Problem 72
Question
Factor the expression completely, if possible. \(25-x^{4} y^{2}\)
Step-by-Step Solution
Verified Answer
The factored expression is \((5 - x^2 y)(5 + x^2 y)\).
1Step 1: Identify the expression type
The given expression is a subtraction between a perfect square and a polynomial. It resembles the difference of perfect squares, which can sometimes be factored further.
2Step 2: Recognize perfect squares
The expression is \(25 - x^4 y^2\). Here, \(25\) is a perfect square since \(25 = 5^2\). Also, \(x^4 y^2\) can be rewritten as \((x^2 y)^2\). Thus, the expression is in the form \(a^2 - b^2\) where \(a = 5\) and \(b = x^2 y\).
3Step 3: Apply the difference of squares formula
The difference of squares formula states \(a^2 - b^2 = (a-b)(a+b)\). Applying this to the identified perfect squares, we have:\[(5 - x^2 y)(5 + x^2 y)\].
4Step 4: Verify if factors can be further simplified
Check if the factors \(5 - x^2 y\) and \(5 + x^2 y\) can be factored further. Neither \(5\) nor \(x^2 y\) can be written in a way that allows for further simplification using integer algebraic identities.
Key Concepts
Difference of SquaresPerfect SquaresPolynomial Factorization
Difference of Squares
The difference of squares is a valuable tool that aids in factoring expressions, particularly when you notice a subtraction between two terms. This technique is named because the expressions it involves are typically in the form of \(a^2 - b^2\). When dealing with the difference of squares, it can be simplified using the formula:
For example, given the expression \(25 - x^4 y^2\), you first observe that both \(25\) and \(x^4 y^2\) are perfect squares. \(25 = 5^2\), and \(x^4 y^2 = (x^2 y)^2\), positioning the expression perfectly to apply this principle.
Using this property of difference of squares, you would factor it to \((5 - x^2 y)(5 + x^2 y)\). This process simplifies polynomial factorization considerably by breaking down the problem into simpler components.
- \(a^2 - b^2 = (a - b)(a + b)\)
For example, given the expression \(25 - x^4 y^2\), you first observe that both \(25\) and \(x^4 y^2\) are perfect squares. \(25 = 5^2\), and \(x^4 y^2 = (x^2 y)^2\), positioning the expression perfectly to apply this principle.
Using this property of difference of squares, you would factor it to \((5 - x^2 y)(5 + x^2 y)\). This process simplifies polynomial factorization considerably by breaking down the problem into simpler components.
Perfect Squares
Understanding perfect squares is crucial when factoring expressions. A perfect square is a number or expression that is the product of an integer or an algebraic expression with itself. For instance:
Perfect squares simplify calculations, allowing us to decompose and manipulate equations efficiently. This is an invaluable skill for breaking complex polynomials into manageable factors.
- \(16\) is a perfect square because \(16 = 4^2\).
- \((x^2 y)^2\) is a perfect square because it is derived from multiplying \(x^2 y\) by itself.
Perfect squares simplify calculations, allowing us to decompose and manipulate equations efficiently. This is an invaluable skill for breaking complex polynomials into manageable factors.
Polynomial Factorization
Polynomial factorization is the process of expressing a polynomial as a product of simpler polynomials. It is a useful skill in algebra, resolving complex expressions into more workable pieces. For the original expression \(25 - x^4 y^2\), factorization begins with identifying patterns such as the difference of squares.
In the case of \((5 - x^2 y)(5 + x^2 y)\), no further simplification is possible using integer factorization rules. Understanding these core concepts of polynomial factorization strengthens one's ability to handle complex algebraic expressions efficiently.
- Recognize potential factorizable forms: \(a^2 - b^2\) in this case.
- Apply appropriate formulas to open these into simpler expressions: \((a-b)(a+b)\).
In the case of \((5 - x^2 y)(5 + x^2 y)\), no further simplification is possible using integer factorization rules. Understanding these core concepts of polynomial factorization strengthens one's ability to handle complex algebraic expressions efficiently.
Other exercises in this chapter
Problem 72
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