Division Algorithm for Polynomials

Dividing One Polynomial by Another

Just as we divide whole numbers (17 ÷ 5 gives quotient 3, remainder 2), we can divide one polynomial by another. The process is long division, very similar to numbers.

Division Algorithm for Polynomials: If $p(x)$ and $g(x)$ are any two polynomials with $g(x) \neq 0$, then there exist unique polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that $$p(x) = g(x) \cdot q(x) + r(x),$$ where either $r(x) = 0$ or $\deg r(x) < \deg g(x)$.

Here $p(x)$ is the dividend, $g(x)$ the divisor, $q(x)$ the quotient, and $r(x)$ the remainder.

Key Point: The remainder's degree must be less than the divisor's degree — that's when you stop dividing. (Notice the resemblance to $a = bq + r$ with $0 \le r < b$ for numbers.)

[Board Important] A note for CBSE students: the division algorithm was removed from the rationalised CBSE textbook, but it remains in many State Board syllabi and is genuinely useful, so we include it here.

The Long Division Steps

Let's divide $p(x) = 3x^2 + x - 2$ by $g(x) = x + 1$ to see the method.

1. Divide the first term of the dividend by the first term of the divisor: $\dfrac{3x^2}{x} = 3x$. This is the first term of the quotient.
2. Multiply the divisor by $3x$: $3x(x+1) = 3x^2 + 3x$, and subtract: $(3x^2 + x) - (3x^2 + 3x) = -2x$. Bring down $-2$: now we have $-2x - 2$.
3. Repeat: $\dfrac{-2x}{x} = -2$. Multiply: $-2(x+1) = -2x - 2$. Subtract: $(-2x - 2) - (-2x - 2) = 0$.
4. Remainder is 0, quotient is $3x - 2$.

So $3x^2 + x - 2 = (x+1)(3x - 2) + 0$.

Key Point: Always write the polynomials in decreasing powers of $x$ before dividing, and leave gaps (or write $0x$) for missing powers.

[Board Important] Show each subtraction line clearly. Skipping steps is the most common reason for losing marks in division questions.

Verifying the Division

After dividing, you can (and should) check your work using the division algorithm itself:

$$\text{Dividend} = (\text{Divisor} \times \text{Quotient}) + \text{Remainder}$$

If this identity holds when you multiply out, your division is correct.

Degree check

The degrees follow a neat rule when $r(x) \neq 0$: $$\deg p(x) = \deg g(x) + \deg q(x)$$ and $\deg r(x) < \deg g(x)$.

Key Point: Use the verification both to catch mistakes and because some exam questions specifically ask you to verify the division algorithm.

[Board Important] A typical 3-mark question: 'On dividing $p(x)$ by $g(x)$, the quotient and remainder were found. Verify the division algorithm.' Just substitute and expand to confirm.

Using Division to Find All Zeroes

The division algorithm has a powerful application: if you know some zeroes of a polynomial, you can find the rest.

The idea

If $\alpha$ is a zero of $p(x)$, then $(x - \alpha)$ is a factor of $p(x)$. So if two zeroes are known, say $\alpha$ and $\beta$, then $(x - \alpha)(x - \beta)$ is a factor. Divide $p(x)$ by this factor; the quotient gives the remaining zeroes.

Worked outline

Given that $\sqrt{2}$ and $-\sqrt{2}$ are two zeroes of $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$:

• $(x - \sqrt2)(x + \sqrt2) = x^2 - 2$ is a factor.
• Divide $p(x)$ by $x^2 - 2$ to get quotient $2x^2 - 3x + 1$.
• Solve $2x^2 - 3x + 1 = 0$: $(2x - 1)(x - 1) = 0 \Rightarrow x = \tfrac{1}{2}, 1$.
• So all four zeroes are $\sqrt2, -\sqrt2, \tfrac{1}{2}, 1$.

Key Point: Known zeroes give a known factor; dividing by it reduces the problem to a smaller (often quadratic) polynomial you can solve.

[Board Important] This 'find all zeroes given some' question is a classic 3–4 mark problem. The factor from a pair of zeroes $\pm\sqrt{k}$ is always $x^2 - k$.

Solved Examples

Example 1: Basic division

Divide $p(x) = x^2 + 3x + 2$ by $g(x) = x + 1$.

Solution:

1. $\dfrac{x^2}{x} = x$; $x(x+1) = x^2 + x$; subtract: $(x^2+3x) - (x^2+x) = 2x$. Bring down $+2$: $2x + 2$.
2. $\dfrac{2x}{x} = 2$; $2(x+1) = 2x+2$; subtract: $0$.
3. Quotient $= x + 2$, remainder $= 0$.

Final Answer: Quotient $x + 2$, remainder 0.

Takeaway: Remainder 0 means the divisor is a factor.

Example 2: Division with a non-zero remainder

Divide $p(x) = x^2 + 2x + 3$ by $g(x) = x - 1$.

Solution:

1. $\dfrac{x^2}{x} = x$; $x(x-1) = x^2 - x$; subtract: $(x^2+2x) - (x^2-x) = 3x$. Bring down $+3$: $3x + 3$.
2. $\dfrac{3x}{x} = 3$; $3(x-1) = 3x - 3$; subtract: $(3x+3)-(3x-3) = 6$.
3. Quotient $= x + 3$, remainder $= 6$.

Final Answer: Quotient $x + 3$, remainder 6.

Takeaway: Stop when the remainder's degree is less than the divisor's.

Example 3: Verify the division algorithm

For Example 2, verify that dividend $=$ divisor × quotient + remainder.

Solution:

1. Divisor × quotient $= (x-1)(x+3) = x^2 + 2x - 3$.
2. Add remainder: $x^2 + 2x - 3 + 6 = x^2 + 2x + 3$.
3. This equals the dividend $p(x)$. ✓

Final Answer: Verified: $(x-1)(x+3) + 6 = x^2 + 2x + 3$.

Takeaway: Multiply out and add the remainder to confirm.

Example 4: Division by a quadratic

Divide $p(x) = x^3 - 3x^2 + 3x - 5$ by $g(x) = x^2 + 1$.

Solution:

1. $\dfrac{x^3}{x^2} = x$; $x(x^2+1) = x^3 + x$; subtract: $(x^3 - 3x^2 + 3x) - (x^3 + x) = -3x^2 + 2x$. Bring down $-5$.
2. $\dfrac{-3x^2}{x^2} = -3$; $-3(x^2+1) = -3x^2 - 3$; subtract: $(-3x^2 + 2x - 5) - (-3x^2 - 3) = 2x - 2$.
3. Degree of $2x - 2$ (which is 1) < degree of divisor (2), so stop.

Final Answer: Quotient $x - 3$, remainder $2x - 2$.

Takeaway: Stop once the remainder's degree drops below the divisor's degree.

Example 5: Find all zeroes given two

Find all zeroes of $p(x) = x^4 - 5x^2 + 4$, given that 1 and $-1$ are two of them.

Solution:

1. Since 1 and $-1$ are zeroes, $(x-1)(x+1) = x^2 - 1$ is a factor.
2. Divide: $x^4 - 5x^2 + 4 \div (x^2 - 1)$ gives quotient $x^2 - 4$ (remainder 0).
3. Solve $x^2 - 4 = 0 \Rightarrow x = 2, -2$.

Final Answer: All zeroes: $1, -1, 2, -2$.

Takeaway: Known zeroes → known factor → divide → solve the quotient.

Example 6: Find all zeroes with surd zeroes

Find all zeroes of $p(x) = x^3 + 3x^2 - 2x - 6$, given that $\sqrt{2}$ and $-\sqrt{2}$ are two zeroes.

Solution:

1. $(x-\sqrt2)(x+\sqrt2) = x^2 - 2$ is a factor.
2. Divide $x^3 + 3x^2 - 2x - 6$ by $x^2 - 2$: quotient $x + 3$, remainder 0.
3. Solve $x + 3 = 0 \Rightarrow x = -3$.

Final Answer: All zeroes: $\sqrt2, -\sqrt2, -3$.

Takeaway: A pair $\pm\sqrt{k}$ contributes the factor $x^2 - k$.

Example 7: Find values that make the remainder zero

Find $k$ so that $x - 2$ is a factor of $p(x) = x^2 - 3x + k$.

Solution:

1. If $x - 2$ is a factor, then $x = 2$ is a zero, so $p(2) = 0$.
2. $p(2) = 4 - 6 + k = -2 + k = 0$.
3. So $k = 2$.

Final Answer: $k = 2$.

Takeaway: $(x - a)$ is a factor ⇔ $p(a) = 0$ (Factor Theorem).

Example 8: Quotient and remainder, then verify

Divide $p(x) = 2x^2 + 3x + 1$ by $g(x) = x + 2$ and verify.

Solution:

1. $\dfrac{2x^2}{x} = 2x$; $2x(x+2) = 2x^2 + 4x$; subtract: $(2x^2 + 3x) - (2x^2 + 4x) = -x$. Bring down $+1$: $-x + 1$.
2. $\dfrac{-x}{x} = -1$; $-1(x+2) = -x - 2$; subtract: $(-x + 1) - (-x - 2) = 3$.
3. Quotient $2x - 1$, remainder 3. Check: $(x+2)(2x-1) + 3 = 2x^2 + 3x - 2 + 3 = 2x^2 + 3x + 1$. ✓

Final Answer: Quotient $2x - 1$, remainder 3.

Takeaway: Always verify with divisor × quotient + remainder.

Example 9: Degree of quotient and remainder

If a degree-5 polynomial is divided by a degree-2 polynomial, what are the degrees of the quotient and the maximum degree of the remainder?

Solution:

1. $\deg q = \deg p - \deg g = 5 - 2 = 3$.
2. The remainder's degree must be less than the divisor's, so at most degree 1.

Final Answer: Quotient degree 3; remainder degree at most 1.

Takeaway: $\deg q = \deg p - \deg g$; $\deg r < \deg g$.

Example 10: Find remainder using the Remainder Theorem

Find the remainder when $p(x) = x^3 - 2x^2 + x + 1$ is divided by $x - 1$.

Solution:

1. By the Remainder Theorem, the remainder is $p(1)$.
2. $p(1) = 1 - 2 + 1 + 1 = 1$.

Final Answer: Remainder $= 1$.

Takeaway: Dividing by $(x - a)$ leaves remainder $p(a)$ — no long division needed.