Let's break down and solve the questions in the case study provided:

1. What is the length and width of the garden when $x = 3$?

The length of the garden is given by $x + 2$, and the width by $x - 1$. Substituting $x = 3$:

$\text{Length} = 3 + 2 = 5 \text{ meters}$ $\text{Width} = 3 - 1 = 2 \text{ meters}$

Thus, the length is 5 meters and the width is 2 meters.

2. Verify whether $A(3) = 14$ is correct by calculating the area using the given polynomial.

The area polynomial is $A(x) = x^2 + x - 2$.

Substituting $x = 3$ into the polynomial:

$A(3) = 3^2 + 3 - 2 = 9 + 3 - 2 = 10$

Thus, $A(3) = 10$, not 14. So, $A(3) = 14$ is incorrect.

3. Factorize the polynomial $A(x) = x^2 + x - 2$.

We need to factor $x^2 + x - 2$. To do this, we look for two numbers whose product is $-2$ (the constant term) and whose sum is $1$ (the coefficient of $x$).

The correct factorization is:

$A(x) = (x + 2)(x - 1)$

4. Find the dimensions of the garden if the area is 0.

If the area is 0, we set the polynomial equal to 0:

$A(x) = x^2 + x - 2 = 0$ From the factorization, we have:

$(x + 2)(x - 1) = 0$

Setting each factor equal to 0:

$x + 2 = 0 \quad \Rightarrow \quad x = -2$ $x - 1 = 0 \quad \Rightarrow \quad x = 1$

Thus, the values of $x$ for which the area is 0 are $x = -2$ and $x = 1$.

For these values:

• When $x = -2$, the length is $-2 + 2 = 0$ and the width is $-2 - 1 = -3$, so the garden has no length.
• When $x = 1$, the length is $1 + 2 = 3$ and the width is $1 - 1 = 0$, so the garden has no width.

Would you like further details or explanations?

Related Questions:

1. What is the area of the garden when $x = 4$?
2. How would the polynomial change if the width were $x + 1$?
3. What happens to the area as $x$ becomes negative?
4. Solve for the value of $x$ when the area is 20 square meters.
5. How can we graph the area function $A(x) = x^2 + x - 2$?

Tip:

Always check the sign of $x$ when solving real-world problems like dimensions. Negative values for length or width usually don't make sense physically.