x3 + x 2 + 16 is exactly divisible by x, where x is a positive integer. The number of all such possible values of x is

Question

x3 +  x 2 + 16 is exactly divisible by x, where x is a positive integer. The number of all such possible values of x is

Prepp Team · Accepted Answer

Finding Positive Integer Values for Divisibility

The problem asks us to find the number of positive integer values for \(x\) such that the expression \(x^3 + x^2 + 16\) is exactly divisible by \(x\).

For an expression to be exactly divisible by a number, the remainder of the division must be zero. In this case, we need to divide \(x^3 + x^2 + 16\) by \(x\).

We can look at each term in the expression separately:

The first term is \(x^3\). When \(x^3\) is divided by \(x\), the result is \(x^2\). This division is exact for any \(x \neq 0\).
The second term is \(x^2\). When \(x^2\) is divided by \(x\), the result is \(x\). This division is also exact for any \(x \neq 0\).
The third term is \(16\). When \(16\) is divided by \(x\), the division is exact only if \(x\) is a factor (or divisor) of \(16\).

For the entire expression \(x^3 + x^2 + 16\) to be exactly divisible by \(x\), both \(x^3 + x^2\) and \(16\) must be divisible by \(x\). Since \(x^3 + x^2\) is always divisible by \(x\) for \(x \neq 0\), the condition for the entire expression to be divisible by \(x\) is simply that \(16\) must be divisible by \(x\).

The problem states that \(x\) is a positive integer. Therefore, \(x\) must be a positive integer divisor of \(16\).

Let's find the positive integer divisors of \(16\). These are the positive integers that divide \(16\) evenly.

We can list them:

\(16 \div 1 = 16\) (1 is a divisor)
\(16 \div 2 = 8\) (2 is a divisor)
\(16 \div 3\) (not an integer)
\(16 \div 4 = 4\) (4 is a divisor)
\(16 \div 5\) (not an integer)
\(16 \div 6\) (not an integer)
\(16 \div 7\) (not an integer)
\(16 \div 8 = 2\) (8 is a divisor)
\(16 \div 9\) to \(15\) (not integers)
\(16 \div 16 = 1\) (16 is a divisor)

The positive integer values of \(x\) for which \(x^3 + x^2 + 16\) is exactly divisible by \(x\) are the positive divisors of \(16\). These values are \(1, 2, 4, 8, \text{ and } 16\).

Now, we need to find the number of such possible values of \(x\). Counting the values we found:

Value 1: \(1\)
Value 2: \(2\)
Value 3: \(4\)
Value 4: \(8\)
Value 5: \(16\)

There are \(5\) possible positive integer values for \(x\).

The number of all such possible values of \(x\) is \(5\).

Revision Table: Key Concepts

Concept	Explanation	Relevance to Problem
Divisibility	A number \(a\) is divisible by \(b\) if dividing \(a\) by \(b\) leaves a remainder of 0.	The expression must be exactly divisible by \(x\).
Polynomial Division	Dividing a polynomial by a term. If dividing by \(x\), each term in the polynomial must be divisible by \(x\).	Used to determine the condition \(16\) must be divisible by \(x\).
Positive Integer	A whole number greater than 0 (e.g., 1, 2, 3, ...).	The variable \(x\) is restricted to be a positive integer.
Divisors (Factors)	Numbers that divide another number exactly.	\(x\) must be a positive divisor of \(16\).

Additional Information: Understanding Polynomial Divisibility

When you have a polynomial and you are checking for divisibility by a variable like \(x\), you can often simplify the problem by looking at the terms that do not contain \(x\). For example, if you have \(ax^n + bx^{n-1} + \dots + cx + d\), and you want to check if it's divisible by \(x\) (where \(x \neq 0\)), all terms with \(x\) in them (like \(ax^n, bx^{n-1}, \dots, cx\)) are divisible by \(x\). The divisibility of the entire polynomial then depends solely on whether the constant term (\(d\) in this example) is divisible by \(x\).

In our problem, the expression is \(x^3 + x^2 + 16\). The terms \(x^3\) and \(x^2\) are divisible by \(x\). The constant term is \(16\). Thus, the condition for \(x^3 + x^2 + 16\) to be divisible by \(x\) is that \(16\) must be divisible by \(x\).

This principle applies whenever the divisor is the variable itself (like \(x\), \(y\), \(z\), etc.) and the terms in the polynomial have powers of that variable, plus possibly a constant term.

To find the positive integer values of \(x\) that divide \(16\), we list the factors of \(16\): \(1, 2, 4, 8, 16\). Each of these numbers is a positive integer, and they all divide \(16\) exactly. There are \(5\) such numbers.