Polynomial Remainder Theorem Explained

When you're diving into the world of algebra, particularly with polynomials, you'll often encounter problems that involve division. One of the most fundamental concepts here is the Polynomial Remainder Theorem. This theorem provides a neat shortcut for finding the remainder of a polynomial division without actually performing the long division process itself. It's a powerful tool that can save you a lot of time and effort, especially when dealing with complex polynomials. Let's explore what this theorem is all about, how it works, and why it's so useful in mathematics. We'll break down the core idea and then apply it to a specific example to make it crystal clear.

Understanding the Polynomial Remainder Theorem

The Polynomial Remainder Theorem states that if a polynomial $f(x)$ is divided by a linear binomial of the form $(x - c)$, then the remainder is equal to $f(c)$. In simpler terms, you can find the remainder by just plugging the root of the divisor into the polynomial. This is a direct consequence of the division algorithm for polynomials, which states that for any polynomials $f(x)$ and $d(x)$ (where $d(x)$ is not the zero polynomial), there exist unique polynomials $q(x)$ (the quotient) and $r(x)$ (the remainder) such that $f(x) = d(x)q(x) + r(x)$, and the degree of $r(x)$ is strictly less than the degree of $d(x)$. When $d(x) = x - c$, its degree is 1, so the remainder $r(x)$ must have a degree less than 1, meaning it's a constant. If we substitute $x=c$ into the division algorithm equation, we get $f(c) = (c - c)q(c) + r(c)$, which simplifies to $f(c) = 0 imes q(c) + r(c)$, so $f(c) = r(c)$. This is the essence of the theorem. It's a concept that's foundational for understanding more advanced polynomial manipulations and is often introduced early in high school algebra courses. The elegance of the theorem lies in its simplicity and the significant reduction in computational complexity it offers. Instead of laboriously carrying out polynomial long division, which can be prone to arithmetic errors, especially with higher-degree polynomials, you can simply evaluate the polynomial at a single point. This makes it an invaluable tool for both theoretical exploration and practical problem-solving in algebra.

Extending the Concept: Division by Higher Degree Polynomials

While the basic Polynomial Remainder Theorem applies to division by linear binomials $(x-c)$, the underlying principle can be extended to division by polynomials of higher degrees, like quadratic binomials. When dividing $f(x)$ by a polynomial $d(x)$ of degree $n$, the remainder $r(x)$ will have a degree strictly less than $n$. For instance, if we divide by a quadratic polynomial like $d(x) = x^2 + 2$, the remainder $r(x)$ will be at most a linear polynomial, meaning it can be expressed in the form $ax + b$. The key to finding this remainder without long division lies in recognizing the roots of the divisor $d(x)$. For $d(x) = x^2 + 2$, the roots are the values of $x$ for which $x^2 + 2 = 0$. Solving this, we get $x^2 = -2$, so $x = ule{0.4em}{0.4em} ext{2}i$ and $x = - ule{0.4em}{0.4em} ext{2}i$. These are complex roots. The division algorithm still holds: $f(x) = d(x)q(x) + r(x)$. If we substitute the roots of $d(x)$ into this equation, the $d(x)q(x)$ term becomes zero, leaving us with $f(c) = r(c)$, where $c$ is a root of $d(x)$.

Since our remainder $r(x)$ is of the form $ax+b$, we can use the two complex roots to set up a system of equations. Let $ule{0.4em}{0.4em} ext{2}i$ be one root. Then $f( ule{0.4em}{0.4em} ext{2}i) = a( ule{0.4em}{0.4em} ext{2}i) + b$. If we use the other root, $- ule{0.4em}{0.4em} ext{2}i$, we get $f(- ule{0.4em}{0.4em} ext{2}i) = a(- ule{0.4em}{0.4em} ext{2}i) + b$. By evaluating $f(x)$ at these complex roots and equating the results to the form of the remainder, we can solve for the coefficients $a$ and $b$. This method, while involving complex numbers, is mathematically sound and bypasses the need for polynomial long division. It highlights the power of complex numbers in algebraic manipulations and provides a more generalized approach to finding remainders.

Another powerful technique, especially when dealing with divisors that have easily identifiable roots, is to use substitutions based on the divisor. For $d(x) = x^2 + 2$, we know that any term $x^2$ can be replaced by $-2$. This is because if $x^2 + 2 = 0$, then $x^2 = -2$. We can use this substitution repeatedly within the polynomial $f(x)$ to reduce its degree until we obtain a polynomial of degree less than 2. This is often simpler than working with complex roots, especially if the problem doesn't explicitly require them. This substitution method is directly derived from the division algorithm. When $f(x)$ is divided by $d(x)$, we have $f(x) = (x^2+2)q(x) + (ax+b)$. If we consider values of $x$ where $x^2+2=0$, then $f(x) = ax+b$. By using the relation $x^2 = -2$, we are essentially working in a quotient ring where elements are represented by polynomials of degree less than the divisor. This perspective is more advanced but underlies the effectiveness of the substitution technique. It allows us to simplify $f(x)$ by systematically eliminating higher powers of $x$ until we are left with a linear expression, which will be our remainder. This approach is particularly elegant for divisors that are powers of simple expressions or have simple factorizations. It’s a direct application of modular arithmetic principles to polynomials.

Applying the Theorem to a Specific Problem

Let's put these concepts into practice with the given problem: Find the remainder when $f(x)=x^5+3 x^4+2 x^3-x^2+2 x-7$ is divided by $d(x)=x^2+2$. We want to find $r(x)$ such that $f(x) = (x^2+2)q(x) + r(x)$, where the degree of $r(x)$ is less than the degree of $d(x)$ (which is 2). Therefore, $r(x)$ will be of the form $ax + b$.

We can use the substitution method. Since $d(x) = x^2 + 2$, any time we encounter $x^2$, we can replace it with $-2$. Let's simplify $f(x)$ by making these substitutions:

• First, let's handle the $x^5$ term. We can write $x^5 = x imes x^4 = x imes (x^2)^2$. Substituting $x^2 = -2$, we get $x imes (-2)^2 = x imes 4 = 4x$.
• Next, the $x^4$ term. We can write $x^4 = (x^2)^2$. Substituting $x^2 = -2$, we get $(-2)^2 = 4$.
• The $x^3$ term: $x^3 = x imes x^2$. Substituting $x^2 = -2$, we get $x imes (-2) = -2x$.
• The $x^2$ term is simply $-2$ according to our substitution.

Now, let's substitute these simplified forms back into $f(x)$:

$f(x) = (4x) + 3(4) + 2(-2x) - (-2) + 2x - 7$

Let's simplify this expression:

$f(x) = 4x + 12 - 4x + 2 + 2x - 7$

Now, group the terms with $x$ and the constant terms:

$f(x) = (4x - 4x + 2x) + (12 + 2 - 7)$

$f(x) = 2x + (14 - 7)$

$f(x) = 2x + 7$

This resulting polynomial, $2x + 7$, has a degree of 1, which is less than the degree of our divisor $d(x)=x^2+2$. Therefore, $2x + 7$ is the remainder when $f(x)$ is divided by $d(x)$. This method is significantly faster than performing polynomial long division, especially for higher-degree polynomials.

Verification using Polynomial Long Division

To ensure our result is correct, let's perform polynomial long division. While the substitution method is elegant, seeing the long division process can solidify your understanding.

We are dividing $x^5+3 x^4+2 x^3-x^2+2 x-7$ by $x^2+2$.

        x^3 + 3x^2 - 0x  - 1
      ____________________
x^2+2 | x^5 + 3x^4 + 2x^3 -  x^2 + 2x - 7
        -(x^5       + 2x^3)
        ____________________
              3x^4 + 0x^3 -  x^2
            -(3x^4       + 6x^2)
            ____________________
                    -7x^2 + 2x
                  -(-7x^2      - 14)
                  ____________________
                         2x + 7

Let's break down the steps:

1. Divide $x^5$ by $x^2$: This gives $x^3$. Multiply $x^3$ by $(x^2+2)$ to get $x^5 + 2x^3$. Subtract this from the dividend. $(x^5+3 x^4+2 x^3-x^2+2 x-7) - (x^5 + 2x^3) = 3x^4 - x^2 + 2x - 7$.

2. Bring down the next term: We have $3x^4 - x^2 + 2x - 7$. Divide $3x^4$ by $x^2$ to get $3x^2$. Multiply $3x^2$ by $(x^2+2)$ to get $3x^4 + 6x^2$. Subtract this. $(3x^4 - x^2 + 2x - 7) - (3x^4 + 6x^2) = -7x^2 + 2x - 7$.

3. Bring down the next term: We have $-7x^2 + 2x - 7$. Divide $-7x^2$ by $x^2$ to get $-7$. Multiply $-7$ by $(x^2+2)$ to get $-7x^2 - 14$. Subtract this. $(-7x^2 + 2x - 7) - (-7x^2 - 14) = 2x + 7$.

The degree of $2x+7$ (which is 1) is less than the degree of $x^2+2$ (which is 2), so $2x+7$ is our remainder.

The quotient is $x^3 + 3x^2 - 7$. Let me re-do the long division, as there was a slight error in my initial scratchpad calculation. Let's correct this.

        x^3 + 3x^2 - 0x  - 7
      ____________________
x^2+2 | x^5 + 3x^4 + 2x^3 -  x^2 + 2x - 7
        -(x^5       + 2x^3)
        ____________________
              3x^4 + 0x^3 -  x^2
            -(3x^4       + 6x^2)
            ____________________
                    -7x^2 + 2x - 7
                  -(-7x^2      - 14)
                  ____________________
                         2x + 7

Revisiting the steps:

1. Divide $x^5$ by $x^2$: Gives $x^3$. Multiply $x^3$ by $(x^2+2)$ to get $x^5 + 2x^3$. Subtract from $f(x)$: $(x^5+3 x^4+2 x^3-x^2+2 x-7) - (x^5 + 2x^3) = 3x^4 - x^2 + 2x - 7$.

2. Next term of quotient: Divide $3x^4$ by $x^2$ to get $3x^2$. Multiply $3x^2$ by $(x^2+2)$ to get $3x^4 + 6x^2$. Subtract from the current polynomial: $(3x^4 - x^2 + 2x - 7) - (3x^4 + 6x^2) = -7x^2 + 2x - 7$.

3. Next term of quotient: Divide $-7x^2$ by $x^2$ to get $-7$. Multiply $-7$ by $(x^2+2)$ to get $-7x^2 - 14$. Subtract from the current polynomial: $(-7x^2 + 2x - 7) - (-7x^2 - 14) = 2x + 7$.

The remainder is indeed $2x + 7$. The quotient is $x^3 + 3x^2 - 7$. This matches the result obtained using the substitution method. This verification confirms the correctness of our application of the generalized remainder theorem principle.

Conclusion

Understanding how to find the remainder when dividing polynomials is a cornerstone of algebra. The Polynomial Remainder Theorem, in its basic form and extended to higher-degree divisors, offers an elegant and efficient alternative to lengthy polynomial long division. By utilizing substitutions based on the roots of the divisor, or by directly substituting $x^2 = -2$ in our example, we were able to swiftly determine the remainder $2x+7$. This method not only simplifies calculations but also deepens our appreciation for the underlying structure and properties of polynomials. Whether you're tackling homework problems or preparing for exams, mastering these techniques will undoubtedly make your algebraic journey smoother and more insightful. Remember, the key is to relate the remainder to the roots of the divisor and exploit the properties of polynomial equality.

For further exploration into polynomial algebra and the Remainder Theorem, you can refer to resources like Khan Academy's Algebra Section, which offers comprehensive explanations and practice problems on these topics.