Let's factorize the expression \( 4x(y+1)^2 - x(y-1)^2 \).

### Step 1: Distribute the common factor \( x \)
We can factor out \( x \) from both terms:
\[
4x(y+1)^2 - x(y-1)^2 = x \left[ 4(y+1)^2 - (y-1)^2 \right]
\]

### Step 2: Simplify the expression inside the brackets
Now, let's simplify the expression inside the brackets:
\[
4(y+1)^2 - (y-1)^2
\]

### Step 3: Expand the squares
Expand both squares:
\[
4(y+1)^2 = 4(y^2 + 2y + 1) = 4y^2 + 8y + 4
\]
\[
(y-1)^2 = y^2 - 2y + 1
\]

### Step 4: Substitute the expanded forms back into the expression
Substituting back, we get:
\[
4y^2 + 8y + 4 - (y^2 - 2y + 1)
\]

### Step 5: Simplify the expression
Distribute the negative sign and combine like terms:
\[
4y^2 + 8y + 4 - y^2 + 2y - 1 = 3y^2 + 10y + 3
\]

So the expression simplifies to:
\[
x(3y^2 + 10y + 3)
\]

### Step 6: Factor the quadratic expression
Now, we factor the quadratic \( 3y^2 + 10y + 3 \).

To factor \( 3y^2 + 10y + 3 \), we look for two numbers that multiply to \( 3 \times 3 = 9 \) and add to \( 10 \). These numbers are \( 9 \) and \( 1 \).

So we can rewrite the middle term \( 10y \) as \( 9y + y \):
\[
3y^2 + 9y + y + 3
\]

Next, we factor by grouping:
\[
3y(y + 3) + 1(y + 3)
\]

Finally, factor out the common factor \( (y + 3) \):
\[
(3y + 1)(y + 3)
\]

### Final Answer
The fully factorized form of the original expression is:
\[
x(3y + 1)(y + 3)
\]

Would you like further details or explanations on any step?

---

Here are 5 related questions:
1. How do you factor quadratics with a leading coefficient greater than 1?
2. What are some methods for factoring polynomials?
3. How can the difference of squares be applied in factoring?
4. What is the importance of factoring in solving equations?
5. How do you recognize common factors in algebraic expressions?

**Tip:** Always look for a common factor before trying more complex factoring methods; it simplifies the process significantly.

This guide walks through the process of factoring the algebraic expression 4x(y+1)² - x(y-1)². The solution includes expanding squares, simplifying terms, and applying the difference of squares theorem. Perfect for students in Grades 9-12 looking to strengthen their understanding of factoring quadratic expressions.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation