The area of a rectangle is `12y^(4)+28y^(3)-5y^(2)`. If its length is `6y^(3)-y^(2)` , then its width is

Options

A. `y+5`
B. `-2y+5`
C. `-2y^(2)+5`
D. `2y+5`

Answer

To find the width of the rectangle given its area and length, we can follow these steps:

### Step 1: Write down the formula for the area of a rectangle.
The area \( A \) of a rectangle is given by the formula:
\[ A = \text{Length} \times \text{Width} \]

### Step 2: Substitute the given values into the formula.
We are given:
- Area \( A = 12y^4 + 28y^3 - 5y^2 \)
- Length \( L = 6y^3 - y^2 \)

Using the formula:
\[ 12y^4 + 28y^3 - 5y^2 = (6y^3 - y^2) \times W \]
where \( W \) is the width we need to find.

### Step 3: Solve for the width \( W \).
To find \( W \), we can rearrange the equation:
\[ W = \frac{12y^4 + 28y^3 - 5y^2}{6y^3 - y^2} \]

### Step 4: Factor the numerator and denominator.
First, factor out \( y^2 \) from both the numerator and the denominator:
\[ W = \frac{y^2(12y^2 + 28y - 5)}{y^2(6y - 1)} \]

### Step 5: Cancel the common factor \( y^2 \).
Since \( y^2 \) is common in both the numerator and the denominator, we can cancel it out:
\[ W = \frac{12y^2 + 28y - 5}{6y - 1} \]

### Step 6: Factor the numerator \( 12y^2 + 28y - 5 \).
To factor \( 12y^2 + 28y - 5 \), we can look for two numbers that multiply to \( 12 \times -5 = -60 \) and add to \( 28 \). The numbers \( 30 \) and \( -2 \) work:
\[ 12y^2 + 30y - 2y - 5 \]
Now, group the terms:
\[ = (12y^2 + 30y) + (-2y - 5) \]
Factor by grouping:
\[ = 6y(2y + 5) - 1(2y + 5) \]
This gives us:
\[ = (6y - 1)(2y + 5) \]

### Step 7: Substitute back into the width equation.
Now substitute back into the width equation:
\[ W = \frac{(6y - 1)(2y + 5)}{(6y - 1)} \]

### Step 8: Cancel the common factor \( (6y - 1) \).
Since \( (6y - 1) \) is common in both the numerator and the denominator, we can cancel it out:
\[ W = 2y + 5 \]

### Final Answer:
Thus, the width of the rectangle is:
\[ \boxed{2y + 5} \]