Question
Factor. \( 2 w^{8}-2 f^{8} \) \( \begin{array}{l}2 w^{8}-2 f^{8}=\square \\ (\text { Factor completely.) }\end{array} \)
Ask by Mann Dawson. in the United States
Feb 24,2025
Upstudy AI Solution
Tutor-Verified Answer
Answer
\( 2(w - f)(w + f)(w² + f²)(w⁴ + f⁴) \)
Solution
Step 1. Factor out the common coefficient.
2w⁸ – 2f⁸ = 2(w⁸ – f⁸)
Step 2. Recognize the difference of two eighth powers as a difference of two squares.
w⁸ – f⁸ = (w⁴)² – (f⁴)² = (w⁴ – f⁴)(w⁴ + f⁴)
Step 3. Factor the difference of squares w⁴ – f⁴ further.
w⁴ – f⁴ = (w²)² – (f²)² = (w² – f²)(w² + f²)
Step 4. Factor the difference of squares w² – f².
w² – f² = (w – f)(w + f)
Step 5. Put it all together.
2(w⁸ – f⁸) = 2(w⁴ – f⁴)(w⁴ + f⁴)
= 2[(w² – f²)(w² + f²)](w⁴ + f⁴)
= 2(w – f)(w + f)(w² + f²)(w⁴ + f⁴)
Since w⁴ + f⁴ does not factor further into polynomials with integer coefficients, this is the complete factorization.
Final Answer:
2(w – f)(w + f)(w² + f²)(w⁴ + f⁴)
Answered by UpStudy AI and reviewed by a Professional Tutor
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Simplify this solution The Deep Dive
First, notice that you can factor out a common factor of \(2\) from the expression: \(2(w^{8} - f^{8})\). Now, we can recognize \(w^{8} - f^{8}\) as a difference of squares, since \(w^{8} = (w^{4})^2\) and \(f^{8} = (f^{4})^2\). This gives us: \(2(w^{4} - f^{4})(w^{4} + f^{4})\). Next, \(w^{4} - f^{4}\) is also a difference of squares, allowing us to factor it further: \(2(w^{2} - f^{2})(w^{2} + f^{2})(w^{4} + f^{4})\). The expression is completely factored as: \[ 2(w^{2} - f^{2})(w^{2} + f^{2})(w^{4} + f^{4}). \]
