Problem 55

Question

For the following exercises, factor the polynomials completely. \(16 x^{4}-200 x^{2}+625\)

Step-by-Step Solution

Verified
Answer
The polynomial factors to \((2x - 5)^2 (2x + 5)^2\).
1Step 1: Recognize a Perfect Square Trinomial
Notice that the polynomial has three terms of the form \(16x^4 - 200x^2 + 625\). This resembles a perfect square trinomial, where each term is a perfect square or has a common factor. Recognize that \(16x^4\) is \((4x^2)^2\), \(200x^2\) needs to be factored as part of \((2)(4x^2)(c)\), and \(625\) is \(25^2\).
2Step 2: Identify Proper Value
Find the value that completes the perfect square trinomial: \((a-b)^2 = a^2 - 2ab + b^2\). Set \(a = 4x^2\) and \(b = 25\). Then \(2ab = 2(4x^2)(25) = 200x^2\), which matches the middle term.
3Step 3: Factoring the Polynomial
Now, you can write the polynomial as a perfect square trinomial: \((4x^2 - 25)^2\). This reflects the identity of \(a^2 - 2ab + b^2 = (a-b)^2\).
4Step 4: Recognize Difference of Squares
Observe that \(4x^2 - 25\) is a difference of two squares. The expression \(4x^2 = (2x)^2\) and \(25 = 5^2\) allows you to apply the difference of squares formula: \(a^2 - b^2 = (a - b)(a + b)\).
5Step 5: Factor Again Using Difference of Squares
Apply the difference of squares: \(4x^2 - 25 = (2x)^2 - 5^2\). This becomes \((2x - 5)(2x + 5)\).
6Step 6: Write the Complete Factorization
After factoring \(4x^2 - 25\) as \((2x - 5)(2x + 5)\), the complete factorization of the original polynomial \(16x^4 - 200x^2 + 625\) is \((2x - 5)^2(2x + 5)^2\).

Key Concepts

Perfect Square TrinomialDifference of SquaresAlgebraic Expressions
Perfect Square Trinomial
In algebra, recognizing patterns such as a perfect square trinomial can significantly simplify polynomial factorization. A perfect square trinomial takes the form of
  • \(a^2 - 2ab + b^2 = (a-b)^2\)
  • \(a^2 + 2ab + b^2 = (a+b)^2\)
These expressions show how a trinomial can be written as the square of a binomial. In the given exercise, the polynomial \(16x^4 - 200x^2 + 625\) is structured as a perfect square trinomial. Here, each term aligns perfectly; \(16x^4\) is \((4x^2)^2\), and \(625\) is \(25^2\). Recognizing the relationships allows us to simplify it into \((4x^2 - 25)^2\). Understanding this transformation is crucial because it converts the polynomial into two identical binomials squared, setting the stage for further simplification.
Difference of Squares
The difference of squares is a powerful algebraic tool to factor expressions further. The identity \(a^2 - b^2 = (a-b)(a+b)\) allows quick simplification of two perfect squares separated by a subtraction sign. This is what we apply in Step 4 of the solution.
  • Identify each part: \(4x^2\) is \((2x)^2\) and \(25\) is \(5^2\).
  • Apply the identity: \((2x)^2 - 5^2 = (2x + 5)(2x - 5)\).
This step breaks down the polynomial \(4x^2 - 25\), evident from the previous step, into simpler binomials. Recognizing the difference of squares speeds up the process because it turns a complex expression into a product of two simpler expressions. This concept is especially useful in recognizing patterns in algebraic expressions that might otherwise look complicated.
Algebraic Expressions
Algebraic expressions are fundamental in mathematics as they represent numbers, operations, and variables. Understanding these expressions paves the way for effective manipulation and simplification. In the context of polynomial factorization, expressions comprise several terms that adhere to certain patterns. Identifying these patterns, as seen in perfect square trinomials or differences of squares, streamlines the factoring process.
  • Terms like \(4x^2\), which can be written as a square, simplify factoring.
  • Recognizable products, such as \(a^2 - 2ab + b^2\), develop problem-solving intuition.
Knowing how to rewrite these expressions through factorization makes complex algebra manageable and efficient. This exercise utilizes the identification of perfectly squared terms and a clear understanding of differences between square values, showcasing the beauty and utility of algebra in dissecting polynomial expressions.