Factor Theorem: Solving a^5-32

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In summary, the conversation discusses how to factor the expression a^5-32 using the factor theorem. It is mentioned that a-2 is a factor of the function, and the rule for factoring a^n-b^n is also provided. The individual asking the question initially thought it was only possible to use f(x) and not f(a), but now understands how to factor the expression.
  • #1
Rowah
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Simple Factoring Problem

How do I factor [tex]a^5-32[/tex]?

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

By the way, we're currently learning the factor theorem.
 
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  • #2
let f(a)= [tex]a^5-32[/tex] .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?
 
  • #3
Use the rule [tex]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}[/tex].
 
  • #4
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).
 
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Related to Factor Theorem: Solving a^5-32

1. What is the Factor Theorem?

The Factor Theorem is a mathematical theorem that helps to find the roots or solutions of a polynomial equation. It states that if a number, x, is a root of a polynomial equation, then (x-a) is a factor of the equation.

2. How is the Factor Theorem used to solve equations?

The Factor Theorem is used to simplify and solve polynomial equations by factoring them into smaller, simpler expressions. This allows us to find the roots or solutions of the equation, which are the values of x that make the equation equal to 0.

3. What is the general form of a polynomial equation?

The general form of a polynomial equation is ax^n + bx^(n-1) + cx^(n-2) + ... + k, where a, b, c, and k are constants, x is the variable, and n is the highest degree of the polynomial.

4. How can I apply the Factor Theorem to solve a^5-32=0?

To solve a^5-32=0 using the Factor Theorem, we need to first rearrange the equation to the general form of a polynomial equation, which is a^5-32=0. Then, we can factor out (a-2), since 2 is a root of the equation. This gives us (a-2)(a^4+2a^3+4a^2+8a+16)=0. Therefore, the solutions to the equation are a=2 and the four other roots of the second factor.

5. Can the Factor Theorem be used to solve equations with other variables besides x?

Yes, the Factor Theorem can be used to solve equations with any variable, as long as it is in the form of a polynomial equation. The variable does not have to be x, it can be any letter or symbol. The key is to identify the roots or solutions of the equation and then use the Factor Theorem to factor out the corresponding expressions.

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