Solving $y^2 + 4y - 32 = 0$ With The Zero Product Property

The zero product property is a cornerstone in solving quadratic equations. It elegantly states that if the product of two factors is zero, then at least one of the factors must be zero. This seemingly simple principle unlocks a powerful method for finding the roots, or solutions, of polynomial equations, particularly quadratic equations, which are equations of the form $ax^2 + bx + c = 0$. In this comprehensive guide, we will delve into the application of the zero product property to solve the specific quadratic equation $y^2 + 4y - 32 = 0$. We will dissect the equation, factorize it meticulously, and then apply the zero product property to arrive at the solutions. This step-by-step approach will not only provide the answer but also illuminate the underlying concepts, making it easier to tackle similar problems in the future.

Understanding the Zero Product Property

At its core, the zero product property is a logical deduction. It hinges on the fundamental nature of multiplication: any number multiplied by zero results in zero. Mathematically, this is expressed as: If $A \cdot B = 0$, then either $A = 0$ or $B = 0$ (or both). This property is not limited to just two factors; it extends to any number of factors. If the product of several factors is zero, at least one of them must be zero. This property is not just a mathematical curiosity; it is an indispensable tool in algebra, particularly in solving equations. It allows us to transform a single equation into a set of simpler equations that are easier to solve. In the context of quadratic equations, the zero product property becomes especially potent when we can factorize the quadratic expression into two linear factors. This factorization is the key to unlocking the solutions, as it allows us to isolate the variable and find its values that make the equation true.

Factoring the Quadratic Equation $y^2 + 4y - 32 = 0$

Before we can apply the zero product property, we need to factor the quadratic expression $y^2 + 4y - 32$. Factoring is the process of expressing a polynomial as a product of simpler polynomials. In this case, we want to find two binomials that, when multiplied together, give us $y^2 + 4y - 32$. The general approach to factoring a quadratic expression of the form $ax^2 + bx + c$ involves finding two numbers that multiply to $ac$ and add up to $b$. In our case, $a = 1$, $b = 4$, and $c = -32$. So, we need to find two numbers that multiply to $-32$ and add up to $4$. By systematically considering the factors of $-32$, we can identify the pair $8$ and $-4$. These numbers satisfy our conditions: $8 \cdot (-4) = -32$ and $8 + (-4) = 4$. Now that we have these numbers, we can rewrite the middle term of the quadratic expression ($4y$) as the sum of $8y$ and $-4y$:

$y^2 + 4y - 32 = y^2 + 8y - 4y - 32$

Next, we use a technique called factoring by grouping. We group the first two terms and the last two terms: $(y^2 + 8y) + (-4y - 32)$. Then, we factor out the greatest common factor (GCF) from each group. From the first group, the GCF is $y$, and from the second group, the GCF is $-4$:

$y(y + 8) - 4(y + 8)$

Notice that both terms now have a common factor of $(y + 8)$. We can factor this out:

$(y + 8)(y - 4)$

Thus, we have successfully factored the quadratic expression: $y^2 + 4y - 32 = (y + 8)(y - 4)$. This factorization is the crucial step that allows us to apply the zero product property and solve for $y$.

Applying the Zero Product Property

Now that we have factored the quadratic equation as $(y + 8)(y - 4) = 0$, we can apply the zero product property. This property tells us that if the product of two factors is zero, then at least one of the factors must be zero. Therefore, either $(y + 8) = 0$ or $(y - 4) = 0$. We now have two simple linear equations to solve. Let's solve each one separately.

Solving $y + 8 = 0$

To solve for $y$ in the equation $y + 8 = 0$, we subtract $8$ from both sides of the equation:

$y + 8 - 8 = 0 - 8$

This simplifies to:

$y = -8$

So, one solution to the quadratic equation is $y = -8$.

Solving $y - 4 = 0$

To solve for $y$ in the equation $y - 4 = 0$, we add $4$ to both sides of the equation:

$y - 4 + 4 = 0 + 4$

This simplifies to:

$y = 4$

So, the other solution to the quadratic equation is $y = 4$.

The Solutions

We have found two solutions for the quadratic equation $y^2 + 4y - 32 = 0$: $y = -8$ and $y = 4$. These are the values of $y$ that make the equation true. We can write the solution set as $\{-8, 4\}$. This means that if we substitute either $-8$ or $4$ for $y$ in the original equation, the equation will hold true. Let's verify this.

Verification

Let's substitute $y = -8$ into the original equation:

$(-8)^2 + 4(-8) - 32 = 64 - 32 - 32 = 0$

The equation holds true. Now let's substitute $y = 4$ into the original equation:

$(4)^2 + 4(4) - 32 = 16 + 16 - 32 = 0$

Again, the equation holds true. This confirms that our solutions are correct. The solutions $y = -8$ and $y = 4$ are also the x-intercepts of the parabola represented by the equation $y = x^2 + 4x - 32$. These points are where the parabola crosses the x-axis.

Conclusion

In this guide, we have demonstrated how to solve the quadratic equation $y^2 + 4y - 32 = 0$ using the zero product property. We began by understanding the fundamental principle behind the property, which states that if the product of factors is zero, at least one of the factors must be zero. We then meticulously factored the quadratic expression into two binomial factors: $(y + 8)(y - 4)$. Applying the zero product property, we set each factor equal to zero and solved the resulting linear equations, obtaining the solutions $y = -8$ and $y = 4$. Finally, we verified our solutions by substituting them back into the original equation. This step-by-step process not only provides the answer but also reinforces the understanding of the underlying concepts and techniques involved in solving quadratic equations. The zero product property is a powerful tool in algebra, and mastering its application is essential for solving a wide range of polynomial equations.

Therefore, the correct answer is C) $y = -8, 4$.