Problem 92
Question
Factor using the formula for the sum or difference of two cubes. $$128-250 y^{3}$$
Step-by-Step Solution
Verified Answer
The factored form of \(128 - 250y^3\) is \((4 - 5y)(16 + 20y + 25y^2)\).
1Step 1: Identify a and b
The given equation is \(128-250y^3\). Here, \(128\) can be written as \(4^3\) and \(250y^3\) can be written as \((5y)^3\). Thus, \(a=4\) and \(b=5y\).
2Step 2: Apply a and b to formula for the difference of cubes
The formula for the difference of two cubes is \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\). Substitute \(a=4\) and \(b=5y\) to the formula: \((4-5y)((4)^2 + (4)(5y) + (5y)^2)\).
3Step 3: Simplify the equation
Simplify the equation to get the factored form: \((4-5y)(16 + 20y + 25y^2)\).
Key Concepts
Understanding the Difference of CubesExploring Different Factoring TechniquesThe Nature of Algebraic Expressions
Understanding the Difference of Cubes
The difference of cubes is a specific form of polynomial that can be factored using a particular formula. It involves expressions that fit the pattern \(a^3 - b^3\), where both terms are perfect cubes. A perfect cube means that the term can be expressed as another number raised to the power of three. For example, \(8\) is a perfect cube of \(2\), as \(2^3 = 8\).
In our exercise, \(128\) and \(250y^3\) are restructured to represent cubes, \(128 = 4^3\) and \(250y^3 = (5y)^3\). Once identified, the formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) is used to factor the expression.
The key steps include: identifying the cube roots \(a\) and \(b\) and applying them into the formula. This simplification turns what may seem like a complex polynomial into a neatly factored expression.
In our exercise, \(128\) and \(250y^3\) are restructured to represent cubes, \(128 = 4^3\) and \(250y^3 = (5y)^3\). Once identified, the formula \(a^3 - b^3 = (a-b)(a^2 + ab + b^2)\) is used to factor the expression.
The key steps include: identifying the cube roots \(a\) and \(b\) and applying them into the formula. This simplification turns what may seem like a complex polynomial into a neatly factored expression.
Exploring Different Factoring Techniques
Factoring polynomials is an essential skill in algebra that involves breaking down expressions into simpler factors or components. The process can vary based on the type of polynomials you have.Different techniques include:
\[(a-b)(a^2 + ab + b^2)\]
Using the exercises where we had \(a = 4\) and \(b = 5y\), the expression can be efficiently resolved using this formula, showcasing how structured processes can make complex algebra easier.
- Factoring out the greatest common factor (GCF)
- Factoring by grouping
- Using special formulas like the difference of squares or cubes
- Applying the quadratic formula in some cases
\[(a-b)(a^2 + ab + b^2)\]
Using the exercises where we had \(a = 4\) and \(b = 5y\), the expression can be efficiently resolved using this formula, showcasing how structured processes can make complex algebra easier.
The Nature of Algebraic Expressions
Algebraic expressions are combinations of variables, constants, and arithmetic operations. They represent quantities and relationships and are foundational in solving equations.
In the given exercise, algebraic expressions include both the constant term \(128\) and the term \(250y^3\) which involves a variable. Understanding how to handle these expressions is crucial in algebra.
Algebraic expressions often need to be transformed through operations such as factoring, which simplifies expressions and solves equations more effortlessly. Breaking down expressions into their components, as demonstrated, shows how we manipulate these forms to achieve a solution.
Thus, dealing with expressions requires familiarity with methods like factoring techniques, allowing for efficient problem-solving in numerous mathematical situations.
In the given exercise, algebraic expressions include both the constant term \(128\) and the term \(250y^3\) which involves a variable. Understanding how to handle these expressions is crucial in algebra.
Algebraic expressions often need to be transformed through operations such as factoring, which simplifies expressions and solves equations more effortlessly. Breaking down expressions into their components, as demonstrated, shows how we manipulate these forms to achieve a solution.
Thus, dealing with expressions requires familiarity with methods like factoring techniques, allowing for efficient problem-solving in numerous mathematical situations.
Other exercises in this chapter
Problem 92
Determine whether each statement "makes sense" or "does not make sense" and explain your reasoning. I factor \(x^{2}+b x+c\) by finding two numbers that have a
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a. Factor \(3 x^{2}+5 x-2\) b. Use the factorization in part (a) to factor$$3(y+1)^{2}+5(y+1)-2$$ Then simplify each factor.
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Contain polynomials in several variables. Factor each polynomial completely and check using multiplication. $$16 a^{2}-32 a b+12 b^{2}$$
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