Polynomial Factored Form Calculation With A=-5 D=42 And X=-12

Hey guys! Today, we're diving deep into the fascinating world of polynomials, specifically focusing on how to express them in factored form. Factoring polynomials can sometimes feel like solving a puzzle, but once you understand the key principles, it becomes a whole lot easier and even kind of fun! We're going to tackle a specific example where we have some initial values: a = -5, d = 42, and x = -12. Our main goal is to complete a polynomial and write it in its factored form. Let's get started!

Understanding the Basics of Polynomials and Factored Form

Before we jump into the nitty-gritty details, let's quickly recap what polynomials and factored forms are all about. Polynomials are algebraic expressions that consist of variables and coefficients, combined using addition, subtraction, and non-negative integer exponents. Think of them as a chain of terms, where each term is a constant multiplied by a variable raised to a power. For example, 3x^2 + 2x - 1 is a polynomial.

Now, what about the factored form? Well, the factored form of a polynomial is essentially expressing it as a product of its factors. Factors are smaller polynomials that, when multiplied together, give you the original polynomial. It's like breaking down a number into its prime factors – but with polynomials! Factoring helps us understand the roots (or zeros) of the polynomial, which are the values of x that make the polynomial equal to zero. These roots are super important because they tell us where the polynomial crosses the x-axis on a graph.

Factoring polynomials can seem daunting at first, but it's a fundamental skill in algebra and calculus. It allows us to simplify complex expressions, solve equations, and analyze the behavior of functions. There are several techniques for factoring, including finding common factors, using special factoring patterns (like the difference of squares), and employing the quadratic formula. We'll touch on some of these techniques as we work through our example.

Why Factored Form Matters

You might be wondering, “Why bother with factored form at all?” Great question! Factored form is incredibly useful for several reasons. First and foremost, it makes it easy to find the roots or zeros of the polynomial. Remember, the roots are the values of x that make the polynomial equal to zero. In factored form, each factor corresponds to a root. For example, if we have a factor (x - a), then x = a is a root. Knowing the roots helps us graph the polynomial, solve equations, and analyze its behavior.

Another advantage of factored form is that it simplifies many algebraic manipulations. When we have a polynomial in factored form, it's often easier to perform operations like division, simplification, and finding common denominators. This is especially useful in calculus, where we often need to work with complex rational expressions. So, mastering the art of factoring polynomials is a valuable investment in your mathematical toolkit.

Setting Up the Problem: p(x) = (x + 5)(x - □)(x + □)

Alright, let's get back to our specific problem. We're given the polynomial in a partially factored form: p(x) = (x + 5)(x - □)(x + □). Our mission, should we choose to accept it (and we do!), is to fill in the blanks. We know that a = -5, d = 42, and x = -12. These values are crucial clues that will help us crack the code. The a value is directly related to the first factor, (x + 5). If we set this factor equal to zero, we get x + 5 = 0, which means x = -5. This tells us that -5 is one of the roots of the polynomial.

The d value and the x value are a bit more mysterious at this stage, but don't worry, we'll figure out how they fit into the puzzle. The key is to understand that the d value likely relates to the other roots of the polynomial, and the x value might be a test point or a specific value we'll use to find the remaining factors. The factored form suggests that we're dealing with a cubic polynomial (a polynomial with a highest degree of 3), since we have three factors. This means we're looking for three roots in total.

Now, let's think strategically about how to find the missing values. We know one root is -5. We need to find two more. The d value, which is 42, could be related to the sum or product of the other roots, depending on the structure of the polynomial. The x value, which is -12, might be a value we can plug into the polynomial to get more information about the coefficients and roots. Let's start by exploring the relationship between the roots and the coefficients of a polynomial.

The Relationship Between Roots and Coefficients

One of the most powerful tools in our polynomial-solving arsenal is the relationship between the roots and the coefficients of a polynomial. For a cubic polynomial of the form ax^3 + bx^2 + cx + d, there are specific relationships between the roots (let's call them r1, r2, and r3) and the coefficients:

• Sum of the roots: r1 + r2 + r3 = -b/a
• Sum of the products of the roots taken two at a time: r1r2 + r1r3 + r2r3 = c/a
• Product of the roots: r1r2r3 = -d/a

These relationships, known as Vieta's formulas, provide a direct link between the roots and the coefficients of the polynomial. They're like a secret code that helps us decode the polynomial's structure. In our case, we have one root (r1 = -5) and some other information (d = 42, x = -12). We can use Vieta's formulas to help us find the other roots and fill in the blanks in our factored form.

Cracking the Code: Finding the Missing Factors

Okay, time to put on our detective hats and solve this puzzle! We know p(x) = (x + 5)(x - □)(x + □), and we want to find the values that go in the boxes. Let's call the missing roots r2 and r3. So, our factors will be (x - r2) and (x - r3). Remember, if we have a root r, the corresponding factor is (x - r). This is because when x = r, the factor becomes zero, making the whole polynomial zero.

We also know that one of our roots is -5, so the factor (x + 5) makes sense. Now, how do we find r2 and r3? This is where our given values of d and x come into play. The value d = 42 likely corresponds to the constant term in the polynomial when it's expanded. In the general cubic polynomial ax^3 + bx^2 + cx + d, the constant term d is related to the product of the roots. Let's think about this.

If we expand our factored form, we'll get a cubic polynomial. The constant term will be the product of the constant terms in each factor. In our case, the constant terms are 5, -r2, and -r3. So, the constant term in the expanded polynomial will be 5(-r2)(-r3) = 5r2r3. Now, we need to figure out how this relates to our given value of d = 42. We need to consider the leading coefficient of our polynomial as well.

Using the Value of d

Let's assume for a moment that the leading coefficient of our polynomial is 1. This means that when we expand our factored form, the coefficient of the x^3 term will be 1. In this case, the constant term d will be the negative of the product of the roots, according to Vieta's formulas. So, if d = 42, then the product of the roots (-5)(r2)(r3) should be equal to -42.

This gives us the equation (-5)(r2)(r3) = -42. If we divide both sides by -5, we get r2r3 = 42/5. This is a clue, but it's not quite enough to solve for r2 and r3 directly. We need another piece of information. This is where the value x = -12 comes in.

The Role of x = -12

The value x = -12 is likely a test point. This means we can plug x = -12 into our polynomial p(x) and see what happens. If we knew the value of p(-12), we could set up an equation and potentially solve for r2 and r3. However, we don't have the value of p(-12). So, we need to think a bit more creatively.

Let's go back to our factored form: p(x) = (x + 5)(x - r2)(x - r3). We know the roots are -5, r2, and r3. The factors (x - r2) and (x - r3) are the key to unlocking the final answer. We've used the value of d to get a relationship between r2 and r3: r2r3 = 42/5. Now, we need to find another relationship or use some trial and error to figure out the values of r2 and r3.

Solving for r2 and r3: A Bit of Trial and Error

Since we don't have a direct equation to solve for r2 and r3, let's try a bit of educated guessing and checking. We know that r2r3 = 42/5, which is approximately 8.4. This means r2 and r3 could be a pair of numbers that multiply to 8.4. Let's think about some possibilities. They could be fractions or decimals, but let's start by trying some simple whole numbers or fractions.

If we consider the original problem, it's likely that the missing values are relatively simple numbers. So, let's try some fractions that could give us 8.4 when multiplied. One possibility is r2 = 6 and r3 = 7/5 because 6 * (7/5) = 42/5. However, this would give us a slightly more complex factored form. Let's try something simpler. Since we have r2r3 = 42/5, we can rewrite this as 5r2r3 = 42. This gives us a hint that we might be dealing with a factor of 5 somewhere.

Another possibility is to consider fractions with a denominator of 5. Let's try r2 = -6 and see if we can find a corresponding r3. If r2 = -6, then r3 = (42/5) / (-6) = -7/5. This gives us the factors (x + 6) and (x + 7/5). However, this doesn't quite fit the form we have in the problem: (x - □)(x + □). We need one factor with a subtraction and one with an addition.

Let's think about another pair of factors that multiply to 42/5. What about r2 = 7 and r3 = 6/5? Then one factor would be (x-7). If r3=-6/5, then another factor would be (x+6/5). This looks better! We have one factor with a subtraction and one with an addition.

The Final Factored Form

After a bit of detective work, we've cracked the code! We found that the missing roots are likely 7 and -6/5. This means our factors are (x - 7) and (x + 6/5). So, the complete factored form of the polynomial is:

p(x) = (x + 5)(x - 7)(x + 6/5)

We can also rewrite the last factor to avoid fractions by multiplying it by 5/5, giving us:

p(x) = (x + 5)(x - 7)(5x + 6)

This is our final answer! We've successfully filled in the blanks and expressed the polynomial in factored form. Remember, this involved using the given values, understanding the relationship between roots and coefficients, and a bit of trial and error. Factoring polynomials can be a challenge, but it's also a rewarding experience that strengthens our algebraic skills. Great job, guys! Keep practicing, and you'll become factoring pros in no time!