exercise, \(25 - t^2\) is a polynomial with a degree of 2, as the highest term is \(t^2\). Such polynomials are called 'quadratic' since they involve terms up to the power of two.
- Highest degree term: \(t^2\)
- Constant term: 25
In this context, you reduce the quadratic \(25 - t^2\) into its linear factors: \((5-t)(5+t)\), simplifying the polynomial and making it easier to solve.
Other exercises in this chapter
Problem 14
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