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**Simplified Expressions:**
1. **Problem 4:**
\[
\frac{x^{2}(x + 2) - 4(x + 2)}{(x + 2)^{2}} = x - 2 \quad \text{where } x \neq -2
\]
2. **Problem 7:**
\[
120^{5} y + 3xy = y(120^{5} + 3x) \quad \text{or} \quad y(24,\!883,\!200,\!000 + 3x)
\]
3. **Problem 3:**
\[
\frac{4x^{2} - 1}{3x^{2} + 10x + 3} \div \frac{6x^{2} + 5x + 1}{4x^{2} + 11} \times \frac{9x^{2} + 6x + 1}{8x^{2} - 6x + 1} = \frac{4x^{2} + 11}{(x + 3)(4x - 1)}
\]
**Note:** Ensure that \(x \neq -2\) in Problem 4 to avoid division by zero.
Solución
Sure, let's simplify each of the given algebraic expressions step by step.
---
### **Problem 4**
Simplify the expression:
\[
\frac{x^{2}(x + 2) - 4(x + 2)}{(x + 2)^{2}}
\]
**Step 1: Factor the Numerator**
Notice that \((x + 2)\) is a common factor in the numerator:
\[
x^{2}(x + 2) - 4(x + 2) = (x + 2)(x^{2} - 4)
\]
**Step 2: Simplify the Expression**
Now, the expression becomes:
\[
\frac{(x + 2)(x^{2} - 4)}{(x + 2)^{2}} = \frac{x^{2} - 4}{x + 2}
\]
**Step 3: Factor Further if Possible**
Factor \(x^{2} - 4\) as a difference of squares:
\[
x^{2} - 4 = (x - 2)(x + 2)
\]
So,
\[
\frac{(x - 2)(x + 2)}{x + 2} = x - 2 \quad \text{(for } x \neq -2\text{)}
\]
**Final Simplified Form:**
\[
x - 2 \quad \text{where } x \neq -2
\]
---
### **Problem 7**
Simplify the expression:
\[
120^{5} y + 3xy
\]
**Step 1: Factor Out the Common Term**
Both terms have a common factor of \(y\):
\[
120^{5} y + 3xy = y(120^{5} + 3x)
\]
**Step 2: (Optional) Evaluate \(120^{5}\) if Needed**
If you need to express \(120^{5}\) numerically:
\[
120^{5} = 120 \times 120 \times 120 \times 120 \times 120 = 24,\!883,\!200,\!000
\]
So,
\[
y(24,\!883,\!200,\!000 + 3x)
\]
**Final Simplified Form:**
\[
y(120^{5} + 3x) \quad \text{or} \quad y(24,\!883,\!200,\!000 + 3x)
\]
---
### **Problem 3**
Simplify the expression:
\[
\frac{4x^{2} - 1}{3x^{2} + 10x + 3} \div \frac{6x^{2} + 5x + 1}{4x^{2} + 11} \times \frac{9x^{2} + 6x + 1}{8x^{2} - 6x + 1}
\]
**Note:** It seems there was a typo in the denominator of the last fraction. Assuming it's \(8x^{2} - 6x + 1\).
**Step 1: Factor Each Polynomial**
- **Numerator of the first fraction:**
\[
4x^{2} - 1 = (2x - 1)(2x + 1)
\]
- **Denominator of the first fraction:**
\[
3x^{2} + 10x + 3 = (3x + 1)(x + 3)
\]
- **Numerator of the second fraction (divisor):**
\[
6x^{2} + 5x + 1 = (2x + 1)(3x + 1)
\]
- **Denominator of the second fraction:**
\[
4x^{2} + 11 \quad (\text{Cannot be factored further})
\]
- **Numerator of the third fraction:**
\[
9x^{2} + 6x + 1 = (3x + 1)^2
\]
- **Denominator of the third fraction:**
\[
8x^{2} - 6x + 1 = (2x - 1)(4x - 1)
\]
**Step 2: Rewrite the Entire Expression with Factored Forms**
\[
\frac{(2x - 1)(2x + 1)}{(3x + 1)(x + 3)} \div \frac{(2x + 1)(3x + 1)}{4x^{2} + 11} \times \frac{(3x + 1)^2}{(2x - 1)(4x - 1)}
\]
**Step 3: Change Division to Multiplication by Reciprocal**
\[
\frac{(2x - 1)(2x + 1)}{(3x + 1)(x + 3)} \times \frac{4x^{2} + 11}{(2x + 1)(3x + 1)} \times \frac{(3x + 1)^2}{(2x - 1)(4x - 1)}
\]
**Step 4: Cancel Out Common Factors**
- \(2x - 1\) cancels with \(2x - 1\)
- \(2x + 1\) cancels with \(2x + 1\)
- \((3x + 1)^2\) cancels with \((3x + 1)(3x + 1)\)
**Remaining Terms:**
\[
\frac{4x^{2} + 11}{(x + 3)(4x - 1)}
\]
**Final Simplified Form:**
\[
\frac{4x^{2} + 11}{(x + 3)(4x - 1)}
\]
---
If there's a different intended form or if the last denominator in **Problem 3** was different, please let me know so I can adjust the solution accordingly!
Revisado y aprobado por el equipo de tutoría de UpStudy
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