Factor the following expression completely. x^22+x

To factor the expression \(x^{22} + x\) completely, we can start by factoring out the greatest common factor. In this case, both terms in the expression have a common factor of \(x\). Thus, we can factor out \(x\):<br /><br />\[<br />x^{22} + x = x(x^{21} + 1)<br />\]<br /><br />Now, we need to check if the expression inside the parentheses, \(x^{21} + 1\), can be factored further. Notice that \(x^{21} + 1\) is a sum of powers, and it can be expressed as:<br /><br />\[<br />x^{21} + 1 = x^{21} + 1^1<br />\]<br /><br />This expression can be factored using the sum of cubes formula, which states that for any \(a\) and \(b\),<br /><br />\[<br />a^3 + b^3 = (a + b)(a^2 - ab + b^2)<br />\]<br /><br />However, \(x^{21} + 1\) is not directly a sum of cubes, but we can express it as a sum of two terms where one term is a power of 3:<br /><br />\[<br />x^{21} + 1 = (x^7)^3 + 1^3<br />\]<br /><br />Applying the sum of cubes formula with \(a = x^7\) and \(b = 1\), we get:<br /><br />\[<br />(x^7)^3 + 1^3 = (x^7 + 1)((x^7)^2 - x^7 \cdot 1 + 1^2)<br />\]<br /><br />Simplifying the second factor:<br /><br />\[<br />(x^7)^2 - x^7 \cdot 1 + 1^2 = x^{14} - x^7 + 1<br />\]<br /><br />Thus, the factorization becomes:<br /><br />\[<br />x(x^7 + 1)(x^{14} - x^7 + 1)<br />\]<br /><br />The expression \(x^7 + 1\) cannot be factored further over the integers, and neither can \(x^{14} - x^7 + 1\).<br /><br />Therefore, the complete factorization of the original expression \(x^{22} + x\) is:<br /><br />\[<br />x(x^7 + 1)(x^{14} - x^7 + 1)<br />\]