Factor Theorem: Solving a^5-32

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SUMMARY

The discussion focuses on factoring the polynomial expression a^5 - 32 using the Factor Theorem. The user identifies that f(a) = a^5 - 32 has a root at a = 2, confirming that a - 2 is a factor. The conversation highlights the application of the formula for factoring differences of powers, specifically a^n - b^n = (a - b)(a^{n-1} + a^{n-2}b + ... + b^{n-1}), where n is a natural number. This method effectively simplifies the polynomial expression.

PREREQUISITES
  • Understanding of the Factor Theorem
  • Familiarity with polynomial functions
  • Knowledge of factoring techniques for differences of powers
  • Basic algebraic manipulation skills
NEXT STEPS
  • Study the Factor Theorem in depth
  • Learn about polynomial long division
  • Explore the application of the Rational Root Theorem
  • Practice factoring higher-degree polynomials
USEFUL FOR

Students learning algebra, educators teaching polynomial functions, and anyone looking to enhance their skills in factoring polynomials using the Factor Theorem.

Rowah
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Simple Factoring Problem

How do I factor a^5-32?

It isn't a difference of cubes or even a difference of squares :(

By the way, we're currently learning the factor theorem.
 
Last edited:
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let f(a)= a^5-32 .when x=2,f(a)=0 so a-2 is a factor of the funtion. suppose you know how to carry on since you learned factor theorem?
 
Use the rule a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+a^{n-3}b^2+\cdots+ab^{n-2}+b^{n-1}), n \in \textbf{N}.
 
Cool, I understand now, thanks, "a" messed me up, I didn't know you could do f(a), I thought it was only f(x).
 
Last edited:

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