Question

multiple choice 1 point simplify the following expression using the rules of exponents: (x² y³)(x⁴ y z⁸) options: x⁶ y⁴ z⁸, x⁵ y³ z⁸, x⁶ y⁴ z⁸ (note: likely a typo in options, but original ocr captures the problem of simplifying (x² y³)(x⁴ y z⁸) with multiple - choice options)

Explanation:

Step1: Multiply like bases (x terms)

For \( x \) terms: \( x^2 \cdot x^4 = x^{2 + 4} = x^6 \)? Wait, no, wait—wait, maybe I misread. Wait, the first term is \( (x^2 y^3) \) and the second is \( (x^4 y z^8) \)? Wait, no, maybe the exponents: let's check again. Wait, maybe the first \( x \) is \( x^2 \), second \( x \) is \( x^4 \), so \( x^{2 + 4} = x^6 \)? No, wait, maybe the options have \( x^6 \)? Wait, no, the options: first option is \( x^6 y^4 z^8 \), second \( x^6 y^3 z^8 \), third \( x^6 y^4 z^8 \)? Wait, maybe I made a mistake. Wait, let's do it properly.

When multiplying monomials, we use the product rule for exponents: \( a^m \cdot a^n = a^{m + n} \), and we multiply coefficients (but here coefficients are 1) and combine like bases.

So for \( x \): \( x^2 \cdot x^4 = x^{2 + 4} = x^6 \)

For \( y \): \( y^3 \cdot y^1 = y^{3 + 1} = y^4 \) (since the second term has \( y^1 \), as \( y = y^1 \))

For \( z \): \( z^0 \cdot z^8 = z^{0 + 8} = z^8 \) (since the first term has no \( z \), so \( z^0 \))

Wait, but let's check the original expression: \( (x^2 y^3)(x^4 y z^8) \)

So breaking down each variable:

• \( x \): \( x^2 \times x^4 = x^{2 + 4} = x^6 \)
• \( y \): \( y^3 \times y^1 = y^{3 + 1} = y^4 \) (because \( y = y^1 \))
• \( z \): \( z^0 \times z^8 = z^{0 + 8} = z^8 \) (because the first term has no \( z \), so exponent 0)

So the simplified form is \( x^6 y^4 z^8 \). Now check the options:

First option: \( x^6 y^4 z^8 \) – that matches.

Wait, but maybe the options were typed with some errors, but assuming the first option is \( x^6 y^4 z^8 \), that's the correct one.

Wait, let's re-express the steps:

1. Identify like bases: \( x \), \( y \), \( z \).
2. For \( x \): \( x^2 \cdot x^4 = x^{2 + 4} = x^6 \) (product rule: \( a^m \cdot a^n = a^{m + n} \))
3. For \( y \): \( y^3 \cdot y^1 = y^{3 + 1} = y^4 \) (since \( y = y^1 \))
4. For \( z \): \( z^0 \cdot z^8 = z^{8} \) (since first term has no \( z \), so \( z^0 \), and \( 0 + 8 = 8 \))
5. Combine the results: \( x^6 y^4 z^8 \)

Answer:

A. \( x^6 y^4 z^8 \) (assuming the first option is this; if the options are labeled with circles, the first circle's option is correct)