Division Algorithm for Polynomials and Finding Zeroes for CBSE Class 10

In polynomial algebra, finding the roots (zeroes) of a linear polynomial ($ax + b = 0$) takes one step; finding the roots of a quadratic polynomial ($ax^2 + bx + c = 0$) is solved via splitting the middle term or using the quadratic formula. But what if you are confronted with a cubic polynomial of degree $3$, or a towering bi-quadratic polynomial of degree $4$? How do mathematicians crack high-degree algebraic equations?

The secret weapon is the Division Algorithm for Polynomials. If you are given two zeroes of a polynomial, you can convert them into a quadratic divisor, divide the higher-degree polynomial using long division, and factorize the resulting quotient to uncover all remaining zeroes!

In CBSE Class 10 Mathematics, Chapter 2 (Polynomials) concludes with the Division Algorithm and the celebrated 4-mark board exam problem: finding all zeroes of a polynomial when two irrational zeroes are given.

What You Will Learn

• Statement of the Division Algorithm for Polynomials: $p(x) = g(x) \cdot q(x) + r(x)$
• The condition on the remainder: $r(x) = 0$ or $\deg r(x) < \deg g(x)$
• Step-by-step methodology for polynomial long division
• How to check whether polynomial $g(x)$ is a factor of polynomial $p(x)$
• The benchmark 4-mark board problem: Finding all zeroes of a degree-4 polynomial when two roots (e.g., $\sqrt{2}$ and $-\sqrt{2}$, or $2 \pm \sqrt{3}$) are given
• Board exam tips, algebraic layout standards, and common subtraction traps

1. Statement of the Division Algorithm for Polynomials

The division algorithm for polynomials mirrors Euclid's Division Lemma for integers: $\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$

Formal Theorem

If $p(x)$ and $g(x)$ are any two polynomials with $g(x) \ne 0$, then we can find unique polynomials $q(x)$ and $r(x)$ such that: $\mathbf{p(x) = g(x) \times q(x) + r(x)}$ where either $\mathbf{r(x) = 0}$ or $\mathbf{\deg r(x) < \deg g(x)}$.

• $p(x)$: The Dividend polynomial
• $g(x)$: The Divisor polynomial
• $q(x)$: The Quotient polynomial
• $r(x)$: The Remainder polynomial

The Factor Theorem Corollary:

If upon dividing $p(x)$ by $g(x)$, the remainder is zero ($r(x) = 0$), then:

<u>$g(x)$ is a factor of $p(x)$, and $q(x)$ is also a factor of $p(x)$!</u>

2. Polynomial Long Division Protocol

When dividing $p(x)$ by $g(x)$:

1. Arrange in Standard Form: Arrange the terms of both the dividend $p(x)$ and divisor $g(x)$ in descending order of their degrees (highest power first).
2. Find the First Term of the Quotient: Divide the highest-degree term of the dividend by the highest-degree term of the divisor.
3. Multiply and Subtract: Multiply the entire divisor by this quotient term, write the result below like powers of the dividend, and subtract (change signs of all subtracted terms).
4. Repeat: Treat the resulting remainder as the new dividend and repeat until the remainder is $0$ or its degree is strictly less than the degree of the divisor.

3. The 4-Mark Benchmark Board Exam Problem

Let us solve the most famous polynomial problem in the CBSE curriculum:

Solved Example: Finding All Zeroes of a Bi-Quadratic Polynomial

Problem: Find all the zeroes of $2x^4 - 3x^3 - 3x^2 + 6x - 2$, if you know that two of its zeroes are $\sqrt{2}$ and $-\sqrt{2}$.

Solution:

Step 1: Form the Divisor from the Given Zeroes

Since $x = \sqrt{2}$ and $x = -\sqrt{2}$ are zeroes:

• $(x - \sqrt{2})$ is a factor.
• $(x + \sqrt{2})$ is a factor. Multiplying these two linear factors gives a quadratic factor: $g(x) = (x - \sqrt{2})(x + \sqrt{2}) = x^2 - (\sqrt{2})^2 = \mathbf{x^2 - 2}$ Therefore, $x^2 - 2$ is a factor of the given polynomial $p(x)$!

Step 2: Divide $p(x)$ by $g(x)$ Using Long Division

Divide $2x^4 - 3x^3 - 3x^2 + 6x - 2$ by $x^2 - 2$:

                      2x² - 3x + 1  <-- Quotient q(x)
                 +-------------------------------------
         x² - 2  | 2x⁴ - 3x³ - 3x² + 6x - 2
                 - (2x⁴      - 4x²)             [Divide 2x⁴ by x² = 2x²]
                 -------------------------
                       - 3x³ +  x² + 6x - 2
                     - (-3x³       + 6x)        [Divide -3x³ by x² = -3x]
                 -------------------------
                                x²      - 2
                              - (x²     - 2)    [Divide x² by x² = +1]
                 -------------------------
                                         0      <-- Remainder r(x) = 0!

The division yields: $\text{Quotient } q(x) = \mathbf{2x^2 - 3x + 1}$ $\text{Remainder } r(x) = 0$

Step 3: Factorize the Quotient to Find the Remaining Zeroes

By the Division Algorithm: $p(x) = (x^2 - 2)(2x^2 - 3x + 1)$ To find the remaining zeroes, set the quotient $q(x) = 0$: $2x^2 - 3x + 1 = 0$ Split the middle term (product $= 2 \times 1 = 2$, sum $= -3 \implies -2$ and $-1$): $2x^2 - 2x - x + 1 = 0$ $2x(x - 1) - 1(x - 1) = 0$ $(2x - 1)(x - 1) = 0$ This gives: $2x - 1 = 0 \implies \mathbf{x = \frac{1}{2}}$ $x - 1 = 0 \implies \mathbf{x = 1}$

Step 4: State the Final Answer

<u>The four zeroes of the given bi-quadratic polynomial are $\mathbf{\sqrt{2}, -\sqrt{2}, 1,}$ and \mathbf{rac{1}{2}}.</u>

4. When Zeroes Are Given in Conjugate Form ($a \pm \sqrt{b}$)

Sometimes, board exams test zeroes of the form $2 \pm \sqrt{3}$:

• Factors are $[x - (2 + \sqrt{3})]$ and $[x - (2 - \sqrt{3})]$.
• Group as: $[(x - 2) - \sqrt{3}][(x - 2) + \sqrt{3}]$
• Apply $(A - B)(A + B) = A^2 - B^2$: $g(x) = (x - 2)^2 - (\sqrt{3})^2 = (x^2 - 4x + 4) - 3 = \mathbf{x^2 - 4x + 1}$
• Divide $p(x)$ by $x^2 - 4x + 1$ to find the remaining zeroes!

5. Summary and Examination Tips

Exam Tip: In polynomial long division, when multiplying by $(x^2 - 2)$, write the product $-4x^2$ directly under the $-3x^2$ term, NOT under the $-3x^3$ term! Leaving space for missing powers prevents subtraction confusion.

Common Mistake: Forgetting to change signs during subtraction in long division. When subtracting, every $+$ becomes $-$ and every $-$ becomes $+$!