(x n- a n) is divisible by (x-a), where x ≠ a, for every _____.

Question

Prepp Team · Accepted Answer

Understanding Divisibility of $(x^n - a^n)$ The question asks us to determine for which values of the natural number exponent $n$ the expression $(x^n - a^n)$ is perfectly divisible by $(x-a)$, assuming $x \neq a$. Divisibility in this context means that when you divide $(x^n - a^n)$ by $(x-a)$, the remainder is zero. Using the Polynomial Remainder Theorem for Divisibility A fundamental concept in algebra is the Polynomial Remainder Theorem. This theorem states that if a polynomial $P(x)$ is divided by a linear factor $(x-a)$, the remainder of the division is equal to $P(a)$. In our case, the polynomial is $P(x) = x^n - a^n$. We are dividing this polynomial by $(x-a)$. According to the Remainder Theorem, the remainder is $P(a)$. Let's calculate $P(a)$: $P(a) = a^n - a^n = 0$ Since the remainder is $0$, this means that $(x^n - a^n)$ is perfectly divisible by $(x-a)$ for any value of $n$ for which $x^n - a^n$ is a valid polynomial. The expression $x^n - a^n$ is a standard polynomial form when $n$ is a positive integer (a natural number). Alternative Method: Factorization of the Difference of Powers We can also look at the factorization of the difference of two powers. The general formula for $x^n - a^n$ for any positive integer $n$ is: $x^n - a^n = (x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + \dots + xa^{n-2} + a^{n-1})$ Let's look at this formula for a few values of $n$: For $n=1$: $x^1 - a^1 = (x-a)$, which is clearly divisible by $(x-a)$. For $n=2$: $x^2 - a^2 = (x-a)(x+a)$, which is divisible by $(x-a)$. For $n=3$: $x^3 - a^3 = (x-a)(x^2 + xa + a^2)$, which is divisible by $(x-a)$. For $n=4$: $x^4 - a^4 = (x-a)(x^3 + x^2a + xa^2 + a^3)$, which is divisible by $(x-a)$. The factorization formula shows that for any natural number $n \ge 1$, $(x^n - a^n)$ can be expressed as the product of $(x-a)$ and another polynomial $(x^{n-1} + x^{n-2}a + \dots + a^{n-1})$. This confirms that $(x^n - a^n)$ is divisible by $(x-a)$ for all natural numbers $n$. Evaluating the Options Based on the Remainder Theorem and the factorization formula, we can assess the given options: natural number n: Our analysis shows that $(x^n - a^n)$ is divisible by $(x-a)$ for all natural numbers $n$ (1, 2, 3, ...). This option aligns with our findings. even natural number n only: This is incorrect because divisibility also holds for odd natural numbers (like $n=1, 3, 5, \dots$). odd natural number n only: This is incorrect because divisibility also holds for even natural numbers (like $n=2, 4, 6, \dots$). prime number n only: This is incorrect because divisibility holds for all natural numbers, including non-prime natural numbers like 1, 4, 6, 8, 9, 10, 12, etc. Therefore, the expression $(x^n - a^n)$ is divisible by $(x-a)$ for every natural number $n$, provided $x \neq a$. Conclusion The condition that $(x^n - a^n)$ is divisible by $(x-a)$ holds true for every natural number $n$. This is a fundamental result derived from polynomial division and factorization. Exponent $n$ Expression $(x^n - a^n)$ Factorization Divisible by $(x-a)$? 1 (Natural) $x-a$ $(x-a) \times 1$ Yes 2 (Natural, Even) $x^2-a^2$ $(x-a)(x+a)$ Yes 3 (Natural, Odd) $x^3-a^3$ $(x-a)(x^2+xa+a^2)$ Yes 4 (Natural, Even) $x^4-a^4$ $(x-a)(x^3+x^2a+xa^2+a^3)$ Yes Any Natural $n$ $x^n-a^n$ $(x-a)(x^{n-1} + \dots + a^{n-1})$ Yes Revision Table: Divisibility Rules Expression Divisible by Condition on $n$ $x^n - a^n$ $(x-a)$ Every natural number $n$ $x^n - a^n$ $(x+a)$ Every even natural number $n$ $x^n + a^n$ $(x+a)$ Every odd natural number $n$ $x^n + a^n$ $(x-a)$ Never (unless $x=a$ or $n=0$, but question states $x \neq a$) Additional Information: Polynomial Factorization and Roots The fact that $(x^n - a^n)$ is divisible by $(x-a)$ for all natural numbers $n$ is directly related to the roots of the polynomial $P(x) = x^n - a^n$. A polynomial $P(x)$ is divisible by $(x-a)$ if and only if $a$ is a root of the polynomial, i.e., $P(a) = 0$. As shown earlier, $P(a) = a^n - a^n = 0$, regardless of the natural number $n$. This confirms $a$ is always a root of $x^n - a^n$, hence it is always divisible by $(x-a)$. This property is widely used in algebra, particularly in simplifying expressions and solving equations.

Prepp Team · Answer

even natural number n only

Prepp Team · Answer

odd natural number n only

Prepp Team · Answer

prime number n only

Exponent \(n\)	Expression \((x^n - a^n)\)	Factorization	Divisible by \((x-a)\)?
1 (Natural)	\(x-a\)	\((x-a) \times 1\)	Yes
2 (Natural, Even)	\(x^2-a^2\)	\((x-a)(x+a)\)	Yes
3 (Natural, Odd)	\(x^3-a^3\)	\((x-a)(x^2+xa+a^2)\)	Yes
4 (Natural, Even)	\(x^4-a^4\)	\((x-a)(x^3+x^2a+xa^2+a^3)\)	Yes
Any Natural \(n\)	\(x^n-a^n\)	\((x-a)(x^{n-1} + \dots + a^{n-1})\)	Yes

Expression	Divisible by	Condition on \(n\)
\(x^n - a^n\)	\((x-a)\)	Every natural number \(n\)
\(x^n - a^n\)	\((x+a)\)	Every even natural number \(n\)
\(x^n + a^n\)	\((x+a)\)	Every odd natural number \(n\)
\(x^n + a^n\)	\((x-a)\)	Never (unless \(x=a\) or \(n=0\), but question states \(x \neq a\))

Understanding Divisibility of \((x^n - a^n)\)

Using the Polynomial Remainder Theorem for Divisibility

Alternative Method: Factorization of the Difference of Powers

Evaluating the Options

Conclusion

Revision Table: Divisibility Rules

Additional Information: Polynomial Factorization and Roots

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