Polynomial Division & Partial Fractions: Solving $\frac{x^3-2 X^2-67 X-24}{2 X^2-x-21}$

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Hey guys, let's dive into a super cool math problem today, tackling polynomial division and partial fractions! We're looking at how to represent the quotient and remainder of a complex fraction, specifically $\frac{x^3-2 x^2-67 x-24}{2 x^2-x-21}$, in its correct partial fraction form. This isn't just about crunching numbers; it's about understanding the structure of rational functions and how to break them down into simpler, more manageable pieces. When you're dealing with fractions where the numerator and denominator are polynomials, especially when the degree of the numerator is higher than or equal to the degree of the denominator, polynomial long division is your best friend. It helps us separate the 'whole part' (the quotient) from the 'fractional part' (the remainder over the original denominator). But the journey doesn't stop there! The real magic happens when we express that fractional part using partial fractions. This technique is fundamental in calculus for integration, in differential equations, and in signal processing. So, let's get our hands dirty and figure out which of the given expressions truly represents the correct form. We'll be breaking down the denominator, performing the division, and then reconstructing the expression in its partial fraction form. Pay close attention to the steps, guys, because understanding this process will unlock a whole new level of mathematical comprehension for you all. It’s all about building a solid foundation, and this problem is a fantastic way to reinforce those key concepts. We want to make sure we're not just getting the right answer, but why it's the right answer, and how this method applies more broadly to other similar problems you might encounter.

Understanding the Problem: Quotient and Remainder as Partial Fractions

Alright, team, let's get into the nitty-gritty of what we're actually trying to achieve with this problem. We've got a rational expression, $\frac{x^3-2 x^2-67 x-24}{2 x^2-x-21}$, and we need to express it in a specific format involving a quotient and remainders represented as partial fractions. First off, why do we even care about this? Well, when the degree of the polynomial in the numerator is greater than or equal to the degree of the polynomial in the denominator, we can't directly apply standard partial fraction decomposition. We first need to perform polynomial long division. This division gives us a polynomial part (the quotient) and a proper rational function part (the remainder over the original denominator). It's like saying, 'This big, messy fraction is actually a simple polynomial plus a smaller, more manageable fraction.' Once we have that smaller, proper fraction, then we can decompose it into partial fractions. The general form of partial fraction decomposition depends on the factors of the denominator. If the denominator has distinct linear factors, like $(x-a)(x-b)$, the decomposition looks like $\frac{A}{x-a} + \frac{B}{x-b}$. If it has repeated linear factors, like $(x-a)^2$, it looks like $\frac{A}{x-a} + \frac{B}{(x-a)^2}$. And if it has irreducible quadratic factors, things get a bit more complex with terms like $\frac{Ax+B}{x^2+bx+c}$.

Our specific problem involves the expression $\frac{x^3-2 x^2-67 x-24}{2 x^2-x-21}$. Notice that the degree of the numerator (3) is greater than the degree of the denominator (2). This immediately tells us that polynomial long division is our first step. After the division, we'll get an expression in the form $Q(x) + \frac{R(x)}{2 x^2-x-21}$, where $Q(x)$ is the quotient (a polynomial) and $R(x)$ is the remainder (a polynomial with a degree less than 2). The crucial part is how this $Q(x)$ and the decomposed form of $\frac{R(x)}{2 x^2-x-21}$ are represented. The options given are:

A. $A+\frac{B}{x+3}+\frac{C}{2 x-7}$ B. $A+\frac{B r+C}{x+3}+\frac{B r+E}{2 x-7}$ C. $A D i s c u s s i o n c a t e g o r y : m a t h e m a t i c s$

Let's break down what these options imply. The term '$A$' in each option likely represents the polynomial quotient $Q(x)$ obtained from the long division. Since the division is of a cubic polynomial by a quadratic polynomial, the quotient will be a linear polynomial, typically of the form $ax+b$. If $A$ represents this quotient, it should be a polynomial, not just a constant. However, sometimes in these contexts, especially if the leading coefficients align nicely, the 'constant' A might stand in for a linear term if the degrees work out perfectly, or it might be a simplification where the problem setter implies A is the form of the quotient. Let's assume for now that 'A' represents the polynomial quotient.

Now, let's look at the fractional parts. The denominator $2x^2 - x - 21$ needs to be factored. If we can factor it into linear terms, say $(px+q)(rx+s)$, then the remainder term $\frac{R(x)}{2 x^2-x-21}$ will be decomposed into partial fractions based on these factors. Option A suggests that the denominator factors into $(x+3)$ and $(2x-7)$ and the decomposition of the remainder is of the form $\frac{B}{x+3} + \frac{C}{2x-7}$. This is the standard form for distinct linear factors. Option B presents a more complex structure for the numerators of the partial fractions: $\frac{Br+C}{x+3}$ and $\frac{Br+E}{2x-7}$. This form is unusual and generally not the standard way to represent partial fractions for linear factors. Typically, for distinct linear factors, the numerators are constants.

So, our mission is clear:

1. Factor the denominator $2x^2 - x - 21$.
2. Perform polynomial long division of $x^3 - 2x^2 - 67x - 24$ by $2x^2 - x - 21$ to find the quotient $Q(x)$ and the remainder $R(x)$.
3. Decompose the remainder fraction $\frac{R(x)}{2x^2 - x - 21}$ into partial fractions based on the factored denominator.
4. Compare the result with the given options to find the correct representation.

This is where the real mathematical detective work begins, guys! Let's break it down piece by piece.

Step 1: Factoring the Denominator

Alright, first things first, we need to get our hands on the factors of the denominator, $2x^2 - x - 21$. This is a quadratic expression, and factoring it will tell us what kind of partial fraction decomposition we'll be dealing with. For a quadratic of the form $ax^2 + bx + c$, we're looking for two binomials $(px+q)(rx+s)$ such that their product equals the original quadratic. In our case, $a=2$, $b=-1$, and $c=-21$. We need to find two numbers that multiply to $a imes c = 2 imes (-21) = -42$ and add up to $b = -1$. Let's brainstorm pairs of factors for -42:

• 1 and -42 (sum = -41)
• -1 and 42 (sum = 41)
• 2 and -21 (sum = -19)
• -2 and 21 (sum = 19)
• 3 and -14 (sum = -11)
• -3 and 14 (sum = 11)
• 6 and -7 (sum = -1)
• -6 and 7 (sum = 1)

Bingo! The pair 6 and -7 adds up to -1. Now, we use these numbers to split the middle term $(-x)$: $2x^2 - x - 21 = 2x^2 + 6x - 7x - 21$

Now, we can factor by grouping:

• Group the first two terms: $2x^2 + 6x = 2x(x + 3)$
• Group the last two terms: $-7x - 21 = -7(x + 3)$

Notice that we have a common factor of $(x+3)$. So, we can rewrite the expression as:

$2x(x + 3) - 7(x + 3) = (2x - 7)(x + 3)$

Fantastic! We've successfully factored the denominator into two distinct linear factors: $(2x - 7)$ and $(x + 3)$. This tells us that our partial fraction decomposition for the remainder term will involve denominators of $(2x-7)$ and $(x+3)$. This is a crucial piece of information that helps us start evaluating the options. We see these factors appearing in options A and B, which is a good sign. Option C seems like a non-starter as it's gibberish. Let's keep this factorization in mind as we move on to the next, and arguably most intensive, step: polynomial long division.

Step 2: Performing Polynomial Long Division

Now for the main event, guys – polynomial long division! We need to divide the numerator, $x^3 - 2x^2 - 67x - 24$, by the denominator, $2x^2 - x - 21$. This process is very similar to regular long division with numbers, but we're dealing with algebraic terms. Remember, the goal here is to find the quotient $Q(x)$ and the remainder $R(x)$ such that:

$\frac{x^3 - 2x^2 - 67x - 24}{2x^2 - x - 21} = Q(x) + \frac{R(x)}{2x^2 - x - 21}$

Let's set it up:

        _____________
2x^2-x-21 | x^3 - 2x^2 - 67x - 24

First Step: We look at the leading terms of both the dividend ($x^3$) and the divisor ($2x^2$). We ask, 'What do we need to multiply $2x^2$ by to get $x^3$?' The answer is $\frac{x^3}{2x^2} = \frac{1}{2}x$. This is the first term of our quotient.

        1/2 x ________
2x^2-x-21 | x^3 - 2x^2 - 67x - 24

Second Step: Multiply the first term of the quotient ($\frac{1}{2}x$) by the entire divisor ($2x^2 - x - 21$): $(\frac{1}{2}x)(2x^2 - x - 21) = x^3 - \frac{1}{2}x^2 - \frac{21}{2}x$

Third Step: Subtract this result from the dividend: $(x^3 - 2x^2 - 67x - 24) - (x^3 - \frac{1}{2}x^2 - \frac{21}{2}x)$

Remember to distribute the minus sign: $x^3 - 2x^2 - 67x - 24 - x^3 + \frac{1}{2}x^2 + \frac{21}{2}x$

Combine like terms: $(-2x^2 + \frac{1}{2}x^2) + (-67x + \frac{21}{2}x) - 24$

To combine the $x^2$ terms: $-2 + \frac{1}{2} = -\frac{4}{2} + \frac{1}{2} = -\frac{3}{2}x^2$ To combine the $x$ terms: $-67 + \frac{21}{2} = -\frac{134}{2} + \frac{21}{2} = -\frac{113}{2}x$

So, the result of the subtraction is: $-\frac{3}{2}x^2 - \frac{113}{2}x - 24$

Now, we bring down the next term (which is already there, -24):

        1/2 x ________
2x^2-x-21 | x^3 - 2x^2 - 67x - 24
        -(x^3 - 1/2 x^2 - 21/2 x)
        ---------------------
              -3/2 x^2 - 113/2 x - 24

Fourth Step: Repeat the process with the new polynomial ($-\frac{3}{2}x^2 - \frac{113}{2}x - 24$). Look at the leading terms: $-\frac{3}{2}x^2$ and $2x^2$. What do we multiply $2x^2$ by to get $-\frac{3}{2}x^2$? The answer is $\frac{-3/2 x^2}{2x^2} = -\frac{3}{4}$. This is the second term of our quotient.

        1/2 x - 3/4 ____
2x^2-x-21 | x^3 - 2x^2 - 67x - 24
        -(x^3 - 1/2 x^2 - 21/2 x)
        ---------------------
              -3/2 x^2 - 113/2 x - 24

Fifth Step: Multiply the second term of the quotient ($-\frac{3}{4}$) by the entire divisor ($2x^2 - x - 21$): $(\frac{-3}{4})(2x^2 - x - 21) = -\frac{6}{4}x^2 + \frac{3}{4}x + \frac{63}{4}$

Simplify: $-\frac{3}{2}x^2 + \frac{3}{4}x + \frac{63}{4}$

Sixth Step: Subtract this result from the current polynomial:

$(-\frac{3}{2}x^2 - \frac{113}{2}x - 24) - (-\frac{3}{2}x^2 + \frac{3}{4}x + \frac{63}{4})$

Distribute the minus sign: $-\frac{3}{2}x^2 - \frac{113}{2}x - 24 + \frac{3}{2}x^2 - \frac{3}{4}x - \frac{63}{4}$

Combine like terms: $(-\frac{3}{2}x^2 + \frac{3}{2}x^2) + (-\frac{113}{2}x - \frac{3}{4}x) + (-24 - \frac{63}{4})$

The $x^2$ terms cancel out, which is good!

Now, let's combine the $x$ terms: $-\frac{113}{2} - \frac{3}{4}$. We need a common denominator of 4: $-\frac{113 imes 2}{2 imes 2} - \frac{3}{4} = -\frac{226}{4} - \frac{3}{4} = -\frac{229}{4}x$

And combine the constant terms: $-24 - \frac{63}{4}$. Common denominator is 4: $-\frac{24 imes 4}{4} - \frac{63}{4} = -\frac{96}{4} - \frac{63}{4} = -\frac{159}{4}$

So, the remainder is: $-\frac{229}{4}x - \frac{159}{4}$

        1/2 x - 3/4 ____
2x^2-x-21 | x^3 - 2x^2 - 67x - 24
        -(x^3 - 1/2 x^2 - 21/2 x)
        ---------------------
              -3/2 x^2 - 113/2 x - 24
            -(-3/2 x^2 + 3/4 x + 63/4)
            -------------------------
                    -229/4 x - 159/4

Since the degree of the remainder ($-\frac{229}{4}x - \frac{159}{4}$) is 1, which is less than the degree of the divisor (2), we stop here.

Our result from the long division is: Quotient $Q(x) = \frac{1}{2}x - \frac{3}{4}$ Remainder $R(x) = -\frac{229}{4}x - \frac{159}{4}$

So, we can write the original expression as: $\frac{x^3 - 2x^2 - 67x - 24}{2x^2 - x - 21} = (\frac{1}{2}x - \frac{3}{4}) + \frac{-\frac{229}{4}x - \frac{159}{4}}{2x^2 - x - 21}$

Now, let's look at the options again. The quotient $Q(x) = \frac{1}{2}x - \frac{3}{4}$ is a linear polynomial. If option 'A' uses '$A$' to represent the quotient, it implies '$A$' should be a linear term, not just a constant. If '$A$' represents the entire polynomial quotient, then our calculation matches the structure of option A, assuming '$A$' stands for $\frac{1}{2}x - \frac{3}{4}$, and the remainder is decomposed into $\frac{B}{x+3} + \frac{C}{2x-7}$. Option B suggests a more complicated form for the numerators of the partial fractions, which is not standard for distinct linear factors.

Step 3: Decomposing the Remainder into Partial Fractions

We've done the heavy lifting with the long division, and now we need to decompose the remainder part: $\frac{-\frac{229}{4}x - \frac{159}{4}}{2x^2 - x - 21}$. We already know from Step 1 that the denominator factors into $(x+3)(2x-7)$. So, we want to find constants $B$ and $C$ such that:

$\frac{-\frac{229}{4}x - \frac{159}{4}}{(x+3)(2x-7)} = \frac{B}{x+3} + \frac{C}{2x-7}$

This is the standard form for partial fraction decomposition with distinct linear factors. This matches the fractional part of Option A. Let's go ahead and find $B$ and $C$ to be thorough, although the form itself is often enough to answer the question.

To find $B$ and $C$, we first combine the right side over a common denominator:

$\frac{B(2x-7) + C(x+3)}{(x+3)(2x-7)}$

Now, we equate the numerators:

$-\frac{229}{4}x - \frac{159}{4} = B(2x-7) + C(x+3)$

We can solve for $B$ and $C$ using two methods: equating coefficients or the Heaviside cover-up method.

Method 1: Equating Coefficients

Expand the right side: $-\frac{229}{4}x - \frac{159}{4} = 2Bx - 7B + Cx + 3C$

Group terms by powers of $x$: $-\frac{229}{4}x - \frac{159}{4} = (2B + C)x + (-7B + 3C)$

Now, equate the coefficients of $x$ and the constant terms:

1. Coefficient of $x$: $2B + C = -\frac{229}{4}$
2. Constant term: $-7B + 3C = -\frac{159}{4}$

We have a system of two linear equations with two variables. Let's solve for $C$ in the first equation: $C = -\frac{229}{4} - 2B$. Substitute this into the second equation: $-7B + 3(-\frac{229}{4} - 2B) = -\frac{159}{4}$ $-7B - \frac{687}{4} - 6B = -\frac{159}{4}$ Combine $B$ terms: $-13B - \frac{687}{4} = -\frac{159}{4}$ Add $\frac{687}{4}$ to both sides: $-13B = -\frac{159}{4} + \frac{687}{4}$ $-13B = \frac{528}{4}$ $-13B = 132$ $B = \frac{132}{-13} = -\frac{132}{13}$

Now substitute $B$ back into the equation for $C$: $C = -\frac{229}{4} - 2(-\frac{132}{13})$ $C = -\frac{229}{4} + \frac{264}{13}$ Find a common denominator (52): $C = -\frac{229 imes 13}{4 imes 13} + \frac{264 imes 4}{13 imes 4}$ $C = -\frac{2977}{52} + \frac{1056}{52}$ $C = -\frac{1921}{52}$

So, $B = -\frac{132}{13}$ and $C = -\frac{1921}{52}$. These are the specific constants, but notice the form of the decomposition: $\frac{B}{x+3} + \frac{C}{2x-7}$

This perfectly matches the fractional part of Option A.

Method 2: Heaviside Cover-Up Method

This method is often quicker for distinct linear factors. To find $B$, cover up the $(x+3)$ factor in the original fraction and substitute $x = -3$ (the root of $x+3$):

$B = \frac{-\frac{229}{4}(-3) - \frac{159}{4}}{2(-3)-7} = \frac{\frac{687}{4} - \frac{159}{4}}{-6-7} = \frac{\frac{528}{4}}{-13} = \frac{132}{-13} = -\frac{132}{13}$

To find $C$, cover up the $(2x-7)$ factor and substitute $x = \frac{7}{2}$ (the root of $2x-7$):

$C = \frac{-\frac{229}{4}(\frac{7}{2}) - \frac{159}{4}}{(\frac{7}{2})+3} = \frac{-\frac{1503}{8} - \frac{159 imes 2}{8}}{\frac{7}{2}+\frac{6}{2}} = \frac{-\frac{1503}{8} - \frac{318}{8}}{\frac{13}{2}} = \frac{-\frac{1821}{8}}{\frac{13}{2}}$

$C = -\frac{1821}{8} imes \frac{2}{13} = -\frac{1821}{4 imes 13} = -\frac{1821}{52}$

Wait, there's a discrepancy between the two methods! Let's re-check the calculation for C using the cover-up method.

$C = \frac{-\frac{229}{4}(\frac{7}{2}) - \frac{159}{4}}{(\frac{7}{2})+3}$ Numerator: $-\frac{1603}{8} - \frac{159 imes 2}{8} = -\frac{1603}{8} - \frac{318}{8} = -\frac{1921}{8}$ Denominator: $\frac{7}{2} + 3 = \frac{7}{2} + \frac{6}{2} = \frac{13}{2}$

$C = \frac{-\frac{1921}{8}}{\frac{13}{2}} = -\frac{1921}{8} imes \frac{2}{13} = -\frac{1921}{4 imes 13} = -\frac{1921}{52}$

Okay, the cover-up method result for C is $C = -\frac{1921}{52}$. Let's check the equating coefficients method again for $C$. Ah, I found the error in the equating coefficients calculation. Let's re-calculate $C$ using $B = -\frac{132}{13}$ and the first equation $2B + C = -\frac{229}{4}$.

$C = -\frac{229}{4} - 2B = -\frac{229}{4} - 2(-\frac{132}{13})$ $C = -\frac{229}{4} + \frac{264}{13}$ Common denominator is 52: $C = \frac{-229 imes 13}{52} + \frac{264 imes 4}{52} = \frac{-2977}{52} + \frac{1056}{52} = \frac{-1921}{52}$

Both methods now agree: $B = -\frac{132}{13}$ and $C = -\frac{1921}{52}$. This confirms that the partial fraction decomposition of the remainder is indeed $\frac{-132/13}{x+3} + \frac{-1921/52}{2x-7}$.

Step 4: Determining the Correct Expression

We've successfully performed the polynomial long division and analyzed the structure of the partial fraction decomposition. Let's consolidate our findings:

1. Polynomial Long Division: We found the quotient $Q(x) = \frac{1}{2}x - \frac{3}{4}$ and the remainder $R(x) = -\frac{229}{4}x - \frac{159}{4}$. So, $\frac{x^3 - 2x^2 - 67x - 24}{2x^2 - x - 21} = (\frac{1}{2}x - \frac{3}{4}) + \frac{-\frac{229}{4}x - \frac{159}{4}}{2x^2 - x - 21}$.

2. Factoring the Denominator: We found $2x^2 - x - 21 = (x+3)(2x-7)$.

3. Partial Fraction Decomposition of Remainder: We confirmed that the remainder fraction decomposes into the form $\frac{B}{x+3} + \frac{C}{2x-7}$, where $B = -\frac{132}{13}$ and $C = -\frac{1921}{52}$.

Now, let's examine the given options:

• A. $A+\frac{B}{x+3}+\frac{C}{2 x-7}$ This option suggests that the expression consists of a term '$A$', plus the sum of two partial fractions with constant numerators ($B$ and $C$) corresponding to the linear factors of the denominator ($x+3$ and $2x-7$). If '$A$' represents the polynomial quotient $Q(x) = \frac{1}{2}x - \frac{3}{4}$, and $B$ and $C$ represent the constants we found for the partial fractions of the remainder, then this form perfectly matches our result. The use of '$A$' is a bit loose if it's meant to represent a linear polynomial, but in multiple-choice questions, '$A$' often stands for the polynomial quotient, and the structure is what matters. The structure is a polynomial (quotient) plus the sum of simple fractions based on the denominator's factors.

• B. $A+\frac{B r+C}{x+3}+\frac{B r+E}{2 x-7}$ This option has a structure where the numerators of the partial fractions are themselves linear expressions ($Br+C$ and $Br+E$). This is not the standard form for decomposing a proper rational function with distinct linear denominators. The standard form requires constant numerators (like $B$ and $C$ in Option A). This option implies a more complex relationship or a misunderstanding of partial fraction decomposition rules.

• C. $A D i s c u s s i o n c a t e g o r y : m a t h e m a t i c s$ This option is clearly nonsensical and does not represent any mathematical expression.

Conclusion:

Based on our detailed analysis, Option A provides the correct form for the expression. The term '$A$' represents the polynomial quotient obtained from the long division (which is $\frac{1}{2}x - \frac{3}{4}$ in this case), and the remaining terms represent the partial fraction decomposition of the remainder, $\frac{B}{x+3} + \frac{C}{2x-7}$. The fact that the denominator factored into distinct linear terms dictates this standard partial fraction form.

Therefore, the expression that represents the correct form for the quotient and remainder, written as partial fractions, of $\frac{x^3-2 x^2-67 x-24}{2 x^2-x-21}$ is Option A.

Keep practicing these techniques, guys! Polynomial long division and partial fraction decomposition are fundamental tools in higher mathematics, and mastering them will make tackling complex problems feel way less intimidating. Great job working through this one!