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Punjab Board Class 11 Mathematics

Chapter 9: Division of Polynomials — Class 11 Maths Notes

Division of polynomials: long division, remainder and factor theorems with explicit algebraic working. This page has the theory, solved textbook examples, every exercise solution, MCQs and a formula sheet for Chapter 9.

Exercise solutions — Chapter 9

Every exercise of Division of Polynomials solved step by step.

Mathematics 11 (Punjab Board)

Chapter 9: Division of Polynomials — Theory

Just as numbers can be divided with a quotient and a remainder, so can polynomials. This chapter builds the tools — long division, the remainder theorem, the factor theorem, and synthetic division — and then uses them on real-life problems.

Introduction: The Division Algorithm

When a polynomial \(f(x)\) (the dividend) is divided by a non-zero polynomial \(g(x)\) (the divisor), there exist unique polynomials \(q(x)\) (the quotient) and \(r(x)\) (the remainder) such that:

\[ f(x) = g(x)\cdot q(x) + r(x), \qquad \text{where } r(x) = 0 \text{ or } \deg r(x) \lt \deg g(x) \]

Matlab: remainder ki degree hamesha divisor ki degree se chhoti hoti hai. Agar divisor \(x - a\) jaisa linear hai, to remainder sirf aik constant (degree 0) hoga.

Key point: Always check your division by multiplying back: divisor × quotient + remainder must equal the dividend.

9.1 Polynomial Function

A polynomial function has the form \(f(x) = a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0\), where the \(a_i\) are constants and \(n\) is a non-negative integer called the degree. A number \(c\) with \(f(c) = 0\) is called a zero (or root) of the polynomial.

Polynomials are divided by long division: divide the leading terms, multiply back, subtract, bring down the next term, and repeat — exactly like numerical long division.

9.1.1 Remainder Theorem

The remainder theorem finds the remainder without doing the division:

\[ \text{When } f(x) \text{ is divided by } x - a, \text{ the remainder is } f(a). \]

Proof idea: write \(f(x) = (x-a)q(x) + r\) where \(r\) is a constant; putting \(x = a\) kills the first term, leaving \(f(a) = r\).

For a divisor like \(x + 3\), write it as \(x - (-3)\) and evaluate \(f(-3)\).

Key point: Remainder theorem sirf \((x - a)\) jaisay linear divisors par lagta hai — sign ka khayal rakhen: \(x + 3\) ka matlab \(a = -3\).

9.1.2 Factor Theorem

The factor theorem is the remainder theorem with remainder zero:

\[ x - a \text{ is a factor of } f(x) \iff f(a) = 0. \]

Is se do kaam hote hain: (1) check karna ke koi expression factor hai ya nahi — bas value put karke dekhen; (2) unknown coefficients nikalna — factor hone ki shart se equations banti hain.

Key point: "Is \(x+1\) a factor of \(f(x)\)?" ka jawab sirf \(f(-1)\) nikal kar mil jata hai — poori division ki zaroorat nahi.

9.1.3 Synthetic Division

Synthetic division is a shortcut for dividing by \(x - a\): write down the coefficients, bring down the first, multiply by \(a\), add to the next coefficient, and repeat. The last number in the bottom row is the remainder; the rest are the quotient's coefficients.

Example: dividing \(x^3 - 7x + 6\) by \(x - 2\) (coefficients \(1, 0, -7, 6\)):

\[ \begin{array}{r|rrrr} 2 & 1 & 0 & -7 & 6 \\ & & 2 & 4 & -6 \\ \hline & 1 & 2 & -3 & 0 \end{array} \]

Remainder 0, quotient \(x^2 + 2x - 3\). Missing powers ke liye coefficient 0 likhna na bhoolen (yahan \(x^2\) ka coefficient 0 tha).

Key point: Synthetic division ke baad quotient ko mazeed factorize karke polynomial ko completely factor kiya ja sakta hai.

9.2 Real-Life Applications of the Remainder and Factor Theorems

Polynomial models appear in data fitting (polynomial regression), where \(P(x)\) predicts a value and the percentage error against an observed value is:

\[ \text{Percentage error} = \left|\frac{\text{observed} - \text{predicted}}{\text{observed}}\right|\times 100 \]

In signal processing, a transfer function like \(H(z) = \frac{N(z)}{D(z)}\) has zeros where \(N(z) = 0\) and poles where \(D(z) = 0\), both found with the factor theorem. A discrete system is stable when every pole lies inside the unit circle, i.e. \(|z| \lt 1\) for each pole.

Key point: Zeros numerator se, poles denominator se — aur stability ka test sirf poles ki absolute values ka hai.

Mathematics 11 (Punjab Board)

Solved Examples — Chapter 9: Division of Polynomials

Textbook solved examples for Punjab Board Class 11 Mathematics, Chapter 9, worked out in full — long division, remainder and factor theorems, synthetic division, and real-life applications.

Example 1: Divide the cubic polynomial \(3x^3 - 10x^2 + 13x - 6\) by the linear polynomial \(x - 2\). Also find the quotient and the remainder.

Long division karenge: leading terms divide karo, multiply karke subtract karo, next term lao. Aakhir mein verify karenge ke divisor × quotient + remainder = dividend.

Step 1: \(3x^3 \div x = 3x^2\). Multiply: \(3x^2(x-2) = 3x^3-6x^2\). Subtract from dividend: \(-4x^2+13x-6\).

Step 2: \(-4x^2 \div x = -4x\). Multiply: \(-4x(x-2) = -4x^2+8x\). Subtract: \(5x-6\).

Step 3: \(5x \div x = 5\). Multiply: \(5(x-2) = 5x-10\). Subtract: \(4\).

\[ 3x^3-10x^2+13x-6 = (x-2)(3x^2-4x+5) + 4 \]
Answer: Quotient \(= 3x^2-4x+5\), remainder \(= 4\).

Example 2: Divide the polynomial \(x^4 - 3x^3 + 5x^2 - 7x + 2\) by \(x^2 - x + 1\). Also find the quotient and the remainder.

Divisor degree 2 ka hai, to quotient degree 2 ka aur remainder degree 1 se kam (ya zero) hoga. Wese hi long division: \(x^4 \div x^2 = x^2\) se start karenge.

Step 1: \(x^4 \div x^2 = x^2\). Multiply: \(x^2(x^2-x+1) = x^4-x^3+x^2\). Subtract: \(-2x^3+4x^2-7x+2\).

Step 2: \(-2x^3 \div x^2 = -2x\). Multiply: \(-2x(x^2-x+1) = -2x^3+2x^2-2x\). Subtract: \(2x^2-5x+2\).

Step 3: \(2x^2 \div x^2 = 2\). Multiply: \(2(x^2-x+1) = 2x^2-2x+2\). Subtract: \(-3x\).

\[ x^4-3x^3+5x^2-7x+2 = (x^2-x+1)(x^2-2x+2) + (-3x) \]
Answer: Quotient \(= x^2-2x+2\), remainder \(= -3x\).

Example 3: Find the remainder without performing division when \(f(x) = x^4 + x^3 + x^2 + 1\) is divided by \(x + 1\).

Remainder theorem: \((x-a)\) se divide karne par remainder \(f(a)\) hota hai. \(x+1 = x-(-1)\), to \(a = -1\) rakh kar \(f(-1)\) nikalenge.

\[ f(-1) = (-1)^4 + (-1)^3 + (-1)^2 + 1 = 1 - 1 + 1 + 1 = 2 \]
Answer: The remainder is \(2\).

Example 4: Find the value of \(k\) if the polynomial \(x^3 + kx^2 - 7x + 6\) has remainder \(-4\) when divided by \(x + 2\).

Remainder theorem se \(f(-2) = -4\). Is equation mein \(k\) unknown hai; solve karke \(k\) nikalenge.

\[ f(-2) = (-2)^3 + k(-2)^2 - 7(-2) + 6 = -8 + 4k + 14 + 6 = 4k + 12 \]
\[ 4k + 12 = -4 \Rightarrow 4k = -16 \Rightarrow k = -4 \]
Answer: \(k = -4\).

Example 5: Show that \(x - 2\) is a factor of \(f(x) = x^3 - 7x + 6\) without factorizing.

Factor theorem: \(x-a\) factor hai agar \(f(a) = 0\). Bas \(f(2)\) nikal kar dekhenge.

\[ f(2) = (2)^3 - 7(2) + 6 = 8 - 14 + 6 = 0 \]

Since \(f(2) = 0\), by the factor theorem \(x - 2\) is a factor of \(f(x)\).

Answer: \(x - 2\) is a factor of \(x^3 - 7x + 6\) (proved).

Example 6: If \(x + 1\) and \(x - 2\) are factors of \(x^3 + px^2 + qx + 2\), find the values of \(p\) and \(q\).

Factor theorem se \(f(-1) = 0\) aur \(f(2) = 0\) — do equations, do unknowns. Solve karke \(p, q\) nikalenge.

\[ f(-1) = -1 + p - q + 2 = 0 \Rightarrow p - q = -1 \quad \text{(i)} \]
\[ f(2) = 8 + 4p + 2q + 2 = 0 \Rightarrow 4p + 2q = -10 \Rightarrow 2p + q = -5 \quad \text{(ii)} \]

Adding (i) and (ii): \(3p = -6 \Rightarrow p = -2\). From (i): \(-2 - q = -1 \Rightarrow q = -1\).

Answer: \(p = -2\), \(q = -1\).

Example 7: If \((x-2)\) and \((x+2)\) are factors of \(x^4 - 13x^2 + 36\), use synthetic division to find the other two factors.

Pehle \(x = 2\) se synthetic division karke aik factor nikalenge, phir bachi hui cubic (actually quadratic quotient) ko \(x = -2\) se divide karenge. Missing powers ke liye 0 coefficients likhenge.

Synthetic division by 2 (coefficients \(1, 0, -13, 0, 36\)):

\[ \begin{array}{r|rrrrr} 2 & 1 & 0 & -13 & 0 & 36 \\ & & 2 & 4 & -18 & -36 \\ \hline & 1 & 2 & -9 & -18 & 0 \end{array} \]

Quotient: \(x^3 + 2x^2 - 9x - 18\). Synthetic division of this by \(-2\):

\[ \begin{array}{r|rrrr} -2 & 1 & 2 & -9 & -18 \\ & & -2 & 0 & 18 \\ \hline & 1 & 0 & -9 & 0 \end{array} \]

New quotient: \(x^2 - 9 = (x-3)(x+3)\). So \(x^4-13x^2+36 = (x-2)(x+2)(x-3)(x+3)\).

Answer: The other two factors are \((x-3)\) and \((x+3)\).

Example 8: A polynomial regression model \(P(x) = x^3 + x^2 + 2x + 1\) is fitted to monthly sales data. If the observed sales in the 3rd month are 40 units, find the percentage error.

Pehle \(x = 3\) rakh kar predicted sales \(P(3)\) nikalenge (remainder theorem jaisa evaluation). Phir percentage error \(= \left|\frac{\text{observed}-\text{predicted}}{\text{observed}}\right|\times 100\).

\[ P(3) = 3^3 + 3^2 + 2(3) + 1 = 27 + 9 + 6 + 1 = 43 \]
\[ \text{Percentage error} = \left|\frac{40-43}{40}\right|\times 100 = \frac{3}{40}\times 100 = 7.5\% \]
Answer: \(7.5\%\).

Example 9: Suppose a polynomial regression model \(P(x) = 3x^3 - 4x^2 + 2x - 5\). If the data point at \(x = -1\) is missing, what should be its predicted value?

Missing value ka best estimate model ki prediction hai: \(P(-1)\) evaluate karenge. Negative \(x\) ki even/odd powers ka khayal rakhen.

\[ P(-1) = 3(-1)^3 - 4(-1)^2 + 2(-1) - 5 = -3 - 4 - 2 - 5 = -14 \]
Answer: \(-14\) units (predicted value).

Example 10: A signal processing system has a transfer function with denominator \(A(z) = z^2 - 0.25\). Use the factor theorem to find the poles of the system and determine if the system is stable.

Poles wahan hote hain jahan denominator zero ho: \(A(z) = 0\). \(z^2 - 0.25 = (z-0.5)(z+0.5)\) factor karke dono poles nikalenge. Stability ke liye har pole ka absolute value 1 se kam hona chahiye.

\[ z^2 - 0.25 = 0 \Rightarrow z^2 - (0.5)^2 = 0 \Rightarrow (z-0.5)(z+0.5) = 0 \]

By the factor theorem, \(A(0.5) = 0\) and \(A(-0.5) = 0\), so the poles are \(z = 0.5\) and \(z = -0.5\). Since \(|0.5| \lt 1\) and \(|-0.5| \lt 1\), both poles lie inside the unit circle.

Answer: Poles at \(z = 0.5\) and \(z = -0.5\); the system is stable.

Mathematics 11 (Punjab Board)

MCQs: Chapter 9 – Division of Polynomials

22 multiple-choice questions covering long division, the remainder theorem, the factor theorem, synthetic division, and real-life applications.

  1. When \(f(x)\) is divided by \(x-a\), the remainder is:

    1. \(f(-a)\)
    2. \(f(a)\)
    3. \(0\)
    4. \(f(x)-a\)
    Remainder theorem: \((x-a)\) se divide karne par remainder \(f(a)\) hota hai.
  2. \(x-a\) is a factor of \(f(x)\) if and only if:

    1. \(f(a) = 1\)
    2. \(f(-a) = 0\)
    3. \(f(a) = 0\)
    4. \(f(0) = a\)
    Factor theorem: factor hone ka matlab remainder zero, yani \(f(a) = 0\).
  3. The degree of the remainder when a polynomial is divided by \(x-a\) is:

    1. \(0\)
    2. \(1\)
    3. Equal to the dividend's degree
    4. Undefined
    Divisor degree 1 ka hai, to remainder ki degree is se kam — yani constant (degree 0).
  4. The remainder when \(x^2+5x+6\) is divided by \(x-2\) is:

    1. \(10\)
    2. \(12\)
    3. \(18\)
    4. \(20\)
    \(f(2) = 4 + 10 + 6 = 20\).
  5. Is \(x+1\) a factor of \(x^2-1\)?

    1. Yes
    2. No
    3. Only for \(x \gt 0\)
    4. Cannot be determined
    \(f(-1) = 1 - 1 = 0\), to factor theorem ke mutabiq \(x+1\) factor hai.
  6. In synthetic division, the last number in the bottom row is the:

    1. Leading coefficient of the quotient
    2. Divisor
    3. Remainder
    4. Degree of the polynomial
    Aakhri entry remainder hoti hai; baqi entries quotient ke coefficients.
  7. If \(-2\) is a root of \(f(x)\), then a factor of \(f(x)\) is:

    1. \(x-2\)
    2. \(x+2\)
    3. \(x+\frac{1}{2}\)
    4. \(2x\)
    Root \(-2\) ka matlab \(f(-2) = 0\), to factor \(x-(-2) = x+2\) hai.
  8. When a degree-4 polynomial is divided by a degree-2 polynomial, the quotient has degree:

    1. \(4\)
    2. \(3\)
    3. \(2\)
    4. \(1\)
    Quotient ki degree = dividend ki degree − divisor ki degree = \(4-2 = 2\).
  9. The remainder when \(x^4+x^3+x^2+1\) is divided by \(x+1\) is:

    1. \(0\)
    2. \(2\)
    3. \(4\)
    4. \(-2\)
    \(f(-1) = 1 - 1 + 1 + 1 = 2\).
  10. If \(f(3) = 0\), then a factor of \(f(x)\) is:

    1. \(x-3\)
    2. \(x+3\)
    3. \(3x\)
    4. \(\frac{x}{3}\)
    \(f(3) = 0\) ka matlab \(x - 3\) factor hai (factor theorem).
  11. The zeros of \(z^2-z-2\) are:

    1. \(1, -2\)
    2. \(-1, -2\)
    3. \(1, 2\)
    4. \(2, -1\)
    \(z^2-z-2 = (z-2)(z+1)\), to zeros \(2\) aur \(-1\) hain.
  12. The poles of a system with denominator \(z^2-0.25\) are:

    1. \(0.25, -0.25\)
    2. \(0.5, -0.5\)
    3. \(0.5\) only
    4. None
    \(z^2 - 0.25 = (z-0.5)(z+0.5) = 0\) se poles \(0.5\) aur \(-0.5\).
  13. A discrete-time system is stable if:

    1. All poles lie outside the unit circle
    2. All zeros lie inside the unit circle
    3. Every pole satisfies \(|z| \lt 1\)
    4. Every zero satisfies \(|z| \gt 1\)
    Stability ka test poles par hai: sab poles unit circle ke andar (\(|z| \lt 1\)) hone chahiye.
  14. Since \(f(2) = 0\) for \(f(x) = x^3-7x+6\), a factor of \(f(x)\) is:

    1. \(x-2\)
    2. \(x+2\)
    3. \(x-1\)
    4. \(x+1\)
    \(f(2) = 8 - 14 + 6 = 0\), to \(x - 2\) factor hai.
  15. The remainder when \(x^4-10x^2-2x+4\) is divided by \(x+3\) is:

    1. \(0\)
    2. \(1\)
    3. \(4\)
    4. \(10\)
    \(f(-3) = 81 - 90 + 6 + 4 = 1\).
  16. If \(x^3+kx^2-7x+6\) leaves remainder \(-4\) when divided by \(x+2\), then \(k\) equals:

    1. \(4\)
    2. \(-2\)
    3. \(-4\)
    4. \(2\)
    \(f(-2) = -8+4k+14+6 = 4k+12 = -4\), to \(k = -4\).
  17. Dividing \(x^2-1\) by \(x-1\) gives quotient and remainder:

    1. \(x-1\), remainder \(0\)
    2. \(x+1\), remainder \(1\)
    3. \(x-1\), remainder \(1\)
    4. \(x+1\), remainder \(0\)
    \(x^2-1 = (x-1)(x+1) + 0\); quotient \(x+1\), remainder \(0\).
  18. If the remainder of a polynomial division is 0, then the divisor is:

    1. Greater than the dividend
    2. A factor of the dividend
    3. Equal to zero
    4. Unrelated to the dividend
    Zero remainder ka matlab divisor dividend ko poori tarah divide karta hai — yani factor hai.
  19. Synthetic division of \(x^3-28x-48\) by \(-4\) gives the quotient:

    1. \(x^2-4x-12\)
    2. \(x^2+4x+12\)
    3. \(x^2-4x+12\)
    4. \(x^2+4x-12\)
    Bottom row \(1, -4, -12, 0\) — quotient \(x^2-4x-12\), remainder \(0\).
  20. Long division of \((3x^2-x+2)\) by \((x-1)\) gives the quotient:

    1. \(3x-2\)
    2. \(3x+1\)
    3. \(3x+2\)
    4. \(3x-1\)
    \(3x^2 \div x = 3x\), phir \(2x \div x = 2\); quotient \(3x+2\), remainder \(4\).
  21. In \(f(x) = (x-a)q(x) + r\), the remainder \(r\) is:

    1. A constant
    2. Always of degree 1
    3. Equal to \(q(x)\)
    4. Always zero
    Divisor degree 1 ka hai, to remainder ki degree is se kam — sirf aik constant.
  22. For \(H(z) = \frac{N(z)}{D(z)}\), the zeros of the transfer function come from the:

    1. Denominator
    2. Quotient
    3. Remainder
    4. Numerator
    Zeros wahan jahan numerator zero ho; poles denominator se aate hain.

Mathematics 11 (Punjab Board)

Chapter 9: Division of Polynomials — Formula Sheet

All the key results of Chapter 9 in one place — the division algorithm, remainder and factor theorems, synthetic division, and the application formulas.

Division Algorithm

\[ f(x) = g(x)\cdot q(x) + r(x), \quad r(x) = 0 \text{ or } \deg r(x) \lt \deg g(x) \]
\[ \text{Dividend} = \text{Divisor} \times \text{Quotient} + \text{Remainder} \]

Remainder Theorem

\[ f(x) \div (x-a) \Rightarrow \text{remainder} = f(a) \]
\[ f(x) \div (x+a) \Rightarrow \text{remainder} = f(-a) \]

Factor Theorem

\[ (x-a) \text{ is a factor of } f(x) \iff f(a) = 0 \]
\[ (x+a) \text{ is a factor of } f(x) \iff f(-a) = 0 \]

Synthetic Division (by \(x - a\))

\[ \text{Coefficients: } c_n, c_{n-1}, \dots, c_0 \quad (\text{use } 0 \text{ for missing powers}) \]
\[ \text{Bring down, multiply by } a, \text{ add — repeat. Last entry } = \text{ remainder.} \]

Applications

\[ \text{Percentage error} = \left|\frac{\text{observed}-\text{predicted}}{\text{observed}}\right|\times 100 \]
\[ H(z) = \frac{N(z)}{D(z)}: \text{ zeros from } N(z) = 0, \text{ poles from } D(z) = 0 \]
\[ \text{Stable system} \iff |z| \lt 1 \text{ for every pole} \]

Chapter 9 FAQs

What is covered in Chapter 9 (Division of Polynomials)?
This chapter covers Polynomial division, remainder theorem, factor theorem, quotient and remainder. The page includes theory and definitions, solved textbook examples, complete exercise solutions, MCQs and a formula sheet.
How many marks does Chapter 9 carry in the Punjab Board paper?
Chapter 9 (Division of Polynomials) carries approximately 12 marks in the Punjab Board Class 11 Mathematics paper.
Are these notes for the new PCTB syllabus?
Yes. These notes follow the PCTB 2025 new book for Class 11 Mathematics, used by BISE Lahore, Gujranwala, Rawalpindi, Faisalabad, Multan, Sahiwal, Sargodha, Bahawalpur and D.G. Khan.
Where are the exercise solutions for Chapter 9?
Use the exercise index on this page — every exercise of Chapter 9 has its own page with complete step-by-step solutions.