Problem 93
Question
Factor completely. Identify any prime polynomials. $$ n^{3}-64 p^{3} $$
Step-by-Step Solution
Verified Answer
Factor: \( n^{3} - 64p^{3} = (n - 4p)(n^{2} + 4np + 16p^{2}) \). The polynomial is not prime.
1Step 1: Recognize the form of the polynomial
The polynomial given is in the form of a difference of cubes: \[ n^{3} - 64p^{3} \].
2Step 2: Write the difference of cubes formula
The difference of cubes formula is: \[ a^{3} - b^{3} = (a - b)(a^{2} + ab + b^{2}) \].
3Step 3: Identify the terms
In this case, \( n^{3} \) is \( a^{3} \) and \( 64p^{3} \) is \( b^{3} \). So, \( a = n \) and \( b = 4p \).
4Step 4: Apply the difference of cubes formula
Substitute \( a = n \) and \( b = 4p \) into the formula:\[ n^{3} - 64p^{3} = (n - 4p)(n^{2} + n(4p) + (4p)^{2}) \].
5Step 5: Simplify the expression inside the parentheses
Simplify the terms:\[ n^{3} - 64p^{3} = (n - 4p)(n^{2} + 4np + 16p^{2}) \].
6Step 6: Confirm the factorization
Check the factorization:\[ n^{3} - 64p^{3} = (n - 4p)(n^{2} + 4np + 16p^{2}) \].
Key Concepts
Difference of CubesPolynomial FactorizationAlgebraic Expressions
Difference of Cubes
When dealing with polynomials, one powerful method is recognizing special forms like the difference of cubes.
A polynomial in the form of \(a^3 - b^3\) is known as a 'difference of cubes'.
There is a specific formula to factor such an expression: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
To use this formula, identify the cube roots of each term in the polynomial. In our example, \(n^3 - 64p^3\), we recognize that \(n^3\) and \(64p^3\) are both cubes.
Here, \(n\) is the cube root of \(n^3\) and \(4p\) is the cube root of \(64p^3\). Applying the formula becomes straightforward once these terms are identified. This method simplifies polynomials and highlights underlying patterns in algebraic expressions.
A polynomial in the form of \(a^3 - b^3\) is known as a 'difference of cubes'.
There is a specific formula to factor such an expression: \[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \]
To use this formula, identify the cube roots of each term in the polynomial. In our example, \(n^3 - 64p^3\), we recognize that \(n^3\) and \(64p^3\) are both cubes.
Here, \(n\) is the cube root of \(n^3\) and \(4p\) is the cube root of \(64p^3\). Applying the formula becomes straightforward once these terms are identified. This method simplifies polynomials and highlights underlying patterns in algebraic expressions.
Polynomial Factorization
Polynomial factorization is the process of breaking down a polynomial into simpler 'factor' polynomials that, when multiplied together, give the original polynomial.
This is particularly useful for solving equations and simplifying expressions.
In our exercise, we use the difference of cubes formula to factorize \(n^3 - 64p^3\).
We require recognizing the terms, substituting them into the formula, and simplifying the result.
Breaking down a polynomial into factors helps in various fields including algebra, calculus, and evolutionary algorithms because it simplifies complex problems into more manageable parts.
The factors often reveal important properties about the polynomial itself.
This is particularly useful for solving equations and simplifying expressions.
In our exercise, we use the difference of cubes formula to factorize \(n^3 - 64p^3\).
We require recognizing the terms, substituting them into the formula, and simplifying the result.
Breaking down a polynomial into factors helps in various fields including algebra, calculus, and evolutionary algorithms because it simplifies complex problems into more manageable parts.
The factors often reveal important properties about the polynomial itself.
Algebraic Expressions
Algebraic expressions involve variables, constants, and arithmetic operations.
These expressions form the basis of algebra.
For example, the expression \(n^3 - 64p^3\) includes the variables \(n\) and \(p\), a constant (64), and the operation of subtraction.
Understanding how to manipulate and factorize algebraic expressions is crucial for solving equations and understanding mathematical relationships.
By learning to handle expressions like \(n^3 - 64p^3\), students gain skills in recognizing patterns and applying formulas, which are fundamental in algebra.
The capability to simplify and factor expressions makes algebraic problems more approachable and less intimidating.
These expressions form the basis of algebra.
For example, the expression \(n^3 - 64p^3\) includes the variables \(n\) and \(p\), a constant (64), and the operation of subtraction.
Understanding how to manipulate and factorize algebraic expressions is crucial for solving equations and understanding mathematical relationships.
By learning to handle expressions like \(n^3 - 64p^3\), students gain skills in recognizing patterns and applying formulas, which are fundamental in algebra.
The capability to simplify and factor expressions makes algebraic problems more approachable and less intimidating.
Other exercises in this chapter
Problem 92
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For exercises 93-96, the completed problem has one mistake. (a) Describe the mistake in words, or copy down the whole problem and highlight or circle the mistak
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