📝 Zero property of multiplication (28 MCQs)

📖 Explanation: Multiplication by zero makes the entire product equal to zero, regardless of the other factor. The student incorrectly treats zero as though it affects only one part of the expression instead of the complete multiplication operation.

📖 Explanation: Area is found by multiplying length and width. Since the width is $0$, the multiplication becomes $x(0)$, which equals $0$. The physical model therefore supports the algebraic result rather than the student's claim.

📖 Explanation: Because $a$ is explicitly nonzero, it cannot be the factor responsible for making the product zero. Therefore $b$ must equal zero. The reasoning requires using the given condition before applying the zero-product relationship.

📖 Explanation: Since $5$ is nonzero, dividing both sides by $5$ is valid and gives $x-3=0$. Solving this equation produces $x=3$. The zero refers to the product's value, not automatically to the variable.

📖 Explanation: For $y=4x$, substituting $x=0$ gives $y=4(0)=0$. Therefore the graph passes through $(0,0)$. This visual interpretation connects the zero property with the coordinate behavior of a linear function.

📖 Explanation: The expression contains two separate terms. The canceled charge contributes $8(0)=0$, while production contributes $15(20)=300$. Adding these gives $300$. Careful grouping prevents the zero term from being incorrectly applied to the entire expression.

📖 Explanation: Both $p(0)$ and $q(0)$ equal zero independently, so their sum is zero no matter what values $p$ and $q$ have. Thus the equation provides no evidence that $p+q=0$; the student's inference loses information.

📖 Explanation: The student is confusing multiplication with addition. Adding zero preserves a number, but multiplying a number by zero produces zero. The distinction between these operations is essential for interpreting expressions correctly.

📖 Explanation: Substituting $n=0$ gives $P=25(0)=0$. The model assumes production is directly proportional to the number of operating machines, so when the number of operating machines is zero, predicted production is also zero.

📖 Explanation: Both methods lead to the same value. Since $x+0=x$, the expression becomes $6x$. Writing $6x+0$ is not wrong because adding zero changes nothing, although $6x$ is the simpler final form.

📖 Explanation: Each product on the left equals zero, so the entire left side is $0+0=0$. The right side is $9(4)=36$. The student's error comes from rearranging factors and terms without preserving the original operations.

📖 Explanation: At $x=0$, the function gives $y=7(0)=0$. Therefore the graph contains the point $(0,0)$. This provides a visual representation of how multiplying a number by zero produces an output of zero.

📖 Explanation: The area of a rectangle is found by multiplying its dimensions. Here the dimensions are $x+5$ and $0$, so the product is $(x+5)(0)=0$. The unknown value of $x$ does not affect this result.

📖 Explanation: A product equals zero when at least one factor is zero. Thus either $k=0$ or $k-5=0$, giving $k=5$. The student's mistake is considering only the first factor and overlooking the second factor.

📖 Explanation: Dividing zero by a nonzero number gives zero. This follows because $8\times0=0$, so zero is the number that satisfies the division relationship. The student's mistake is confusing the roles of dividend and divisor.

📖 Explanation: The numerator represents completed deliveries, while $12$ is a nonzero divisor. Since no deliveries were completed, the rate is $\frac{0}{12}=0$. A zero numerator does not make division undefined when the denominator is nonzero.

📖 Explanation: Student A is correct. Division by zero is undefined, but here the divisor is $15$, not zero. Because $15$ is nonzero and $15\times0=0$, the quotient is $0$.

📖 Explanation: For every value of $x$, $0x=0$. Since the denominator $7$ is nonzero, dividing gives $0\div7=0$. The student incorrectly treats zero as a factor that can be canceled without considering the entire expression.

📖 Explanation: For every nonzero $x$, $\frac{0}{x}=0$, so every allowed point has $y=0$. However, $x=0$ is excluded because division by zero is undefined. Thus the graph is the $x$-axis with a missing point at the origin.

📖 Explanation: Average rate is calculated by dividing the transferred amount by the elapsed time. Since $0$ liters were transferred over $20$ minutes, the rate is $\frac{0}{20}=0$ liters per minute. The nonzero denominator makes the division valid.

📖 Explanation: The position of zero matters in division. With $a\neq0$, $\frac{0}{a}=0$ because $a(0)=0$. But $\frac{a}{0}$ has no value because no number multiplied by zero can produce the nonzero value $a$.

📖 Explanation: Division asks for a number that, when multiplied by the divisor, produces the dividend. There is no number $q$ satisfying $0q=14$, so $14\div0$ has no defined numerical value.

📖 Explanation: The model requires dividing completed units by operating time. With $t=0$, the expression becomes $\frac{50}{0}$, which is undefined. It cannot be interpreted as a finite efficiency because no numerical multiplier of zero produces $50$.

📖 Explanation: The divisor cannot be treated like $1$. If $\frac{x}{0}=q$, then $0q=x$. For nonzero $x$, this is impossible because every product involving zero equals zero. Therefore the expression is undefined.

📖 Explanation: When $x\neq0$, $\frac{6x}{3x}=2$ after canceling the common nonzero factor. At $x=0$, however, the original expression becomes $\frac{0}{0}$, which is undefined. The simplification therefore requires the restriction $x\neq0$.

📖 Explanation: The expression $y=\frac{1}{x}$ requires a nonzero denominator. At $x=0$, it becomes $\frac{1}{0}$, which is undefined. The graph's missing vertical line visually reflects this restriction on the domain.

📖 Explanation: The location of zero matters. Since $7$ is nonzero, $0\div7=0$. In contrast, $7\div0$ asks for a number whose product with zero equals $7$, which is impossible. Therefore only the second expression is undefined.

📖 Explanation: Values such as $\frac{a}{0.1}$ can grow very large as the denominator approaches zero, but this does not make $\frac{a}{0}$ a real number. At exactly zero, no real quotient satisfies $0q=a$ when $a\neq0$.