Factoring Quadratics: Which Statement Holds True?

Hey guys! Today, we're diving into the world of quadratic expressions and specifically tackling the expression $-2h^2 - 15h - 7$. Our mission is to figure out which statement about its factors is actually true. We've got four options on the table, and we'll break down each one to find the correct answer. So, let's put on our math hats and get started!

Understanding the Problem

Before we jump into the options, let's make sure we understand what we're dealing with. We have the quadratic expression $-2h^2 - 15h - 7$. Factoring this means we want to rewrite it as a product of two binomials. A binomial, if you recall, is just a polynomial with two terms. For example, $(h + 2)$ or $(2h + 1)$ are binomials.

Our options give us potential factors, and we need to determine which one actually fits. This involves either factoring the quadratic expression ourselves or testing each option to see if it divides evenly into the expression. We'll explore both methods to give you a comprehensive understanding.

Key Concepts: Factoring, Quadratic Expressions, Binomials

Why Factoring Matters

You might be wondering, why do we even care about factoring? Well, factoring is a fundamental skill in algebra and has tons of applications. It helps us:

• Solve quadratic equations: By setting the factored expression equal to zero, we can find the values of h that make the equation true.
• Simplify expressions: Factoring can help us cancel out common factors in fractions or simplify more complex algebraic expressions.
• Graph quadratic functions: The factors of a quadratic can tell us about the x-intercepts of the parabola.

So, mastering factoring is a crucial step in your math journey!

Method 1: Factoring the Quadratic

Let's start by factoring the quadratic expression $-2h^2 - 15h - 7$ ourselves. This way, we can directly see the factors and compare them to the given options. Factoring quadratics can sometimes feel like a puzzle, but with a systematic approach, it becomes much easier.

Step 1: Look for a Common Factor

First, we always check if there's a common factor we can pull out from all the terms. In this case, there isn't a common factor other than 1 (or -1, but we'll deal with the negative sign later).

Step 2: Handle the Leading Coefficient

Our quadratic has a leading coefficient of -2 (the number in front of the $h^2$ term). This makes factoring a bit trickier than when the leading coefficient is 1. We'll use a method that involves multiplying the leading coefficient by the constant term.

Multiply -2 by -7: (-2) * (-7) = 14

Step 3: Find Two Numbers

Now, we need to find two numbers that multiply to 14 and add up to -15 (the coefficient of the h term). These numbers are -14 and -1.

• (-14) * (-1) = 14
• (-14) + (-1) = -15

Step 4: Rewrite the Middle Term

We'll rewrite the middle term (-15h) using the two numbers we found:

$-2h^2 - 15h - 7$ becomes $-2h^2 - 14h - h - 7$

Notice that we've split -15h into -14h and -h. This sets us up for factoring by grouping.

Step 5: Factor by Grouping

Now, we'll group the first two terms and the last two terms and factor out the greatest common factor (GCF) from each group:

• From $-2h^2 - 14h$, we can factor out -2h: -2h(h + 7)
• From $-h - 7$, we can factor out -1: -1(h + 7)

So, our expression looks like this: -2h(h + 7) - 1(h + 7)

Step 6: Factor Out the Common Binomial

We now have a common binomial factor of (h + 7). We'll factor that out:

(h + 7)(-2h - 1)

Step 7: Distribute the Negative Sign (Optional)

To make our factors look more like the options, we can distribute the negative sign from the -2h term into the second binomial:

-(h + 7)(2h + 1)

So, the factored form of $-2h^2 - 15h - 7$ is -(h + 7)(2h + 1).

Method 2: Testing the Options

Another way to approach this problem is to test each of the given options. We can do this by dividing the original quadratic expression by each potential factor. If the division results in a polynomial with no remainder, then that option is indeed a factor.

Option A: (h + 2)

Let's divide $-2h^2 - 15h - 7$ by (h + 2). We can use polynomial long division or synthetic division for this. For brevity, we'll skip the detailed steps here, but if you perform the division, you'll find that there is a remainder. So, (h + 2) is not a factor.

Option B: (3h - 2)

Similarly, dividing $-2h^2 - 15h - 7$ by (3h - 2) will also result in a remainder. This means (3h - 2) is not a factor either.

Option C: (2h + 1)

When we divide $-2h^2 - 15h - 7$ by (2h + 1), we get -h - 7 with no remainder. This indicates that (2h + 1) is a factor of the quadratic expression.

Option D: (h - 7)

Dividing $-2h^2 - 15h - 7$ by (h - 7) also results in a remainder, so (h - 7) is not a factor.

Comparing Results and Answering the Question

Both methods have led us to the same conclusion. Let's recap:

• We factored the quadratic expression $-2h^2 - 15h - 7$ and found its factors to be -(h + 7)(2h + 1).
• We tested each option by dividing the quadratic expression by the potential factor. Only (2h + 1) resulted in no remainder.

Therefore, the correct answer is C. One of the factors is (2h + 1).

Tips and Tricks for Factoring

Before we wrap up, let's share a few tips and tricks that can make factoring easier:

• Always look for a common factor first. This simplifies the expression and makes it easier to factor further.
• Practice, practice, practice! The more you factor, the more comfortable you'll become with the process.
• Use the "ac method" (the method we used earlier) when the leading coefficient is not 1.
• Check your work by multiplying the factors back together to make sure you get the original expression.

Conclusion

Factoring quadratic expressions can seem daunting at first, but with a solid understanding of the concepts and a systematic approach, you can master it. We tackled the problem of finding the correct factor of $-2h^2 - 15h - 7$ using two methods: factoring the quadratic directly and testing the options by division. Both methods confirmed that (2h + 1) is indeed a factor.

Remember, factoring is a fundamental skill in algebra, so keep practicing and building your confidence. You've got this!