Factorise completely: 32x²y - 48x³y³

bendict.light

7 months ago

Alright, let’s carefully factorize and then check each option

We are given:
\[
32x^2y - 48x^3y^3
\]

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### Step 1: Find the Highest Common Factor (HCF)
- Coefficients: HCF of 32 and 48 = **16**
- Variables:
- \(x^2\) and \(x^3\) → common factor = \(x^2\)
- \(y\) and \(y^3\) → common factor = \(y\)

So HCF = \(16x^2y\).

---

### Step 2: Factorize
\[
32x^2y - 48x^3y^3 = 16x^2y(2 - 3xy^2)
\]

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### Step 3: Check each option

**A. \(16x^2y(2 - 3xy^2)\)**
- Expanding: \(16x^2y \cdot 2 = 32x^2y\)
- Expanding: \(16x^2y \cdot (-3xy^2) = -48x^3y^3\)
Correct.

---

**B. \(8xy(4x - 6x^2y^2)\)**
- Expanding: \(8xy \cdot 4x = 32x^2y\)
- Expanding: \(8xy \cdot (-6x^2y^2) = -48x^3y^3\)
This also works, but it is **not fully factorized** because the common factor taken out is smaller (8xy instead of 16x²y).

---

**C. \(8x^2y(4 - 6xy^2)\)**
- Expanding: \(8x^2y \cdot 4 = 32x^2y\)
- Expanding: \(8x^2y \cdot (-6xy^2) = -48x^3y^3\)
This also works, but again **not fully factorized** (HCF is 16x²y, not 8x²y).

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**D. \(16xy(2x - 3x^2y^2)\)**
- Expanding: \(16xy \cdot 2x = 32x^2y\)
- Expanding: \(16xy \cdot (-3x^2y^2) = -48x^3y^3\)
This works, but again **not fully factorized** (HCF is 16x²y, not 16xy).

---

### Final Answer
The **completely factorized form** is:
\[
16x^2y(2 - 3xy^2)
\]

So the correct option is **A**.

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Why others don’t go:
- **B, C, D** are valid but **not complete factorization** because they didn’t take out the **highest common factor (HCF)**.
- Only **A** uses the full HCF \(16x^2y\).