Solving $(x+5)(x-3)=0$: Find The Roots

Dec 10, 2025 by Andrew McMorgan 39 views

Hey guys, today we're diving deep into a super common type of math problem: solving equations. Specifically, we're looking at the equation $(x+5)(x-3)=0$ and figuring out which of the given options are the correct solutions. This is all about understanding the Zero Product Property, which is a fundamental concept in algebra. So, grab your notebooks, and let's get this done!

Understanding the Zero Product Property

The Zero Product Property is the key to unlocking the solutions for equations like this one. It's a fancy name for a very simple idea: if the product of two or more factors is zero, then at least one of the factors must be zero. In our equation, $(x+5)(x-3)=0$ , we have two factors: $(x+5)$ and $(x-3)$ . For their product to be zero, either the first factor $(x+5)$ must equal zero, or the second factor $(x-3)$ must equal zero (or, in some cases, both could be zero, but that's not possible here since $x$ can't be both $-5$ and $3$ simultaneously).

This property is absolutely crucial for solving polynomial equations, especially those in factored form. It simplifies the problem from dealing with a single, potentially complex equation, into two much simpler linear equations. Remember this property, guys, because you'll be using it all the time in your math journey. It’s the bedrock upon which many algebraic techniques are built, including factoring quadratics and finding roots of higher-degree polynomials. When you see an equation set equal to zero and it's presented as a product of terms, your first thought should always be, "Ah, the Zero Product Property!"

So, let's break down our equation: $(x+5)(x-3)=0$ . We have our two factors. We need to set each factor equal to zero and solve for $x$ . This is where the magic happens, and it's surprisingly straightforward. It’s like having a secret code that unlocks the values of $x$ that make the whole equation true. This method works because zero is the only number that, when multiplied by any other number, results in zero. Any other number, when multiplied, will yield a non-zero result. This unique property of zero is what makes the Zero Product Property so powerful and indispensable in algebra.

Setting the First Factor to Zero

Our first factor is $(x+5)$ . According to the Zero Product Property, for the entire equation $(x+5)(x-3)=0$ to be true, this factor must be equal to zero. So, we set up our first simple equation:

$x + 5 = 0$

To solve for $x$ , we need to isolate $x$ on one side of the equation. We can do this by subtracting 5 from both sides:

$x + 5 - 5 = 0 - 5$

This simplifies to:

$x = -5$

So, $x = -5$ is one of the solutions to the original equation. You can mentally check this: if $x = -5$ , then the first factor becomes $(-5 + 5) = 0$ . And $0$ multiplied by anything else is $0$ . Pretty neat, huh? This confirms that $x = -5$ is a valid solution. It’s always a good practice, especially when you’re starting out, to plug your solutions back into the original equation to ensure they work. This habit builds confidence and reduces errors. Don't skip this step, especially in exams!

Setting the Second Factor to Zero

Now, let's look at our second factor, which is $(x-3)$ . Applying the Zero Product Property again, this factor must also be equal to zero for the product to be zero:

$x - 3 = 0$

To solve for $x$ here, we need to isolate $x$ . We do this by adding 3 to both sides of the equation:

$x - 3 + 3 = 0 + 3$

This simplifies to:

$x = 3$

So, $x = 3$ is the other solution to the original equation. Let's check this one too: if $x = 3$ , the second factor becomes $(3 - 3) = 0$ . And anything multiplied by $0$ is $0$ . This confirms $x = 3$ is also a correct solution. These two values, $x = -5$ and $x = 3$ , are the only values that will make the equation $(x+5)(x-3)=0$ true. The power of the Zero Product Property is that it systematically finds all possible solutions when an equation is in this factored form. It guarantees that you won't miss any roots, provided you correctly set each factor to zero and solve.

Checking the Given Options

Now that we've found the solutions ourselves, let's check them against the options provided in the question: A. $x=-15$ , B. $x=-5$ , C. $x=-3$ , D. $x=2$ , E. $x=3$ , F. $x=5$ . Our calculated solutions are $x = -5$ and $x = 3$ . Let's see which options match:

Option B: $x = -5$ - This matches one of our solutions. Correct!
Option E: $x = 3$ - This matches our other solution. Correct!

Let's quickly see why the other options are incorrect. This is super important for understanding, guys. It's not just about finding the right answers, but also about understanding why the wrong answers are wrong.

Option A: $x = -15$ - If $x = -15$ , the equation becomes $(-15 + 5)(-15 - 3) = (-10)(-18) = 180$ . This is definitely not $0$ . So, A is incorrect.
Option C: $x = -3$ - If $x = -3$ , the equation becomes $(-3 + 5)(-3 - 3) = (2)(-6) = -12$ . This is not $0$ . So, C is incorrect.
Option D: $x = 2$ - If $x = 2$ , the equation becomes $(2 + 5)(2 - 3) = (7)(-1) = -7$ . This is not $0$ . So, D is incorrect.
Option F: $x = 5$ - If $x = 5$ , the equation becomes $(5 + 5)(5 - 3) = (10)(2) = 20$ . This is not $0$ . So, F is incorrect.

As you can see, only $x = -5$ and $x = 3$ make the equation equal to zero. So, the correct options are B and E. Remember, when a question says "Check all that apply," you need to be thorough and check every single option against your findings or by plugging them back into the equation. It's a common mistake to only find one solution and stop, but sometimes there are multiple correct answers.

Expanding and Solving (Alternative Method - More Work!)

While the Zero Product Property is the most efficient way to solve this, you could also expand the equation first and then solve the resulting quadratic equation. This is usually more work, but it's good to know how to do it. Let's expand $(x+5)(x-3)=0$ :

Using the FOIL method (First, Outer, Inner, Last):

First: $x * x = x^2$
Outer: $x * -3 = -3x$
Inner: $5 * x = 5x$
Last: $5 * -3 = -15$

Combining these terms, we get:

$x^2 - 3x + 5x - 15 = 0$

Combine the like terms ( $-3x$ and $5x$ ):

$x^2 + 2x - 15 = 0$

Now we have a standard quadratic equation in the form $ax^2 + bx + c = 0$ . We can solve this using a few methods:

Factoring (again!): We need to find two numbers that multiply to -15 and add to 2. Those numbers are 5 and -3. So, we can factor it back into $(x+5)(x-3) = 0$ . This brings us right back to where we started, and we use the Zero Product Property again to find $x = -5$ and $x = 3$ . This shows that the factored form is the easiest starting point!
Quadratic Formula: The quadratic formula is $x = rac{-b inom{ ext{}}{} binom{ ext{}}{} ext{ }}{2a}$ . For our equation $x^2 + 2x - 15 = 0$ , we have $a=1$ , $b=2$ , and $c=-15$ . Plugging these into the formula: $x = rac{-2 inom{ ext{ }}{ } binom{ ext{ }}{ } ext{ }}{2(1)}$ $x = rac{-2 inom{ ext{ }}{ } binom{ ext{ }}{ } ext{ }}{2}$ We need to find the square root of $b^2 - 4ac$ , which is the discriminant. Discriminant $= (2)^2 - 4(1)(-15) = 4 + 60 = 64$ . $inom{ ext{ }}{ } binom{ ext{ }}{ } ext{ } = inom{64}{1} = inom{8}{1}$ . So, $x = rac{-2 inom{ ext{ }}{ } binom{ ext{ }}{ } ext{ }}{2}$ becomes: $x = rac{-2 inom{ ext{ }}{ } ext{ }}{2}$ This gives us two possibilities: $x_1 = rac{-2 + 8}{2} = rac{6}{2} = 3$ $x_2 = rac{-2 - 8}{2} = rac{-10}{2} = -5$ Again, we get $x = 3$ and $x = -5$ . This confirms our previous results and demonstrates the power of the quadratic formula, though it's definitely more steps than using the Zero Product Property on the factored form.

This alternative method, while longer, reinforces the idea that different algebraic techniques can lead to the same correct answers. It's good to have multiple tools in your mathematical toolbox, guys. Knowing when to use each tool is part of becoming a math whiz!

Conclusion: The Winning Solutions

So, to wrap things up, the equation $(x+5)(x-3)=0$ has two solutions. These are found by setting each factor equal to zero due to the Zero Product Property:

$x + 5 = 0

ightarrow x = -5$

$x - 3 = 0

ightarrow x = 3$

When we compare these solutions to the given options, the ones that match are B. $x=-5$ and E. $x=3$ . These are the only values that make the original equation true. Always remember to check all that apply when the question indicates it, and don't hesitate to plug your answers back into the original equation to verify. Keep practicing, and you'll master these types of problems in no time! You guys got this!