Factor Completely: $5s^2 - 65s - 70$

Hey guys! Today we're diving deep into the world of algebra to tackle a factoring problem that might look a little intimidating at first glance: Factor completely the expression $5s^2 - 65s - 70$. Don't worry, though! By the end of this article, you'll be a factoring pro, ready to take on any quadratic expression that comes your way. We'll break down the process step-by-step, explaining each move so you can truly understand why we do what we do. This isn't just about getting the right answer; it's about building your mathematical muscles and confidence. So, grab your pencils, get comfy, and let's get factoring!

Understanding the Goal: What Does "Factor Completely" Mean?

Alright, so what does it actually mean to factor completely an expression like $5s^2 - 65s - 70$? Think of it like taking apart a complex machine into its simplest, individual components. In math, factoring means rewriting an expression as a product of its factors. When we say "completely," we mean we want to break it down as much as possible, ensuring that each factor itself cannot be factored any further. For polynomials, this usually means expressing it as a product of linear factors (like $(ax+b)$) or irreducible quadratic factors. Our goal with $5s^2 - 65s - 70$ is to find simpler expressions that, when multiplied together, give us back our original quadratic. This process is super useful in solving equations, simplifying fractions, and understanding the behavior of functions. It's a fundamental skill in algebra, and mastering it will open up a lot of doors for you in your math journey. So, let's roll up our sleeves and get to the first step of making this happen.

Step 1: Look for a Greatest Common Factor (GCF)

Okay, team, the very first thing you should always do when faced with a factoring problem, especially one involving polynomials like our $5s^2 - 65s - 70$, is to look for a Greatest Common Factor (GCF). This is your secret weapon for simplifying the expression before you even get to the more complex factoring methods. The GCF is the largest number or variable (or combination of both) that divides evenly into all the terms in the expression. In our case, we have three terms: $5s^2$, $-65s$, and $-70$. Let's examine the coefficients first: 5, -65, and -70. Can we find a number that divides into all of these? Absolutely! The number 5 is a factor of 5 ($5 imes 1$), it's a factor of -65 ($-65 = 5 imes -13$), and it's a factor of -70 ($-70 = 5 imes -14$). Since 5 is the smallest coefficient and it divides into the others, it's our numerical GCF. Now, let's look at the variables. We have $s^2$, $s$, and no $s$ in the last term. Since not all terms have an 's', the variable 's' is not part of our GCF. Therefore, the Greatest Common Factor (GCF) for $5s^2 - 65s - 70$ is just 5. This step is crucial because factoring out the GCF first makes the remaining expression much simpler to handle. It's like getting rid of the big, obvious pieces before you start sorting the smaller, fiddlier ones. So, let's pull that 5 out! We rewrite our expression as $5(s^2 - 13s - 14)$. See? Already looks a bit friendlier, right? This initial step is a game-changer, so never skip it!

Step 2: Factoring the Trinomial

Alright, now that we've successfully factored out the GCF of 5, we're left with the trinomial inside the parentheses: $s^2 - 13s - 14$. This is where the real fun begins! Our next mission is to factor the trinomial. Since this is a quadratic trinomial in the form $as^2 + bs + c$, where $a=1$, we're looking for two binomials of the form $(s+p)(s+q)$ that multiply back to give us $s^2 - 13s - 14$. When we multiply $(s+p)(s+q)$, we get $s^2 + (p+q)s + pq$. Comparing this to our trinomial, we need to find two numbers, $p$ and $q$, such that:

1. Their product ($pq$) equals the constant term, which is -14.
2. Their sum ($p+q$) equals the coefficient of the middle term, which is -13.

This is the core of factoring trinomials when $a=1$. We need to think of pairs of numbers that multiply to -14. Let's list them out:

• 1 and -14
• -1 and 14
• 2 and -7
• -2 and 7

Now, for each pair, let's check their sum to see if it equals -13:

• $1 + (-14) = -13$
• $-1 + 14 = 13$
• $2 + (-7) = -5$
• $-2 + 7 = 5$

Bingo! The pair $1$ and $-14$ gives us a sum of -13. So, our $p$ and $q$ are 1 and -14 (or vice versa, it doesn't matter). This means we can factor the trinomial $s^2 - 13s - 14$ into $(s+1)(s-14)$.

Let's quickly double-check this by multiplying it out: $(s+1)(s-14) = s(s-14) + 1(s-14) = s^2 - 14s + s - 14 = s^2 - 13s - 14$. Perfect! We've successfully factored the trinomial. Now we just need to combine this with the GCF we found earlier.

Step 3: Putting It All Together: The Complete Factorization

Now for the grand finale, guys! We've done the heavy lifting: we found the GCF and we factored the remaining trinomial. The final step is to put it all together to get the complete factorization of our original expression, $5s^2 - 65s - 70$. Remember, we started by factoring out the GCF, which was 5. This gave us $5(s^2 - 13s - 14)$. Then, we factored the trinomial inside the parentheses, $s^2 - 13s - 14$, into $(s+1)(s-14)$. So, to get the complete factorization, we simply combine these parts. The complete factored form of $5s^2 - 65s - 70$ is $5(s+1)(s-14)$.

To be absolutely sure we've got it right, let's perform one final check by multiplying everything out. We'll start with the binomials:

$(s+1)(s-14) = s^2 - 14s + s - 14 = s^2 - 13s - 14$

Now, multiply this result by the GCF, which is 5:

$5(s^2 - 13s - 14) = 5 imes s^2 - 5 imes 13s - 5 imes 14 = 5s^2 - 65s - 70$

And voilà! We've arrived back at our original expression. This confirms that our complete factorization $5(s+1)(s-14)$ is absolutely correct. Each of the factors, 5, $(s+1)$, and $(s-14)$, cannot be factored any further. The number 5 is a prime number, and $(s+1)$ and $(s-14)$ are linear binomials, which are as simple as they get. So, we've successfully factored completely!

Why is Factoring Completely So Important?

So, why do we bother with this whole process of factoring completely? It might seem like just another hoop to jump through in math class, but trust me, it's a super powerful tool that unlocks many other mathematical concepts. One of the most direct applications is in solving quadratic equations. If you have an equation like $5s^2 - 65s - 70 = 0$, setting it equal to zero means you can use the Zero Product Property. Once factored, $5(s+1)(s-14) = 0$, you can set each factor (or potential factor) equal to zero. For example, $(s+1)=0$ gives $s=-1$, and $(s-14)=0$ gives $s=14$. The factor of 5 doesn't change the solutions since $5 eq 0$. This makes finding the roots of polynomial equations significantly easier than trying to solve them through other methods, especially for higher-degree polynomials.

Another key area where factoring completely shines is in simplifying rational expressions (fractions involving polynomials). Imagine you have a complex fraction where both the numerator and the denominator are polynomials. By factoring both completely, you can often cancel out common factors, drastically simplifying the expression. This is akin to simplifying regular numerical fractions, like $10/20$ becoming $1/2$ by canceling out the common factor of 10. Simplifying rational expressions makes them easier to analyze, graph, and manipulate.

Furthermore, understanding how to factor completely builds a strong foundation for more advanced algebra topics like calculus and trigonometry. Concepts like finding limits, analyzing function behavior, and solving differential equations often rely on the ability to manipulate and simplify algebraic expressions, with factoring being a primary technique. So, while it might feel like a specific skill for quadratic equations, its importance ripples throughout your entire mathematical journey, making you a more versatile and capable problem-solver. Keep practicing, guys, and you'll see just how valuable this skill truly is!