Introduction The expression x² – 8xy + 15y² may look like a simple algebraic term, but understanding its factors reveals the hidden structure that makes many later calculations—whether in pure math, physics, or engineering—much more manageable. In this article we will explore what it means to factor this quadratic, why the process matters, and how you can confidently break it down step by step. By the end, you’ll not only know the final result but also grasp the underlying principles that apply to any similar binomial.
Detailed Explanation
At its core, x² – 8xy + 15y² is a quadratic form in two variables, x and y. Quadratic forms are expressions where each term is of degree two, and they often appear in equations describing parabolas, conic sections, and optimization problems. The specific combination of the squared term x², the mixed term ‑8xy, and the squared term 15y² suggests that the expression can be rewritten as a product of two linear factors, each containing a single variable multiplied by y. This factorization is possible because the coefficients satisfy a particular relationship that allows the quadratic to be expressed as (x + a y)(x + b y), where a and b are numbers we need to determine.
The process of factoring begins by recognizing the pattern of a perfect square trinomial or, more generally, a quadratic that can be expressed as a product of two binomials. For x² – 8xy + 15y², we look for two numbers whose product equals the constant term 15 (the coefficient of y²) and whose sum equals the middle coefficient ‑8 (the coefficient of xy). Those numbers are ‑3 and ‑5, because (‑3)·(‑5) = 15 and (‑3) + (‑5) = ‑8. This means the expression can be rewritten as (x ‑ 3y)(x ‑ 5y), which is the desired factorization It's one of those things that adds up. Simple as that..
Understanding this factorization is valuable because it simplifies the original expression, making it easier to solve equations, perform partial fraction decompositions, or analyze the sign of the expression for different values of x and y. On top of that, the technique of finding two numbers that meet both the product and sum criteria is a foundational skill that recurs throughout algebra, calculus, and even number theory Less friction, more output..
Step‑by‑Step or Concept Breakdown
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Identify the form – Recognize that x² – 8xy + 15y² is a quadratic in x with y acting as a constant coefficient. Write it as x² + (‑8y)x + 15y².
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Seek two numbers – Look for two numbers, say p and q, such that:
- p · q = 15 (the constant term)
- p + q = ‑8 (the coefficient of the mixed term, remembering that y is factored out).
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Determine p and q – The pair ‑3 and ‑5 satisfies both conditions: (‑3)·(‑5) = 15 and (‑3) + (‑5) = ‑8.
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Rewrite the middle term – Replace ‑8xy with ‑3xy ‑ 5xy, giving x² ‑ 3xy ‑ 5xy + 15y².
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Group and factor – Group the terms as (x² ‑ 3xy) + (‑5xy + 15y²). Factor out the common factor in each group:
- From the first group, factor out x: x(x ‑ 3y)
- From the second group, factor out ‑5y: ‑5y(x ‑ 3y)
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Combine the common binomial – Both groups now contain (x ‑ 3y), so the expression becomes (x ‑ 3y)(x ‑ 5y).
This systematic approach—identifying the correct pair, rewriting, grouping, and extracting common factors—ensures that any quadratic of the form ax² + bxy + cy² can be factored efficiently, provided the discriminant is a perfect square.
Real Examples
Example 1 – Solving an equation:
Consider the equation x² – 8xy + 15y² = 0. Factoring gives (x ‑ 3y)(x ‑ 5y) = 0, which implies either x = 3y or x = 5y. Thus, the solutions are lines through the origin with slopes 3 and 5, illustrating how factorization transforms a quadratic equation into a product of linear factors, each representing a straight line.
Example 2 – Simplifying rational expressions:
If we have the fraction (x² – 8xy + 15y²) / (x ‑ 5y), factoring the numerator yields (x ‑ 3y)(x ‑ 5y) / (x ‑ 5y). Canceling the common factor (x ‑ 5y) (provided x ≠ 5y) simplifies the expression to x ‑ 3y, demonstrating how factorization can reduce complexity in algebraic
