Chapter 10: Problem 59
Factor. $$2 x^{2}-5 x-6 x y+15 y$$
Short Answer
Expert verified
The factoring of \(2x^{2}-5x-6xy+15y\) results in \((2x - 5)(x - 3y)\)
Step by step solution
01
Rearrange the terms
Rearrange the components in the equation to group like terms together. This will result in \(2x^{2}-5x-6xy+15y\) becoming \(2x^{2}-5x-6xy+15y\)
02
Look for common factors
Seek out common factors within grouped terms. In 2x^{2} and -5x, x can be factored out and in -6xy and 15y, y can be factored out. This results in \(x(2x - 5) - y(6x -15)\).
03
Simplify the equation
Upon inspection, it's possible to further factorize the two terms in the parenthesis by pulling out the common factors. The equation simplifies to \(x(2x - 5) - 3y(2x -5)\).
04
Apply the distributive property
The distributive property allows us to factor the common binomial (2x - 5) out of the two terms. This results in \((2x - 5)(x - 3y)\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Distributive Property
The distributive property is a valuable tool in algebra that helps us rewrite expressions and factor polynomials efficiently. It essentially states that for any three numbers, a, b, and c, the equation \(a(b + c) = ab + ac\) holds true. This property is crucial when factoring polynomials because it allows us to "distribute" terms across others or "factor" a common term out.
For example, in the exercise provided, the expression \(x(2x - 5) - 3y(2x - 5)\) implements the distributive property. We notice that \((2x - 5)\) is a common factor in both terms. By applying the distributive property here, we can rewrite the expression as \((2x - 5)(x - 3y)\). This is because we're essentially saying that both initial terms can be multiplied by the same expression, \((2x - 5)\), consolidating them into a single product.
This property not only simplifies equations but also makes it easier to solve them or find their roots.
For example, in the exercise provided, the expression \(x(2x - 5) - 3y(2x - 5)\) implements the distributive property. We notice that \((2x - 5)\) is a common factor in both terms. By applying the distributive property here, we can rewrite the expression as \((2x - 5)(x - 3y)\). This is because we're essentially saying that both initial terms can be multiplied by the same expression, \((2x - 5)\), consolidating them into a single product.
This property not only simplifies equations but also makes it easier to solve them or find their roots.
Common Factors
Finding common factors is a fundamental part of simplifying and factoring polynomials. A common factor is a term that appears in both parts of an expression. Identifying these can significantly simplify the process of factoring a polynomial.
In our original problem, we started with the expression \(2x^2 - 5x - 6xy + 15y\). To begin factoring this polynomial, we looked for common factors in the grouped terms:
In our original problem, we started with the expression \(2x^2 - 5x - 6xy + 15y\). To begin factoring this polynomial, we looked for common factors in the grouped terms:
- In the first pair, \(2x^2 - 5x\), the common factor is \(x\), giving us \(x(2x - 5)\).
- In the second pair, \(-6xy + 15y\), the common factor is \(y\), which simplifies to \(-y(6x - 15)\).
Grouping Like Terms
When factoring polynomials, grouping like terms is a strategy used to identify and extract common factors more easily. By rearranging the terms of a polynomial, it becomes simpler to see which terms can be grouped together based on common factors.
In the initial polynomial \(2x^2 - 5x - 6xy + 15y\), the terms were restructured to make the process of factoring more manageable. Let’s break it down:
In the initial polynomial \(2x^2 - 5x - 6xy + 15y\), the terms were restructured to make the process of factoring more manageable. Let’s break it down:
- We looked at \(2x^2 - 5x\) and \(-6xy + 15y\), separating terms that could share a common factor.
- This grouping highlighted potential factors that each segment of the equation could share, such as \(x\) in the first group and \(y\) in the second.