Polynomial Division Made Easy: Divide 5x^4+5x^3-4x^2-5 By X+2

Dec 10, 2025 by Alex Johnson 62 views

H1: Polynomial Division Made Easy: Divide 5x^4+5x3-4x^2-5 by x+2

Welcome to our deep dive into the fascinating world of polynomial division! Today, we're going to tackle a specific problem: dividing the polynomial $5x^4 + 5x^3 - 4x^2 - 5$ by the binomial $x + 2$ . Don't let those exponents and coefficients intimidate you; we'll break it down step-by-step, making this mathematical task feel like a walk in the park. By the end of this article, you'll be comfortable expressing your answer in the standard form $Q(x) + rac{R(x)}{D(x)}$ , which is crucial for understanding various algebraic concepts. We'll also explore why this process is so important in mathematics and where you might encounter it in real-world applications.

Understanding the Basics of Polynomial Division

Before we jump into our specific example, let's get a solid grasp of what polynomial division actually is. Think of it as an extension of the long division you learned in elementary school, but instead of numbers, we're working with expressions containing variables and exponents. The goal is to divide a polynomial (the dividend) by another polynomial (the divisor) to find a quotient and a remainder. The remainder is what's 'left over' after the division. The standard form $Q(x) + rac{R(x)}{D(x)}$ is a way to express the result of this division, where $Q(x)$ is the quotient polynomial, $R(x)$ is the remainder polynomial (which has a degree less than the divisor), and $D(x)$ is the original divisor.

Why is this skill so important? Polynomial division is a fundamental tool in algebra. It's used to factor polynomials, find roots (or zeros) of polynomial equations, simplify complex rational expressions, and is a building block for more advanced mathematical topics like calculus and abstract algebra. Understanding how to perform polynomial division effectively will not only help you solve problems like the one we're about to tackle but will also build a stronger foundation for your overall mathematical journey. It's like learning to tie your shoelaces before you can run a marathon – a necessary, foundational skill that opens up a world of possibilities.

Step-by-Step: Dividing $5x^4 + 5x^3 - 4x^2 - 5$ by $x + 2$

Now, let's get our hands dirty with the actual division. We are dividing the dividend $5x^4 + 5x^3 - 4x^2 - 5$ by the divisor $x + 2$ . We will use the method of long polynomial division, which mirrors numerical long division. It's essential to write out the dividend with placeholders for any missing terms, even if they have a coefficient of zero. In our case, we have $5x^4 + 5x^3 - 4x^2 + 0x - 5$ . This ensures we align terms correctly.

Step 1: Set up the division.

        _____________
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5

Step 2: Divide the leading term of the dividend by the leading term of the divisor.

Our leading term in the dividend is $5x^4$ , and in the divisor is $x$ . So, $rac{5x^4}{x} = 5x^3$ . This is the first term of our quotient, $Q(x)$ .

        5x^3 _________
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5

Step 3: Multiply the first term of the quotient by the entire divisor.

$5x^3 imes (x + 2) = 5x^4 + 10x^3$ . Write this result below the dividend, aligning like terms.

        5x^3 _________
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)

Step 4: Subtract this result from the corresponding terms in the dividend.

$(5x^4 + 5x^3) - (5x^4 + 10x^3) = 5x^4 + 5x^3 - 5x^4 - 10x^3 = -5x^3$ . Bring down the next term from the dividend ( $-4x^2$ ).

        5x^3 _________
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2

Step 5: Repeat the process.

Now, our new dividend is $-5x^3 - 4x^2$ . Divide the leading term ( $-5x^3$ ) by the leading term of the divisor ( $x$ ): $rac{-5x^3}{x} = -5x^2$ . This is the next term in our quotient.

        5x^3 - 5x^2 _______
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2

Multiply $-5x^2$ by $(x + 2)$ : $-5x^2(x + 2) = -5x^3 - 10x^2$ . Subtract this from $-5x^3 - 4x^2$ .

$(-5x^3 - 4x^2) - (-5x^3 - 10x^2) = -5x^3 - 4x^2 + 5x^3 + 10x^2 = 6x^2$ . Bring down the next term ( $+0x$ ).

        5x^3 - 5x^2 _______
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2
              -(-5x^3 - 10x^2)
              ------------------
                     6x^2 + 0x

Step 6: Continue the process until the degree of the remainder is less than the degree of the divisor.

Divide $6x^2$ by $x$ : $rac{6x^2}{x} = 6x$ . This is the next term in our quotient.

        5x^3 - 5x^2 + 6x ____
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2
              -(-5x^3 - 10x^2)
              ------------------
                     6x^2 + 0x

Multiply $6x$ by $(x + 2)$ : $6x(x + 2) = 6x^2 + 12x$ . Subtract this from $6x^2 + 0x$ .

$(6x^2 + 0x) - (6x^2 + 12x) = 6x^2 + 0x - 6x^2 - 12x = -12x$ . Bring down the next term ( $-5$ ).

        5x^3 - 5x^2 + 6x ____
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2
              -(-5x^3 - 10x^2)
              ------------------
                     6x^2 + 0x
                    -(6x^2 + 12x)
                    -------------
                          -12x - 5

Divide $-12x$ by $x$ : $rac{-12x}{x} = -12$ . This is the last term in our quotient.

        5x^3 - 5x^2 + 6x - 12
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2
              -(-5x^3 - 10x^2)
              ------------------
                     6x^2 + 0x
                    -(6x^2 + 12x)
                    -------------
                          -12x - 5

Multiply $-12$ by $(x + 2)$ : $-12(x + 2) = -12x - 24$ . Subtract this from $-12x - 5$ .

$(-12x - 5) - (-12x - 24) = -12x - 5 + 12x + 24 = 19$ . This is our remainder, $R(x)$ . Since the degree of $19$ (which is 0) is less than the degree of $x + 2$ (which is 1), we stop here.

        5x^3 - 5x^2 + 6x - 12
x + 2 | 5x^4 + 5x^3 - 4x^2 + 0x - 5
        -(5x^4 + 10x^3)
        ----------------
              -5x^3 - 4x^2
              -(-5x^3 - 10x^2)
              ------------------
                     6x^2 + 0x
                    -(6x^2 + 12x)
                    -------------
                          -12x - 5
                         -(-12x - 24)
                         -------------
                                19

The Final Answer: $Q(x) + rac{R(x)}{D(x)}$

We have successfully completed the polynomial division. Our quotient, $Q(x)$ , is $5x^3 - 5x^2 + 6x - 12$ . Our remainder, $R(x)$ , is $19$ . Our divisor, $D(x)$ , is $x + 2$ .

Therefore, the result of dividing $5x^4 + 5x^3 - 4x^2 - 5$ by $x + 2$ in the form $Q(x) + rac{R(x)}{D(x)}$ is:

$5x^3 - 5x^2 + 6x - 12 + rac{19}{x + 2}$

This expression tells us that $5x^4 + 5x^3 - 4x^2 - 5$ is equal to the quotient $(5x^3 - 5x^2 + 6x - 12)$ multiplied by the divisor $(x + 2)$ , plus the remainder $19$ . It's a way of expressing the original polynomial in terms of its factors (or near factors) and the leftover part.

Alternative Method: Synthetic Division

For division by binomials of the form $x - c$ , there's a quicker method called synthetic division. In our case, the divisor is $x + 2$ , which can be written as $x - (-2)$ . So, our value for $c$ is $-2$ . Synthetic division uses only the coefficients of the polynomials.

Step 1: Set up.

Write down the value of $c$ (which is $-2$ ) and the coefficients of the dividend $5x^4 + 5x^3 - 4x^2 + 0x - 5$ , which are $5, 5, -4, 0, -5$ .

-2 | 5   5   -4    0   -5
   |_____________________

Step 2: Bring down the first coefficient.

-2 | 5   5   -4    0   -5
   |_____________________
     5

Step 3: Multiply the number below the line by $c$ and write the result under the next coefficient.

$-2 imes 5 = -10$ . Write $-10$ under the $5$ .

-2 | 5   5   -4    0   -5
   |    -10
   |_____________________
     5

Step 4: Add the numbers in the second column.

$5 + (-10) = -5$ . Write the result below the line.

-2 | 5   5   -4    0   -5
   |    -10
   |_____________________
     5  -5

Step 5: Repeat the multiply and add process.

Multiply $-2 imes -5 = 10$ . Write under $-4$ . Add: $-4 + 10 = 6$ .

-2 | 5   5   -4    0   -5
   |    -10   10
   |_____________________
     5  -5    6

Multiply $-2 imes 6 = -12$ . Write under $0$ . Add: $0 + (-12) = -12$ .

-2 | 5   5   -4    0   -5
   |    -10   10  -12
   |_____________________
     5  -5    6  -12

Multiply $-2 imes -12 = 24$ . Write under $-5$ . Add: $-5 + 24 = 19$ .

-2 | 5   5   -4    0   -5
   |    -10   10  -12  24
   |_____________________
     5  -5    6  -12  19

Step 6: Interpret the results.

The numbers on the bottom row, except for the last one, are the coefficients of the quotient. The last number is the remainder. The degree of the quotient is one less than the degree of the dividend. So, $5, -5, 6, -12$ are the coefficients for $x^3, x^2, x^1,$ and $x^0$ respectively. The remainder is $19$ .

This gives us the same quotient $5x^3 - 5x^2 + 6x - 12$ and remainder $19$ . Synthetic division is often faster and less prone to error once you're comfortable with it!

Why is $Q(x) + rac{R(x)}{D(x)}$ Important?

The form $Q(x) + rac{R(x)}{D(x)}$ is incredibly useful in mathematics. It's a direct application of the Division Algorithm for Polynomials, which states that for any polynomials $P(x)$ (dividend) and $D(x)$ (divisor) with $D(x) eq 0$ , there exist unique polynomials $Q(x)$ (quotient) and $R(x)$ (remainder) such that $P(x) = D(x)Q(x) + R(x)$ , where the degree of $R(x)$ is less than the degree of $D(x)$ . Rewriting this as $rac{P(x)}{D(x)} = Q(x) + rac{R(x)}{D(x)}$ allows us to decompose a complex rational expression into a simpler polynomial part and a proper rational function part.

This form is particularly helpful when:

Analyzing the behavior of functions: The $Q(x)$ part can represent an asymptote (slant or horizontal) for the graph of the rational function $rac{P(x)}{D(x)}$ . This gives us valuable insight into the graph's long-term behavior.
Simplifying integration in calculus: Integrating a rational function is often easier when it's expressed in this form. The polynomial part is straightforward to integrate, and the remainder term can then be dealt with separately, possibly using partial fraction decomposition.
Factoring and finding roots: The Remainder Theorem, which is closely related to polynomial division, states that when a polynomial $P(x)$ is divided by $x - c$ , the remainder is $P(c)$ . If the remainder is 0, then $c$ is a root of the polynomial, and $(x - c)$ is a factor. This process is fundamental to solving polynomial equations.

Understanding and mastering polynomial division, and expressing the result in the form $Q(x) + rac{R(x)}{D(x)}$ , equips you with a powerful algebraic technique that has broad applications across various branches of mathematics.

Conclusion

We've successfully navigated the process of dividing the polynomial $5x^4 + 5x^3 - 4x^2 - 5$ by the binomial $x + 2$ , arriving at the result $5x^3 - 5x^2 + 6x - 12 + rac{19}{x + 2}$ . Whether you prefer the methodical approach of long polynomial division or the streamlined efficiency of synthetic division, the key is to understand each step and how it contributes to the final answer. This skill is not just an isolated mathematical exercise; it's a stepping stone to deeper understanding in algebra, calculus, and beyond. Keep practicing, and you'll find these operations becoming second nature!

For further exploration into polynomial operations and theorems, you can visit ** Khan Academy's Algebra section** or the ** Wolfram MathWorld website**.

H1: Polynomial Division Made Easy: Divide 5x4+5x3-4x^2-5 by x+2