What is the remainder of the following division problem?
$\frac{1}{x \longdiv { x + 1 }}$
$x$
$\overline{0+1}$
The expression provided is written in a traditional long division format: \[\frac{1}{x \overline{) x + 1 }}\] In this notation, the polynomial \(p(x) = x + 1\) is the dividend, and the polynomial \(d(x) = x\) is the divisor. The objective is to determine the remainder of this division. According to the division algorithm for polynomials, for any dividend \(p(x)\) and a non-zero divisor \(d(x)\), there exist a unique quotient \(q(x)\) and a unique remainder \(r(x)\) such that \(p(x) = d(x)q(x) + r(x)\), where either \(r(x) = 0\) or the degree of \(r(x)\) is strictly less than the degree of \(d(x)\).
To perform the division of \(x + 1\) by \(x\), we look for a constant \(q(x) = c\) and a remainder \(r(x) = r\) such that: \[x + 1 = x \cdot c + r\] We can rewrite the left side to isolate terms with \(x\): \[x + 1 = 1 \cdot x + 1\] By inspection, we see that the coefficient \(c = 1\) satisfies the equation for the term containing \(x\), and the constant term \(r = 1\) satisfies the remainder condition. Since the degree of the divisor is 1, the remainder must have a degree less than 1, which means it must be a constant. Here, the remainder is 1, which is a constant and satisfies the degree condition (degree of 1 is 0, which is less than 1).
Comparing our calculated remainder, which is 1, with the given options, we find it corresponds perfectly to the choice labeled '1'. Therefore, the remainder of the division of \(x+1\) by \(x\) is 1.
Consider you are dividing a group of 11 items into boxes that hold 10 items each. If you represent the number of items as \(x+1\) (where \(x=10\)), the division \(x+1\) divided by \(x\) means you fill 1 box (the quotient) and have 1 item left over (the remainder). This is a foundational concept in algebra that helps us understand how remainders function when dividing variables and constants.