Problem 104
Question
Factor completely. (Hint on Exercises \(97-102\) : Factors contain rational numbers.) $$(x+2)^{2}-49$$
Step-by-Step Solution
Verified Answer
The completely factored form of \((x+2)^{2}-49\) is \((x+9)(x-5)\).
1Step 1: Identify \(a\) and \(b\) in the formula
The given expression is the form of \(a^{2}-b^{2}\). Here, \((x+2)\) is acting as \(a\) (so \(a = x+2\)) and \(49\) is a perfect square with \(\sqrt{49}=7\); so \(b = 7\).
2Step 2: Use the formula to factor the expression
Using the formula \(a^2 - b^2 = (a+b)(a-b)\), replace \(a\) with \((x+2)\) and \(b\) with \(7\). This gives: \((x+2+7)(x+2-7)\).
3Step 3: Simplify the factors
Combine like terms within each factor to simplify: the first factor becomes \((x+9)\) and the second factor becomes \((x-5)\).
Key Concepts
Difference of SquaresPerfect SquaresAlgebraic Identities
Difference of Squares
The difference of squares is a special algebraic identity that helps in factoring expressions of the form \(a^2 - b^2\). The key to understanding this concept is recognizing that it is the subtraction of two perfect squares. An expression that follows this pattern can be expressed as the product of two conjugates: \((a + b)(a - b)\). This powerful methodology allows you to factor complex polynomial expressions effortlessly.
In our example, we have \((x+2)^2 - 49\), which fits neatly into the difference of squares format. The first term, \((x + 2)^2\), can be seen as \(a^2\), where \(a = (x + 2)\). The second term, 49, is \(b^2\), where \(b = 7\). Consequently, applying the difference of squares identity enables us to factor the expression as \((x+9)(x-5)\).
Understanding this identity is crucial because it simplifies factoring tasks and reveals connections in algebraic expressions. Recognize them more readily, and these algebraic puzzles become simpler to solve.
In our example, we have \((x+2)^2 - 49\), which fits neatly into the difference of squares format. The first term, \((x + 2)^2\), can be seen as \(a^2\), where \(a = (x + 2)\). The second term, 49, is \(b^2\), where \(b = 7\). Consequently, applying the difference of squares identity enables us to factor the expression as \((x+9)(x-5)\).
Understanding this identity is crucial because it simplifies factoring tasks and reveals connections in algebraic expressions. Recognize them more readily, and these algebraic puzzles become simpler to solve.
Perfect Squares
Perfect squares are numbers or expressions that result from squaring a whole number or expression. So, if a number is a perfect square, its square root is an integer. This concept is fundamental in spotting special algebraic structures, such as those appearing in our expression: \((x+2)^2 - 49\).
In this instance, both \((x+2)^2\) and 49 are perfect squares. The term \((x+2)^2\) means that it comes from squaring the binomial \((x+2)\). Likewise, the number 49 is a perfect square because \(49 = 7^2\). Recognizing perfect squares is the preliminary step in employing the difference of squares identity effectively.
When you're tackling algebraic identities, identifying perfect squares allows you to simplify expressions more efficiently. Spotting these squares is akin to discovering a shortcut in mathematics—understanding them can notably aid in algebraic manipulation and in solving equations smoothly.
In this instance, both \((x+2)^2\) and 49 are perfect squares. The term \((x+2)^2\) means that it comes from squaring the binomial \((x+2)\). Likewise, the number 49 is a perfect square because \(49 = 7^2\). Recognizing perfect squares is the preliminary step in employing the difference of squares identity effectively.
When you're tackling algebraic identities, identifying perfect squares allows you to simplify expressions more efficiently. Spotting these squares is akin to discovering a shortcut in mathematics—understanding them can notably aid in algebraic manipulation and in solving equations smoothly.
Algebraic Identities
Algebraic identities are equations that hold true for all variable values. These are invaluable tools in mathematics because they allow you to simplify complex algebraic expressions into more manageable formats. One such identity is the difference of squares, \(a^2 - b^2 = (a + b)(a - b)\), used in the given exercise.
Using such identities intelligently can reduce trial and error in solving algebra problems, as they provide a reliable guide to breaking down expressions. They serve as the backbone in algebra, providing tried and tested blueprints for restructuring expressions and solving equations. Think of them as universal rules that, once mastered, open up a plethora of solutions with fewer steps.
In the broader scope of mathematics, becoming adept at applying algebraic identities like the difference of squares is akin to learning a secret code. It unlocks the potential to tackle problems with efficiency and elegance, offering a path to mastery in algebra.
Using such identities intelligently can reduce trial and error in solving algebra problems, as they provide a reliable guide to breaking down expressions. They serve as the backbone in algebra, providing tried and tested blueprints for restructuring expressions and solving equations. Think of them as universal rules that, once mastered, open up a plethora of solutions with fewer steps.
In the broader scope of mathematics, becoming adept at applying algebraic identities like the difference of squares is akin to learning a secret code. It unlocks the potential to tackle problems with efficiency and elegance, offering a path to mastery in algebra.
Other exercises in this chapter
Problem 104
Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$-10 a^{4} b^{2}+15 a^{3} b^{3}+25 a^{2} b^{4}$$
View solution Problem 104
Use an example and explain how to factor out the greatest common factor of a polynomial.
View solution Problem 104
My graphing calculator showed the same graphs for \(y_{1}=4 x^{2}-20 x+24 \quad\) and \(\quad y_{2}=4\left(x^{2}-5 x+6\right), \quad\) so I can conclude that th
View solution Problem 105
Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$a^{2} y-b^{2} y-a^{2} x+b^{2} x$$
View solution