Question

name date per unit 1. polynomials 1.1 practice
you must show all work to receive any credit. no work = 0, including for fractions, equations, and steps.
multiplying / adding fractions: $\frac{3}{4} \times \frac{1}{2} = \frac{3}{8}$ $\frac{2}{3} cdot 4 = \frac{2}{3} cdot \frac{4}{1} = \frac{8}{3}$ $\frac{5}{3} + \frac{2}{3} = \frac{7}{3}$
multiplying monomials
$2y^4 cdot 3y^5 =$
$6m^{5/2} cdot m^{1/2} =$
$x^3 cdot 2x =$
power to a power
$(n^3)^6 =$
$(2x^5)^3 =$
$(y^4)^{1/2} =$
power of a product
$(ab^3)^5 =$
$4(x^2y^3)^4 =$
$(-3ab^3)^2 =$
dividing monomials
$\frac{m^5}{m^3} =$
$\frac{a^{4/3}}{a^{1/3}} =$
$\frac{12b^{1/6}c^{1/9}}{6b^{1/3}c^{1/4}} =$
power of a quotient
$left( \frac{3m^3}{n^2}
ight)^2 =$
$left( \frac{x^{3/2}}{y^{5/4}}
ight)^8 =$
$left( \frac{x^2}{y^4}
ight)^{1/2} =$
three students solved the same problem and their steps are recorded below. fill in the sentence statements below:

	student a	student b	student c
step 1	$(8x^6y^3)(3y^4)^2$	$(8x^6y^3)(3y^4)^2$	$(8x^6y^3)(3y^4)^2$
step 2	$(8x^6y^3)(3y^8)$	$(8x^6y^3)(9y^8)$	$(8x^6y^3)(9y^8)$
step 3	$24x^6y^{11}$	$17x^6y^{11}$	$72x^6y^{11}$

student a was ____, their mistake on step , and they should have ____.
student b was ____, their mistake on step , and they should have ____.
student c was ____, their mistake on step , and they should have ____.

Answer:

Analyzing Student A:

Step 1: Analyze the original problem and Step 1

The original problem is \((2x^{2}y)^{3}(3y^{4})^{2}\). In Step 1, Student A correctly applies the power of a product rule to \((2x^{2}y)^{3}\), getting \(8x^{6}y^{3}\). So Step 1 is correct.

Step 2: Analyze Step 2

For \((3y^{4})^{2}\), using the power of a power rule \((a^{m})^{n}=a^{mn}\), we have \(3^{2}y^{4\times2} = 9y^{8}\). But Student A wrote \((3y^{8})\) (incorrectly calculated \(3^{2}\) as 3 instead of 9). So the mistake is in Step 2.

Step 3: Correct Step 2

Student A should have calculated \((3y^{4})^{2}\) as \(9y^{8}\) instead of \(3y^{8}\). Then, multiplying \(8x^{6}y^{3}\) and \(9y^{8}\) (in Step 3) would give \(72x^{6}y^{11}\), but first, the correction in Step 2 is needed.

Analyzing Student B:

Step 1: Analyze Step 1

Student B correctly applies the power of a product rule to \((2x^{2}y)^{3}\), getting \(8x^{6}y^{3}\). Step 1 is correct.

Step 2: Analyze Step 2

Student B correctly calculates \((3y^{4})^{2}=9y^{8}\). Step 2 is correct.

Step 3: Analyze Step 3

Now, multiply \(8x^{6}y^{3}\) and \(9y^{8}\). Using the rule \(a^{m}\cdot a^{n}=a^{m + n}\) for the \(y\)-terms: \(y^{3}\cdot y^{8}=y^{3 + 8}=y^{11}\), and the coefficient is \(8\times9 = 72\). But Student B wrote \(17x^{6}y^{11}\) (incorrectly added \(8 + 9\) for the coefficient instead of multiplying). So the mistake is in Step 3.

Correct Step 3

Student B should have multiplied \(8\) and \(9\) to get \(72\), so the result should be \(72x^{6}y^{11}\) instead of \(17x^{6}y^{11}\).

Analyzing Student C:

Step 1: Analyze the original problem and Step 1

The original problem is \((2x^{2}y)^{3}(3y^{4})^{2}\), but Student C's original problem is written as \((2x^{2}y)^{3}(3y^{3})^{2}\) (a typo, changed \(y^{4}\) to \(y^{3}\)). So the mistake starts in the original problem (or Step 0, but in terms of steps, the first step with the wrong problem). However, assuming the original problem is correct, Student C miswrote the original problem as \((2x^{2}y)^{3}(3y^{3})^{2}\) (so the base of the second term is incorrect). But if we consider the intended original problem \((2x^{2}y)^{3}(3y^{4})^{2}\), Student C's Step 1 has a wrong exponent on \(y\) in the second term (used \(y^{3}\) instead of \(y^{4}\)). But looking at the steps, in Step 1, Student C's \((3y^{3})^{2}\) (if we take the written problem) or the intended \((3y^{4})^{2}\). Wait, looking at the table, Student C's original problem is \((2x^{2}y)^{3}(3y^{3})^{2}\) (typo). But assuming the correct original problem is \((2x^{2}y)^{3}(3y^{4})^{2}\), Student C's Step 1 has a mistake in the exponent of \(y\) in the second term (used \(y^{3}\) instead of \(y^{4}\)). But let's check the steps as per the table:

• Step 1: Student C has \((8x^{6}y^{3})(3y^{3})^{2}\) (wrong, should be \((3y^{4})^{2}\))
• Step 2: Student C calculates \((3y^{3})^{2}=9y^{6}\) (if we take the wrong original problem) or if it's a typo and should be \((3y^{4})^{2}\), then \((3y^{4})^{2}=9y^{8}\). But in the table, Student C's Step 2 is \((8x^{6}y^{3})(9y^{6})\) (assuming the original problem was miswritten as \((3y^{3})^{2}\)). But the key is:

If we consider the correct original problem \((2x^{2}y)^{3}(3y^{4})^{2}\), Student C miswrote the original problem (changed \(y^{4}\) to \(y^{3}\)) in the "Original Problem" row. So the mistake is in the original problem (or Step 0), but in terms of the steps, the first step where the problem is misrepresented. However, looking at the steps:

• Step 1: Student C correctly applies the power of a product to \((2x^{2}y)^{3}\) (gets \(8x^{6}y^{3}\))
• Step 2: If we take the miswritten original problem \((2x^{2}y)^{3}(3y^{3})^{2}\), then \((3y^{3})^{2}=9y^{6}\) (correct for the miswritten problem), but the original problem was \((3y^{4})^{2}\). So the mistake is in the "Original Problem" (typo), but in the steps, the first step where the problem is set up. But assuming the original problem was correct, Student C's mistake is in the original problem (writing \(y^{3}\) instead of \(y^{4}\)), which affects all steps. But if we go by the steps as per the table:

Student C's Step 2: \((8x^{6}y^{3})(9y^{6})\) (if original problem was \((3y^{3})^{2}\)), but the correct original problem has \((3y^{4})^{2}\), so the exponent on \(y\) in the second term is wrong. Then, in Step 3, multiplying \(y^{3}\) and \(y^{6}\) (for the miswritten problem) gives \(y^{9}\), but for the correct problem, it should be \(y^{3}\cdot y^{8}=y^{11}\). But Student C's Step 3 is \(72x^{6}y^{11}\), which is the correct result for the original problem, but the intermediate steps had a miswritten original problem. So the mistake is in the "Original Problem" (typo, changing \(y^{4}\) to \(y^{3}\)), so the first step (original problem) is wrong, and they should have used \(y^{4}\) instead of \(y^{3}\) in the original problem.

Final Answers:

• Student A was incorrect, their mistake on step 2, and they should have calculated \((3y^{4})^{2}\) as \(9y^{8}\) (instead of \(3y^{8}\)).
• Student B was incorrect, their mistake on step 3, and they should have multiplied \(8\) and \(9\) (to get \(72\)) instead of adding them (to get \(17\)).
• Student C was incorrect (due to a typo in the original problem), their mistake on step 1 (original problem setup), and they should have used \(y^{4}\) instead of \(y^{3}\) in \((3y^{4})^{2}\) (in the original problem).

(Note: The analysis for Student C is based on the typo in the original problem as per the table. If we assume the original problem was written correctly for Student C, there may be a different interpretation, but based on the given table, the original problem for Student C is \((2x^{2}y)^{3}(3y^{3})^{2}\) which is a typo from the correct \((2x^{2}y)^{3}(3y^{4})^{2}\).)