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

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.