Factoring Quadratics: Find The Factors Of 2x^2+5x-12

Hey guys! Today, we're diving into the awesome world of quadratic equations, specifically how to find the factors of a given quadratic expression. You know, those times when you're staring at something like $2x^2+5x-12$ and wondering, "What in the world multiplies to make that?" Well, wonder no more! We're going to break down how to find its factors, and by the end of this, you'll be spotting them like a pro. We'll be looking at a few options, and you'll need to figure out which one is the real factor. So, buckle up, grab your thinking caps, and let's get this math party started!

Understanding Quadratic Expressions and Factors

Alright, let's get down to business. What exactly is a quadratic expression, and what does it mean to find its factors? Think of it like breaking down a big number into smaller numbers that multiply together to give you the original. For example, the factors of 12 are 2 and 6, or 3 and 4, because $2 \times 6 = 12$ and $3 \times 4 = 12$. Quadratic expressions are just polynomials with the highest power of the variable being 2. Our expression here, $2x^2+5x-12$, is a perfect example. It's made up of an $x^2$ term, an $x$ term, and a constant term. When we talk about factoring a quadratic expression, we're essentially trying to rewrite it as a product of two simpler expressions, usually binomials (expressions with two terms). So, we're looking for something like $(ax+b)(cx+d)$ that, when multiplied out, equals $2x^2+5x-12$. Finding these factors is super useful in algebra, especially when you're solving quadratic equations because it helps you find the roots (the values of $x$ that make the equation equal to zero). It's like unlocking a secret code for the equation! We'll explore different methods to tackle this, but for this specific problem, we're given a few choices and need to pick the correct one. This often involves a bit of trial and error, or a more systematic approach like the 'ac method' or 'splitting the middle term'. The goal is to reverse the FOIL method (First, Outer, Inner, Last), which is how you multiply two binomials together. So, if $(ax+b)(cx+d)$ is the factored form, then $acx^2 + (ad+bc)x + bd$ is the expanded form. We need to match the coefficients of our target quadratic $2x^2+5x-12$ with the expanded form to find the right $a, b, c, d$ values.

The 'AC Method' for Factoring Quadratics

So, how do we actually find these factors systematically? One of the most popular and effective methods is the 'AC method'. It's a lifesaver, guys! For a quadratic expression in the form $ax^2+bx+c$, you first multiply the coefficient of the $x^2$ term ($a$) by the constant term ($c$). In our case, $a=2$ and $c=-12$, so $a \times c = 2 \times (-12) = -24$. Now, here's the crucial part: you need to find two numbers that multiply to give you this product (-24) and add up to give you the coefficient of the $x$ term ($b$), which is 5 in our expression. So, we're searching for two numbers, let's call them $p$ and $q$, such that $p \times q = -24$ and $p + q = 5$. Let's brainstorm some pairs of numbers that multiply to -24:

• 1 and -24 (sum is -23)
• -1 and 24 (sum is 23)
• 2 and -12 (sum is -10)
• -2 and 12 (sum is 10)
• 3 and -8 (sum is -5)
• -3 and 8 (sum is 5)
• 4 and -6 (sum is -2)
• -4 and 6 (sum is 2)

Boom! We found our pair: -3 and 8! They multiply to -24 and add up to 5. This is the magic key. Now, we use these numbers to rewrite the middle term ($+5x$) of our original quadratic expression. We split $5x$ into $-3x + 8x$ (or $8x - 3x$, the order doesn't matter). So, our expression becomes $2x^2 - 3x + 8x - 12$. The next step is to factor by grouping. We group the first two terms and the last two terms: $(2x^2 - 3x) + (8x - 12)$. Now, find the greatest common factor (GCF) for each group. In the first group, the GCF is $x$, so we get $x(2x - 3)$. In the second group, the GCF is 4, so we get $4(2x - 3)$. Notice something cool? Both groups have the same binomial factor: $(2x - 3)$. This is a great sign that we're on the right track! Now, we factor out this common binomial: $(2x - 3)(x + 4)$. And there you have it! The factors of $2x^2+5x-12$ are $(2x-3)$ and $(x+4)$.

Testing the Given Options

Okay, so we found the factors using the AC method: $(2x-3)$ and $(x+4)$. Now, let's look at the options provided in the question to see which one matches what we found. The options are:

A. $(2x-3)$ B. $(2x-4)$ C. $(2x+3)$ D. $(x-4)$

Comparing our results, we see that option A, $(2x-3)$, is one of the factors we discovered. That's awesome! But just to be thorough and to make sure we didn't miss anything, let's quickly check the other options. We can do this by trying to multiply each option by a potential second factor to see if we get our original quadratic. For instance, let's test option B, $(2x-4)$. If $(2x-4)$ were a factor, then $2x-4$ would divide evenly into $2x^2+5x-12$. Also, notice that $(2x-4)$ can be simplified to $2(x-2)$. If $(x-2)$ were a factor, then plugging $x=2$ into the quadratic should give 0. Let's check: $2(2)^2 + 5(2) - 12 = 2(4) + 10 - 12 = 8 + 10 - 12 = 18 - 12 = 6$. Since it's not 0, $(x-2)$ is not a factor, and thus $(2x-4)$ is not a factor. Now let's look at option C, $(2x+3)$. If this were a factor, we'd be looking for another factor $(x+k)$ such that $(2x+3)(x+k) = 2x^2+5x-12$. Expanding this gives $2x^2 + 2kx + 3x + 3k = 2x^2 + (2k+3)x + 3k$. Comparing coefficients, we need $2k+3 = 5$ (which means $2k=2$, so $k=1$) and $3k = -12$ (which means $k=-4$). Since we got different values for $k$, $(2x+3)$ is not a factor. Finally, let's test option D, $(x-4)$. If $(x-4)$ were a factor, then plugging $x=4$ into the quadratic should give 0. Let's check: $2(4)^2 + 5(4) - 12 = 2(16) + 20 - 12 = 32 + 20 - 12 = 52 - 12 = 40$. Since it's not 0, $(x-4)$ is not a factor. This confirms that our initial finding was correct, and option A is indeed the factor. It's always good practice to double-check your work, especially when dealing with multiple-choice questions. You can also check by multiplying out the factors we found: $(2x-3)(x+4) = 2x(x) + 2x(4) - 3(x) - 3(4) = 2x^2 + 8x - 3x - 12 = 2x^2 + 5x - 12$. Perfect match!

The Importance of Factoring Quadratics

So, why bother with all this factoring jazz? Well, guys, understanding how to factor quadratic expressions is a fundamental skill in mathematics that opens up a whole world of possibilities. Primarily, it's essential for solving quadratic equations. Remember when we found the factors of $2x^2+5x-12$ were $(2x-3)$ and $(x+4)$? If we set the expression equal to zero, $2x^2+5x-12=0$, we can then write it as $(2x-3)(x+4)=0$. The Zero Product Property tells us that if the product of two things is zero, then at least one of those things must be zero. So, either $2x-3=0$ or $x+4=0$. Solving these simple linear equations gives us the roots of the quadratic: $2x=3 ightarrow x = 3/2$ and $x=-4$. These are the specific values of $x$ that make the original quadratic equation true. Without factoring, finding these roots can be much more difficult, often requiring the quadratic formula, which is another powerful tool but factoring can be quicker when possible. Beyond solving equations, factoring is crucial for simplifying rational expressions (fractions with polynomials). Imagine you have a complex fraction like $\frac{2x^2+5x-12}{x^2-16}$. By factoring both the numerator and the denominator, you can simplify it significantly: $\frac{(2x-3)(x+4)}{(x-4)(x+4)}$. You can then cancel out the common factor $(x+4)$, leaving you with a much simpler expression $\frac{2x-3}{x-4}$ (provided $x \neq -4$). This simplification is vital in calculus when dealing with limits and derivatives. Furthermore, factoring helps in graphing quadratic functions. The factors tell you the x-intercepts of the parabola, which are key points for sketching its graph. Knowing the roots helps you understand where the function crosses the x-axis, giving you a better visual representation of the function's behavior. It's also a building block for understanding more advanced algebraic concepts like partial fraction decomposition, which is used extensively in integration in calculus. So, while it might seem like just another math exercise, mastering factoring quadratics is like getting a master key to unlock many more complex mathematical doors. It's a skill that pays off massively as you progress in your math journey!

Conclusion: You've Mastered Factoring!

So, there you have it, folks! We took on the challenge of finding a factor for the quadratic expression $2x^2+5x-12$. We explored the 'AC method', a super reliable technique for breaking down quadratics into their binomial components. We identified that the key was finding two numbers that multiply to -24 (which is $a \times c$) and add up to 5 (which is $b$). Those magic numbers turned out to be 8 and -3. By rewriting the middle term and factoring by grouping, we successfully factored the expression into $(2x-3)(x+4)$. Then, we systematically checked the given options and confirmed that A. $(2x-3)$ is indeed one of the factors. Remember, factoring isn't just about solving one problem; it's a foundational skill that helps you solve quadratic equations, simplify complex expressions, and understand the behavior of functions. Keep practicing, and you'll become a factoring ninja in no time! Math is all about building these skills step-by-step, and you've just conquered another important one. High fives all around!