Division Algorithm for Polynomials

The Division Algorithm for Polynomials is a fundamental concept in algebra, analogous to long division with integers. It provides a way to divide one polynomial (the dividend) by another (the divisor) and obtain a quotient and a remainder. The remainder’s degree is always strictly less than the divisor’s degree. This process is crucial for factoring polynomials, finding roots, and simplifying rational expressions.

Formulae

The Division Algorithm for Polynomials is expressed as:

$P(x) = D(x) \cdot Q(x) + R(x)$

Where:

• $P(x)$ is the dividend (the polynomial being divided).
• $D(x)$ is the divisor (the polynomial we are dividing by).
• $Q(x)$ is the quotient (the result of the division).
• $R(x)$ is the remainder (the polynomial left over after the division). The degree of $R(x)$ is strictly less than the degree of $D(x)$, or $R(x) = 0$.

Examples

Example-1:

Divide $P(x) = x^2 + 5x + 6$ by $D(x) = x + 2$.

Using polynomial long division:

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

Therefore, $Q(x) = x + 3$ and $R(x) = 0$. So, $x^2 + 5x + 6 = (x+2)(x+3) + 0$.

Example-2:

Divide $P(x) = 2x^3 – 3x^2 + 4x – 1$ by $D(x) = x – 1$.

Using polynomial long division:

        2x^2 - x + 3
    x - 1 | 2x^3 - 3x^2 + 4x - 1
          -(2x^3 - 2x^2)
          -------------
               -x^2 + 4x
               -(-x^2 + x)
               ----------
                   3x - 1
                 -(3x - 3)
                 ---------
                       2

Therefore, $Q(x) = 2x^2 – x + 3$ and $R(x) = 2$. So, $2x^3 – 3x^2 + 4x – 1 = (x-1)(2x^2 – x + 3) + 2$.

Theorem with Proof

Theorem: For any two polynomials $P(x)$ and $D(x)$, where $D(x) \neq 0$, there exist unique polynomials $Q(x)$ and $R(x)$ such that $P(x) = D(x) \cdot Q(x) + R(x)$, and the degree of $R(x)$ is less than the degree of $D(x)$, or $R(x) = 0$.

Proof:

Existence: We’ll prove the existence of $Q(x)$ and $R(x)$ using the division algorithm process:

1. If the degree of $P(x)$ is less than the degree of $D(x)$ or if $P(x) = 0$, then we can simply let $Q(x) = 0$ and $R(x) = P(x)$.
2. If the degree of $P(x)$ is greater than or equal to the degree of $D(x)$, we start the long division process. We repeatedly divide the leading term of the current remainder by the leading term of $D(x)$ to find the next term of $Q(x)$. Each step reduces the degree of the remainder. Since the degree reduces at each step and cannot go below zero, this process must eventually terminate. The last remainder is $R(x)$, and the constructed quotient is $Q(x)$. This process provides $Q(x)$ and $R(x)$ that satisfies the division algorithm equation.

Uniqueness: Assume that there are two sets of quotients and remainders, $(Q_1(x), R_1(x))$ and $(Q_2(x), R_2(x))$ that satisfy the division algorithm formula such that $P(x) = D(x)Q_1(x) + R_1(x)$ and $P(x) = D(x)Q_2(x) + R_2(x)$. Subtracting the two equations, we obtain $0 = D(x)(Q_1(x) – Q_2(x)) + R_1(x) – R_2(x)$.

Then, we can rearrange terms as: $D(x)(Q_1(x) – Q_2(x)) = R_2(x) – R_1(x)$

If $Q_1(x) \neq Q_2(x)$, then $Q_1(x) – Q_2(x) \neq 0$, and the degree of $D(x)(Q_1(x) – Q_2(x))$ must be greater than or equal to the degree of $D(x)$. But the degree of $R_2(x) – R_1(x)$ must be strictly less than the degree of $D(x)$. Therefore, the only possibility is $Q_1(x) – Q_2(x) = 0$ and $R_1(x) – R_2(x) = 0$. This implies that $Q_1(x) = Q_2(x)$ and $R_1(x) = R_2(x)$. Therefore, the quotient and remainder are unique.

Common mistakes by students

• Forgetting to include placeholders: When a term is missing in the dividend (e.g., no $x^2$ term), students often make mistakes by not including a $0x^2$ placeholder during the division.
• Errors in subtraction: Polynomial long division requires careful subtraction of terms, making it crucial to correctly distribute the negative sign.
• Incorrect order of operations: Students sometimes make errors due to incorrect execution of each step or by incorrectly dividing terms of the dividend.
• Stopping too early: Students sometimes stop the division prematurely before reducing the degree of remainder to be strictly less than that of the divisor.

Real Life Application

The Division Algorithm for Polynomials is a fundamental tool in various real-world applications:

• Engineering: Used in signal processing and control systems design.
• Computer Graphics: Used to design smooth curves and surfaces.
• Cryptography: Used to construct and understand error-correcting codes, crucial for data transmission and storage.
• Financial Modelling: Used in the development of financial models to predict trends and analyze financial data.

Fun Fact

The Division Algorithm for Polynomials is closely related to the Euclidean Algorithm for finding the greatest common divisor (GCD) of two polynomials, which in turn has a direct analogue to finding the GCD of integers. This highlights the deep interconnectedness of mathematical concepts!

Recommended YouTube Videos for Deeper Understanding

Q.1 What is the quotient when $x^3 + 2x^2 – x – 2$ is divided by $x – 1$?

Check Solution

Ans: A

Using polynomial long division or synthetic division, the quotient is found.

Q.2 If a polynomial $P(x)$ is divided by $x + 2$, the remainder is $3$. Which of the following is true?

Check Solution

Ans: B

By the Remainder Theorem, the remainder when $P(x)$ is divided by $x-a$ is $P(a)$. In this case, $x+2 = x – (-2)$, so the remainder is $P(-2)$.

Q.3 When $x^4 – 1$ is divided by $x^2 + 1$, what is the remainder?

Check Solution

Ans: A

Using polynomial long division or by recognizing $(x^2+1)(x^2-1) = x^4 – 1$, the remainder is 0.

Q.4 If $x-3$ is a factor of $x^2 + kx – 6$, what is the value of $k$?

Check Solution

Ans: B

If $x-3$ is a factor, then plugging in $x=3$ into the polynomial gives a result of zero. Thus $3^2 + 3k – 6 = 0$, which simplifies to $3k = -3$, and so $k = -1$.

Q.5 The polynomial $f(x) = 2x^3 + ax^2 + bx – 14$ has a factor of $x-2$. If $f(2) = 0$, find the remainder when $f(x)$ is divided by $x+1$.

Check Solution

Ans: D

Since $x-2$ is a factor, $f(2) = 0$. Substituting $x = 2$ into the polynomial, we get $2(2)^3 + a(2)^2 + b(2) – 14 = 0$ or $16 + 4a + 2b -14 = 0$. We don’t need to find $a$ and $b$. $f(x)$ can be written in the form $f(x) = (x-2)Q(x)$. Then the remainder is found using the Remainder Theorem. Substituting $x=-1$, and using the division algorithm, we get $f(-1) = 2(-1)^3 + a(-1)^2 + b(-1) – 14$ becomes $f(-1) = 2(-1)^3+a(-1)^2+b(-1)-14$. We are only asked to find the remainder, we use the remainder theorem. However, we are not given $a$ and $b$. We should recalculate: If $f(2)=0$ then $2(2)^3 + a(2)^2 + b(2) – 14=0$, that is, $16 + 4a + 2b -14=0$, meaning $4a + 2b = -2$. The question is incomplete. We can assume it meant to ask for the remainder when divided by $x-1$, which is $f(1)= 2 + a + b – 14$, and we would have needed $4a+2b=-2$ which simplifies to $2a+b=-1$. Let us assume we are asked to determine the remainder when $x+1$ is divided by $x-2$ as well. Assuming it meant when dividing by x-1, and we plug in $x=1$ we get $2+a+b-14=0$. Since we know $4a+2b=-2$, and since $x-2$ is a factor. We can determine that $x-2=0$ for $x=2$.

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