Problem 7
Question
Factor each binomial completely. $$ m^{4}-n^{4} $$
Step-by-Step Solution
Verified Answer
Answer: The completely factored form of the binomial \(m^4 - n^4\) is \((m - n)(m + n)(m^2 + n^2)\).
1Step 1: Identify the difference of two squares
Observe that the given binomial is a difference of two squares, where \(m^4\) represents the square of \(m^2\) and \(n^4\) represents the square of \(n^2\).
$$
m^4 - n^4
$$
2Step 2: Apply the difference of squares formula
Apply the difference of squares formula: \(a^2 - b^2 = (a-b)(a+b)\). In this case, \(a = m^2\) and \(b = n^2\).
$$
(m^2 - n^2)(m^2 + n^2)
$$
3Step 3: Factor again using the difference of squares formula
We can see that \((m^2 - n^2)\) is also a difference of two squares. Apply the difference of squares formula again:
$$
(m - n)(m + n)(m^2 + n^2)
$$
4Step 4: Write the final factored form
The completely factored form of the given binomial is:
$$
(m - n)(m + n)(m^2 + n^2)
$$
Key Concepts
Binomial FactorizationAlgebraic ExpressionsDifference of Squares Formula
Binomial Factorization
Binomial factorization involves expressing a binomial — a polynomial with two terms — as the product of two or more simpler expressions. When you encounter an algebraic expression like the one in our exercise, \( m^4 - n^4 \), it appears complex, but with knowledge of factoring techniques, we can simplify it significantly.
The process typically starts by identifying a pattern that matches a factoring formula, such as the difference of two squares. This identification is crucial since applying the right formula simplifies the binomial into a product of factors, which makes it more manageable for further algebraic manipulation or solving equations.
The process typically starts by identifying a pattern that matches a factoring formula, such as the difference of two squares. This identification is crucial since applying the right formula simplifies the binomial into a product of factors, which makes it more manageable for further algebraic manipulation or solving equations.
Recognizing Patterns
As the exercise highlights, recognizing the underlying structure of the binomial is the first step. We look for squares, cubes, or other patterns such as a common factor that can be pulled out of both terms. Factoring not only makes complex calculations easier but is also essential for functions analysis, solving quadratic equations, and simplifying algebraic fractions.Algebraic Expressions
An algebraic expression is a mathematical phrase that can contain ordinary numbers, variables (like x or y), and operations (such as addition, subtraction, multiplication, and division). The beauty of algebraic expressions lies in their ability to represent relationships and to be manipulated according to established algebraic rules and properties.
In contexts like our exercise \( m^4 - n^4 \), we see a high-power polynomial. Algebra allows us to simplify and understand these expressions by breaking them down into their basic components or factors. This simplification is fundamental when solving equations, understanding function behavior, and finding solutions to various mathematical problems.
In contexts like our exercise \( m^4 - n^4 \), we see a high-power polynomial. Algebra allows us to simplify and understand these expressions by breaking them down into their basic components or factors. This simplification is fundamental when solving equations, understanding function behavior, and finding solutions to various mathematical problems.
Manipulating Expressions
Through various algebraic techniques such as expanding, factoring, and simplifying, we can re-write expressions in different forms to suit the purpose of our mathematical endeavor. Factoring high-power polynomials, as shown in the exercise, is a testament to the power of algebraic manipulation.Difference of Squares Formula
The difference of squares formula is a specific case of binomial factorization. It states that for any two terms \( a \) and \( b \), their squares' difference can be factored into the product of the sum and the difference of the two terms: \( a^2 - b^2 = (a - b)(a + b) \).
In practice, like in our exercise with \( m^4 - n^4 \), the formula can be used repeatedly if the terms themselves are also differences of squares. This nested factoring is particularly useful in algebra for breaking down complex polynomials into simpler, more digestible factors.
In practice, like in our exercise with \( m^4 - n^4 \), the formula can be used repeatedly if the terms themselves are also differences of squares. This nested factoring is particularly useful in algebra for breaking down complex polynomials into simpler, more digestible factors.
Further Insights
The formula not only simplifies calculation but helps understand the symmetrical properties of algebraic expressions. It is a brilliant example of how algebraic identities can be powerful tools in dissecting and solving polynomial expressions. The exercise's solution illustrates a classical application of this formula, turning a formidable high-power polynomial into a product of manageable binomial factors.Other exercises in this chapter
Problem 7
Factor, if possible, the following trinomials. \(x^{2}-x y-6 y^{2}\)
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Factor the following, if possible. $$ 14 p^{2}+31 p q-10 q^{2} $$
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For the following problems, use the grouping method to factor the polynomials. Some polynomials may not. be factorable using the grouping method. $$ 2 a b+3 a+1
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For the following problems, the first quantity represents the product and the second quantity a factor. Find the other factor.
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