Problem 77
Question
In Exercises \(57-84\), factor completely, or state that the polynomial is prime. $$x^{2}-12 x+36-49 y^{2}$$
Step-by-Step Solution
Verified Answer
The factored form of the given polynomial is \((x-6-7y)(x-6+7y)\)
1Step 1: Recognize the structure of the polynomial
The given polynomial \(x^{2}-12 x+36-49 y^{2}\) can be re-written as \((x-6)^{2} - (7y)^{2}\). This is recognized as a difference of squares (A^2 - B^2).
2Step 2: Apply the formula for difference of squares
The difference of squares can be factored using the formula \(A^2-B^2 = (A-B)(A+B)\). Here, A is \((x-6)\) and B is \((7y)\).
3Step 3: Substitute A and B into the formula
By substituting A and B into the formula, we get \((x-6-7y)(x-6+7y)\).
Key Concepts
difference of squarespolynomial structurefactoring formulas
difference of squares
One of the fundamental concepts in algebra is the difference of squares. It represents an expression that can be seen as the subtraction of one square number from another. Recognizing this structure is beneficial, especially when simplifying expressions. In algebraic terms, the difference of squares is represented as \(A^2 - B^2\). This can be easily factored into \((A - B)(A + B)\). This pattern is critical to identify as it allows for simpler manipulation of polynomials.
For instance, in the given problem, the expression \((x-6)^2 - (7y)^2\) perfectly fits the difference of squares format. Here, \((x-6)\) and \((7y)\) are replaced respectively for \(A\) and \(B\). Recognizing this structure allows for the expression to be factored quickly using the formula mentioned above. Understanding this concept not only simplifies current problems but also aids in tackling future algebraic challenges effectively.
For instance, in the given problem, the expression \((x-6)^2 - (7y)^2\) perfectly fits the difference of squares format. Here, \((x-6)\) and \((7y)\) are replaced respectively for \(A\) and \(B\). Recognizing this structure allows for the expression to be factored quickly using the formula mentioned above. Understanding this concept not only simplifies current problems but also aids in tackling future algebraic challenges effectively.
polynomial structure
Polynomials are expressions consisting of variables raised to whole number powers and coefficients. Understanding the structure of a polynomial is crucial because it opens up strategic ways to simplify or factor them. Every polynomial has a certain degree, which is the highest power of the variable in the expression. Observing how the terms are organized within the equation gives insights into potential factorization paths.
In the given problem, \(x^2 - 12x + 36 - 49y^2\), breaking it down is key to easy understanding. Notice how \(x^2 - 12x + 36\) can be transformed into \((x-6)^2\), evident from the perfect square trinomial structure. Meanwhile, \(-49y^2\) remains a perfect square term. When rewritten, it forms a difference of squares scenario. Recognizing and rearranging terms based on their polynomial structure is vital in simplifying problems and finding convenient solutions.
In the given problem, \(x^2 - 12x + 36 - 49y^2\), breaking it down is key to easy understanding. Notice how \(x^2 - 12x + 36\) can be transformed into \((x-6)^2\), evident from the perfect square trinomial structure. Meanwhile, \(-49y^2\) remains a perfect square term. When rewritten, it forms a difference of squares scenario. Recognizing and rearranging terms based on their polynomial structure is vital in simplifying problems and finding convenient solutions.
factoring formulas
Factoring is one of the most powerful techniques in algebra for solving equations and simplifying expressions. There are several key factoring formulas that are widely applicable across various problems. One of these is the formula for factoring the difference of squares, \(A^2 - B^2 = (A - B)(A + B)\), as seen in the original exercise.
Other important formulas include:
Other important formulas include:
- Perfect square trinomial: \(A^2 + 2AB + B^2 = (A + B)^2\)
- Sum of cubes: \(A^3 + B^3 = (A + B)(A^2 - AB + B^2)\)
- Difference of cubes: \(A^3 - B^3 = (A - B)(A^2 + AB + B^2)\)
Other exercises in this chapter
Problem 77
In Exercises \(77-84,\) evaluate each expression without using a calculator. $$36^{1 / 2}$$
View solution Problem 77
Explain how to find the least common denominator for denominators of \(x^{2}-100\) and \(x^{2}-20 x+100\).
View solution Problem 77
Write each number in scientific notation. $$ 0.027 $$
View solution Problem 77
Give an example of two things that you do that are not commutative.
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