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Chapter 7: Problem 39

Explain why the polynomial factorization $$ 1-x^{n}=(1-x)\left(1+x+x^{2}+\cdots+x^{n-1}\right) $$ holds for every integer \(n \geq 2\).

Short Answer

Expert verified
The given equation states that \(1-x^n = (1-x)(1+x+x^2+\cdots+x^{n-1})\). To prove this, we distribute the term (1-x) to the terms in the series and observe the cancellation of intermediate terms, resulting in \(1-x^n = (1 - x^{n})\). Thus, the equation holds for every integer \(n \geq 2\).

Step by step solution

01

Write the general form of the given equation

Let's write the given equation as : \[ (1-x^{n}) = (1-x)(A), \] where \(A = 1+x+x^{2}+\cdots+x^{n-1}\).
02

Multiply out the right side

We will now multiply (1-x) by A and observe the result. \[ (1-x)(A) = (1-x)(1+x+x^{2}+\cdots+x^{n-1}). \]
03

Distribute (1-x) to the terms in A

The next step is to distribute (1-x) to each term in the series A: \begin{align*} (1-x)(1+x+x^{2}+\cdots+x^{n-1}) &= (1)(1+x+x^{2}+\cdots+x^{n-1}) \\ &\quad -(x)(1+x+x^{2}+\cdots+x^{n-1}). \end{align*} We can now see a pattern emerging when we distribute the terms.
04

Write out the subtraction of terms

Now, we will write out the subtraction of terms: \begin{align*} (1-x)(1+x+\cdots+x^{n-1}) &= (1+x+x^{2}+\cdots+x^{n-1}) \\ &\quad - (x+x^{2}+\cdots + x^{n}). \end{align*}
05

Combine the terms

Now, we can combine the terms as follows: \begin{align*} (1-x)(1+x+\cdots+x^{n-1}) &= (1 - x^{n}). \end{align*} Notice that all the intermediate terms cancel each other, leaving only the first and the last term on the right side.
06

Conclusion

So, we have proved the given equation: \[ 1-x^{n}=(1-x)\left(1+x+x^{2}+\cdots+x^{n-1}\right), \] which holds for every integer \(n \geq 2\).

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