Question

01/09/2026 - algebra 2
exp. prop. (zero & quotient prop.) #4
date
review: simplify.
simplificar.

1. \\((3x^4y^0z^2)^3\\) \t\t\t\t\t2) \\(yx^2z^3 \cdot (2zx^3y^4)^0\\)

new: simplify. your answer should contain only positive exponents.
simplificar. tu respuesta debe contener sólo exponentes positivos.

1. \\(\frac{2x^2}{x^4}\\) \t\t\t\t\t4) \\(\frac{4x}{x^4}\\)
2. \\(\frac{2m^2}{4m^2}\\) \t\t\t\t\t6) \\(\frac{4x^3}{3x^2}\\)
3. \\(\frac{2a^2b^3}{3ba^2}\\) \t\t\t\t\t8) \\(\frac{2xy^4}{4x^4y^2}\\)

Explanation:

Problem 1: \((3x^{4}y^{0}z^{2})^{3}\)

Step 1: Apply the power of a product rule \((ab)^n = a^n b^n\) and the zero exponent rule \(a^0 = 1\) (for \(y^0\))

First, recall that any non - zero number to the power of 0 is 1, so \(y^{0}=1\). Then, using the power of a product rule \((abc)^n=a^{n}b^{n}c^{n}\), we have:
\((3x^{4}y^{0}z^{2})^{3}=3^{3}\cdot(x^{4})^{3}\cdot(y^{0})^{3}\cdot(z^{2})^{3}\)

Step 2: Simplify each term

• Calculate \(3^{3}=27\)
• Using the power of a power rule \((a^{m})^{n}=a^{mn}\), \((x^{4})^{3}=x^{4\times3}=x^{12}\)
• Since \(y^{0} = 1\), \((y^{0})^{3}=1^{3}=1\)
• \((z^{2})^{3}=z^{2\times3}=z^{6}\)

Multiply these results together: \(27\times x^{12}\times1\times z^{6}=27x^{12}z^{6}\)

Step 1: Apply the zero exponent rule

Recall that for any non - zero number \(a\), \(a^{0}=1\). Here, \(2zx^{3}y^{4}\) is non - zero (assuming \(x,y,z\) are non - zero), so \((2zx^{3}y^{4})^{0}=1\)

Step 2: Multiply by the remaining term

\(yx^{2}z^{3}\cdot1 = yx^{2}z^{3}\)

Step 1: Apply the quotient rule for exponents

The quotient rule for exponents states that \(\frac{a^{m}}{a^{n}}=a^{m - n}\) (where \(a
eq0\)). Here, \(a = x\), \(m = 2\), and \(n = 4\)
So \(\frac{2x^{2}}{x^{4}}=2\times x^{2-4}\)

Step 2: Simplify the exponent

\(x^{2 - 4}=x^{-2}\), and since we want positive exponents, \(x^{-2}=\frac{1}{x^{2}}\)
So \(2\times\frac{1}{x^{2}}=\frac{2}{x^{2}}\)

Answer:

\(27x^{12}z^{6}\)

Problem 2: \(yx^{2}z^{3}\cdot(2zx^{3}y^{4})^{0}\)