Problem 23
Question
Factor each difference of two squares. $$x^{4}-16$$
Step-by-Step Solution
Verified Answer
Therefore, the factored form of \(x^{4} - 16\) is \((x - 2)(x + 2)(x^{2} + 4)\)
1Step 1: Identifying the Values of a and b
Comparing the equation \(x^{4}-16\) to the standard form \(a^{2}-b^{2}\), we can identify \(a=x^{2}\) and \(b=4\). Verify these values by squaring each of them: \((x^{2})^{2}= x^{4}\) and \(4^{2}=16\), which matches the original equation.
2Step 2: Apply the Difference of Two Squares Formula
Substitute x² for a and 4 for b in the formula. This will be as follows: \(a^{2} - b^{2} = (a - b)(a + b) \Rightarrow (x^{2})^{2} - 4^{2} = (x^{2} - 4)(x^{2} + 4)\)
3Step 3: Repeat Factoring on Derived Binomial Expression
The term (x^2 - 4) is a difference of squares and can be factored again. We'll apply the formula one more time with a = x and b = 2: \((x^{2} - 4) = (x - 2)(x + 2)\), so the final solution becomes \((x - 2)(x + 2)(x^{2} + 4)\)
Key Concepts
FactoringPolynomial ExpressionsAlgebraic Identities
Factoring
Factoring refers to the process of breaking down a more complex expression into simpler pieces called factors, which when multiplied together give back the original expression. This is an essential skill in algebra, as it helps simplify expressions and solve equations.
When you encounter an expression like a difference of squares, such as \(x^4 - 16\), you'll use factoring to rewrite it in a simpler form. First, you identify the parts of the expression that make it a difference of squares. With the given example, you notice that both \(x^4\) and \(16\) are perfect squares. The expression can be factored using the difference of squares formula, \((a^2 - b^2 = (a - b)(a + b))\).
This formula allows you to express \(x^4 - 16\) as \((x^2)^2 - 4^2\), which factors into \((x^2 - 4)(x^2 + 4)\). By factoring each part completely, you decompose the expression into simpler, solvable parts. The ability to factor expressions like these is vital in solving polynomial equations and simplifying rational expressions.
When you encounter an expression like a difference of squares, such as \(x^4 - 16\), you'll use factoring to rewrite it in a simpler form. First, you identify the parts of the expression that make it a difference of squares. With the given example, you notice that both \(x^4\) and \(16\) are perfect squares. The expression can be factored using the difference of squares formula, \((a^2 - b^2 = (a - b)(a + b))\).
This formula allows you to express \(x^4 - 16\) as \((x^2)^2 - 4^2\), which factors into \((x^2 - 4)(x^2 + 4)\). By factoring each part completely, you decompose the expression into simpler, solvable parts. The ability to factor expressions like these is vital in solving polynomial equations and simplifying rational expressions.
Polynomial Expressions
Polynomial expressions consist of variables and constants combined using addition, subtraction, multiplication, and non-negative integer exponents. Polynomials are categorized by their degree, which is the highest power of the variable in the expression.
Consider the polynomial \(x^4 - 16\). This is a quartic polynomial because the highest degree is 4 (coming from \(x^4\)). When dealing with polynomial expressions, you often need to rearrange or simplify them, especially when further problems or solutions depend on having them in factored form.
The expression can be transformed using various algebraic techniques, such as factoring. By identifying patterns, like differences of squares, you can simplify polynomials, resolving expressions into linear or quadratic factors. When simplified, polynomials become easier to manipulate, making it possible to solve complex equations efficiently.
Consider the polynomial \(x^4 - 16\). This is a quartic polynomial because the highest degree is 4 (coming from \(x^4\)). When dealing with polynomial expressions, you often need to rearrange or simplify them, especially when further problems or solutions depend on having them in factored form.
The expression can be transformed using various algebraic techniques, such as factoring. By identifying patterns, like differences of squares, you can simplify polynomials, resolving expressions into linear or quadratic factors. When simplified, polynomials become easier to manipulate, making it possible to solve complex equations efficiently.
Algebraic Identities
Algebraic identities are mathematical expressions that hold true for all values of the variables involved. These identities are like tools in algebra that simplify expressions, equations, and problem-solving processes.
One important algebraic identity is the difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). This identity recognizes that whenever two perfect squares are subtracted, the expression can be factored into a product of two binomials. Utilizing this identity simplifies complex polynomial expressions and is vital for solving many algebraic problems.
When using the identity on \(x^4 - 16\), you efficiently transform it into simpler expressions ready for further manipulation: \((x^2 - 4)(x^2 + 4)\) and further \((x - 2)(x + 2)(x^2 + 4)\). By understanding and applying algebraic identities, you unlock a straightforward approach to handle seemingly complicated expressions, enhancing your algebraic problem-solving skillset.
One important algebraic identity is the difference of squares: \(a^2 - b^2 = (a - b)(a + b)\). This identity recognizes that whenever two perfect squares are subtracted, the expression can be factored into a product of two binomials. Utilizing this identity simplifies complex polynomial expressions and is vital for solving many algebraic problems.
When using the identity on \(x^4 - 16\), you efficiently transform it into simpler expressions ready for further manipulation: \((x^2 - 4)(x^2 + 4)\) and further \((x - 2)(x + 2)(x^2 + 4)\). By understanding and applying algebraic identities, you unlock a straightforward approach to handle seemingly complicated expressions, enhancing your algebraic problem-solving skillset.
Other exercises in this chapter
Problem 23
Factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$y^{2}-16 y+48$$
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Use factoring to solve each quadratic equation. Check by substitution or by using a graphing utility and identifying \(x\) -intercepts. $$2 x^{2}=5 x$$
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Now let's move on to factorizations that may require two or more techniques. Factor completely, or state that the polynomial is prime. Check factorizations usin
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Use the method of your choice to factor each trinomial, or state that the trinomial is prime. Check each factorization using FOIL multiplication. $$5 x^{2}+33 x
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