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Chapter 7: Problem 99

Extend the concepts of \(7.1-7.4\) to factor completely. $$(x+5)^{2}-(x-2)^{2}$$

Short Answer

Expert verified
The given expression \((x+5)^2 - (x-2)^2\) can be factored completely by identifying it as a difference of squares, applying the formula \(a^2 - b^2 = (a + b)(a - b)\), and simplifying the expression. The final factored form is \((2x + 3)(2)\).

Step by step solution

01

Identify the difference of squares

In the given expression \((x+5)^2 - (x-2)^2\), we can identify it as a difference of squares with \(a = x + 5\) and \(b = x - 2\).
02

Apply the formula

Now that we have identified the expression as a difference of squares, we can use the formula for factoring such expressions, which is \(a^2 - b^2 = (a + b)(a - b)\). So, we obtain: \((x+5)^2 - (x-2)^2 = (x + 5 + x - 2)(x + 5 - x + 2)\)
03

Simplify the expression

Now we simplify the expression obtained in Step 2: \((2x + 3)(2)\) Therefore, the completely factored form of the given expression \((x+5)^2 - (x-2)^2\) is \((2x + 3)(2)\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Difference of Squares
When you encounter a mathematical expression like \(a^2 - b^2\), you are dealing with what is known as a difference of squares. This is a specific type of polynomial expression, where two perfect square terms are subtracted from one another. The formula to factor this kind of expression is straightforward and immensely useful: \(a^2 - b^2 = (a + b)(a - b)\).
Understanding and recognizing difference of squares allows you to simplify what could otherwise be complex algebraic expressions. In our initial example, \((x+5)^2 - (x-2)^2\), each part \((x+5)\) and \((x-2)\) is squared, making the whole expression fit perfectly into the difference of squares format.
Algebraic Expressions
Algebraic expressions are mathematical phrases that combine numbers, variables, and arithmetic operations. These expressions are the building blocks in algebra and come in various forms, from very simple to highly complex. They can include addition, subtraction, multiplication, division, and more.
In the problem we solved, \(x+5\) and \(x-2\) are sub-expressions that, when squared, formed a larger expression. Identifying parts of such expressions is crucial, as it helps you apply appropriate methods like factoring. Using algebraic rules to simplify expressions, like recognizing a difference of squares, makes problem-solving more efficient.
Polynomial Factoring
Factoring polynomials involves breaking down a polynomial into a product of simpler polynomials. It is a crucial skill in algebra that helps in simplifying expressions and solving equations. Factoring can involve different formulas, with the difference of squares being one common method.
In our original problem, by identifying \((x+5)^2 - (x-2)^2\) as a difference of squares, we were able to factor it into \(2x + 3\) and \(2\). This process of simplifying helps solve or evaluate expressions more easily and is often a necessary step in calculus and higher algebraic contexts.
Intermediate Algebra
Intermediate algebra serves as a bridge between basic algebra and advanced topics like calculus. At this level, you refine skills in manipulating algebraic expressions, solving equations, and applying various algebraic techniques like factoring.
Understanding concepts such as the difference of squares is essential in intermediate algebra. It builds a foundation for dealing with more involved expressions and equations, as seen in our example with \((x+5)^2 - (x-2)^2\). Mastering these foundational elements prepares students for the complexities of advanced mathematics by solidifying their problem-solving strategies and mathematical reasoning.

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Most popular questions from this chapter

Write an equation and solve. The longer leg of a right triangle is \(7 \mathrm{cm}\) more than the shorter leg. The length of the hypotenuse is \(3 \mathrm{cm}\) more than twice the length of the shorter leg. Find the length of the hypotenuse.

Factor by grouping. What should be the first step in factoring \(3 x y+6 x+15 y+30 ?\)

Factor completely. You may need to begin by taking out the GCF first or by rearranging terms. $$9 f^{2} j^{2}+45 f j+9 f j^{2}+45 f^{2} j$$

Factor completely. $$p^{6}-1$$

The following equations are not quadratic but can be solved by factoring and applying the zero product rule. Solve each equation. $$(4 f+5)\left(f^{2}-3 f-18\right)=0$$
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