Polynomials — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 2  ·  Q–
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Class 9 Mathematics Exercise-2.2 NCERT Solutions Olympiad Board Exam
Chapter 2

Polynomials

Step-by-step NCERT solutions with stress–strain analysis and exam-oriented hints for Boards, JEE & NEET.

4 Questions
10–15 min Ideal time
Q1 Now at
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Q1
NUMERIC2 marks

Find the value of the polynomial \(5x - 4x^2 + 3\) at:

  1. \(x = 0\)
  2. \(x = -1\)
  3. \(x = 2\)
📘 Concept & Theory Theory / Concept Used ›

A polynomial is an algebraic expression consisting of variables and coefficients. To find the value of a polynomial at a particular value of \(x\), we substitute the given value of \(x\) into the polynomial and simplify step-by-step.

The given polynomial is:

\[\small p(x)=5x-4x^2+3 \]

Here:

  • \(5x\) is the linear term.
  • \(-4x^2\) is the quadratic term.
  • \(3\) is the constant term.

While evaluating a polynomial, always follow the order: Substitution → Exponents → Multiplication → Addition/Subtraction.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Write the polynomial properly.
  2. Substitute the given value of \(x\).
  3. Calculate powers first.
  4. Perform multiplication carefully.
  5. Simplify by addition/subtraction.
  6. Write the final value clearly.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  25 steps
  1. Given: \[\small p(x)=5x-4x^2+3\]
  2. Substitute \(x=0\):
    For \(x = 0\), the polynomial simplifies to:
  3. \[\small \begin{aligned} p(0) &=5(0)-4(0)^2+3 \end{aligned} \]
  4. Evaluate the square:
  5. \[\small \begin{aligned} p(0) &=5(0)-4(0)+3 \end{aligned} \]
  6. Perform multiplication:
  7. \[\small \begin{aligned} p(0) &=0-0+3 \end{aligned} \]
  8. Simplify:
  9. \[\small \begin{aligned} p(0) &=3 \end{aligned} \]
    Thus, the value of the polynomial at \(x = 0\) is \(3\).
  10. For \(x = -1\), substitute and simplify similarly:
  11. \[\small \begin{aligned} p(-1) &=5(-1)-4(-1)^2+3 \end{aligned} \]
  12. Evaluate the square:
  13. \[\small \begin{aligned} p(-1) &=5(-1)-4(1)+3 \end{aligned} \]
  14. Perform multiplication:
  15. \[\small \begin{aligned} p(-1) &=-5-4+3 \end{aligned} \]
  16. Simplify:
  17. \[\small \begin{aligned} p(-1) &=-6 \end{aligned} \]
    Thus, the value of the polynomial at \(x = -1\) is \(-6\).
  18. For \(x = 2\), substitute and simplify similarly:
  19. \[\small \begin{aligned} p(2) &=5(2)-4(2)^2+3 \end{aligned} \]
  20. Evaluate the square:
  21. \[\small \begin{aligned} p(2) &=5(2)-4(4)+3 \end{aligned} \]
  22. Perform multiplication:
  23. \[\small \begin{aligned} p(2) &=10-16+3 \end{aligned} \]
  24. Simplify:
  25. \[\small \begin{aligned} p(2) &=-3 \end{aligned} \]
    Thus, the value of the polynomial at \(x = 2\) is \(-3\).
🎯 Exam Significance Exam Significance ›
  • Evaluation of polynomials is one of the most fundamental algebraic skills frequently asked in school examinations.
  • Questions based on substitution and simplification appear regularly in CBSE board exams and internal assessments.
  • Competitive examinations test calculation accuracy and sign handling using such polynomial evaluations.
  • This concept forms the foundation for factorisation, algebraic identities, graphing of functions, and higher algebra in Classes 10, 11 and 12.
  • Careful handling of negative signs and exponents is extremely important for entrance examinations like Olympiads and NTSE.
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1 / 4  ·  25%
Q2 →
Q2
NUMERIC3 marks
Find \(p(0)\), \(p(1)\) and \(p(2)\) for each of the following polynomials.
  1. \(p(y) = y^2 – y + 1\)
  2. \(p(t) = 2 + t + 2t^2 – t\)
  3. \(p(x) = x^3\)
  4. \(p(x) = (x – 1) (x + 1)\)
📘 Concept & Theory Theory / Concept Used ›

The value of a polynomial at a particular number is obtained by substituting the given value of the variable into the polynomial and simplifying carefully.

While solving such questions:

  • Replace the variable by the given value.
  • Evaluate powers first.
  • Perform multiplication carefully.
  • Simplify using addition and subtraction.

Such problems strengthen algebraic simplification and sign handling skills.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Write the polynomial clearly.
  2. Substitute the given value 0, 1 & 2.
  3. Evaluate powers first.
  4. Perform multiplication carefully.
  5. Simplify using addition and subtraction.
  6. Write the final value neatly.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. \(p(y)=y^2-y+1\)
  2. Finding \(p(0)\)
    For \(y = 0\), the polynomial simplifies to:
  3. \[\small \begin{aligned} p(0) &=0^2-0+1 \end{aligned} \]
  4. Evaluate the square:
  5. \[\small \begin{aligned} p(0) &=0-0+1 \end{aligned} \]
    Thus, the value of the polynomial at \(y = 0\) is \(1\).
  6. Finding \(p(1)\)
    For \(y = 1\), substitute and simplify similarly:
  7. \[\small \begin{aligned} p(1) &=1^2-1+1 \end{aligned} \]
  8. Evaluate the square:
  9. \[\small \begin{aligned} p(1) &=1-1+1 \end{aligned} \]
    Thus, the value of the polynomial at \(y = 1\) is \(1\).
  10. Finding \(p(2)\)
    For \(y = 2\), substitute and simplify similarly:
  11. \[\small \begin{aligned} p(2) &=2^2-2+1 \end{aligned} \]
  12. Evaluate the square:
  13. \[\small \begin{aligned} p(2) &=4-2+1 \end{aligned} \]
    Thus, the value of the polynomial at \(y = 2\) is \(3\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. \(p(t)=2+t+2t^2-t\)
  2. Simplify the polynomial first:
  3. \[\small \begin{aligned} p(t) &=2+t+2t^2-t \\ &=2t^2+2 \end{aligned} \]
    Now substitute \(t = 0\), \(1\) and \(2\) to find the respective values.
  4. \[\small \begin{aligned} p(0) &=2(0)^2+2 \\ &=2 \end{aligned} \]
    Thus, the value of the polynomial at \(t = 0\) is \(2\).
  5. \[\small \begin{aligned} p(1) &=2(1)^2+2 \\ &=4 \end{aligned} \]
    Thus, the value of the polynomial at \(t = 1\) is \(4\).
  6. \[\small \begin{aligned} p(2) &=2(2)^2+2 \\ &=10 \end{aligned} \]
    Thus, the value of the polynomial at \(t = 2\) is \(10\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. \(p(x)=x^3\)
  2. Finding \(p(0)\)
    For \(x = 0\), the polynomial simplifies to:
  3. \[\small \begin{aligned} p(0) &=0^3 \end{aligned} \]
    Thus, the value of the polynomial at \(x = 0\) is \(0\).
  4. Finding \(p(1)\)
    For \(x = 1\), substitute and simplify similarly:
  5. \[\small \begin{aligned} p(1) &=1^3 \end{aligned} \]
    Thus, the value of the polynomial at \(x = 1\) is \(1\).
  6. Finding \(p(2)\)
    For \(x = 2\), substitute and simplify similarly:
  7. \[\small \begin{aligned} p(2) &=2^3 \end{aligned} \]
    Thus, the value of the polynomial at \(x = 2\) is \(8\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  7 steps
  1. \(p(x)=(x-1)(x+1)\)
  2. Finding \(p(0)\)
    For \(x = 0\), the polynomial simplifies to:
  3. \[\small \begin{aligned} p(0) &= (0-1)(0+1) \end{aligned} \]
    Thus, the value of the polynomial at \(x = 0\) is \(-1\).
  4. Finding \(p(1)\)
    For \(x = 1\), substitute and simplify similarly:
  5. \[\small \begin{aligned} p(1) &= (1-1)(1+1) \end{aligned} \]
    Thus, the value of the polynomial at \(x = 1\) is \(0\).
  6. Finding \(p(2)\)
    For \(x = 2\), substitute and simplify similarly:
  7. \[\small \begin{aligned} p(2) &= (2-1)(2+1) \end{aligned} \]
    Thus, the value of the polynomial at \(x = 2\) is \(3\).
🎯 Exam Significance Exam Significance ›
  • Polynomial evaluation is one of the most frequently tested algebraic concepts in school examinations.
  • These questions improve substitution accuracy and sign handling.
  • Understanding polynomial values helps in graph plotting, factorisation and algebraic identities.
  • Such simplification skills are useful in Olympiads, NTSE and other competitive aptitude examinations.
  • Stepwise simplification reduces calculation mistakes in board exams.
← Q1
2 / 4  ·  50%
Q3 →
Q3
NUMERIC3 marks
Verify whether the following are zeroes of the polynomial indicated against them.
  1. \(p(x) = 3x + 1, x = -\frac{1}{3}\)
  2. \( p(x) = 5x – π, x = \frac{4}{5}\)
  3. \(p(x) = x^2 – 1, x = 1, –1\)
  4. \( p(x) = (x + 1) (x – 2), x = – 1, 2\)
  5. \( p(x) = x^2, x = 0\)
  6. \(p(x) = lx + m, x = – \frac{m}{l}\)
  7. \(p(x) = 3x^2 – 1, x = \frac{1}{\sqrt{3}}, –\frac{2}{\sqrt{3}}\)
  8. \( p(x) = 2x + 1, x = -\frac{1}{2} \)
📘 Concept & Theory Theory / Concept Used ›

A number \(a\) is called a zero of a polynomial \(p(x)\) if:

\[\small p(a)=0 \]

To verify whether a given number is a zero of a polynomial:

  1. Substitute the given value into the polynomial.
  2. Simplify carefully using algebraic operations.
  3. If the final value is \(0\), then the number is a zero.
  4. If the final value is not \(0\), then it is not a zero.

Zeroes of polynomials are very important because they represent the values where the polynomial becomes zero and the graph intersects the x-axis.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Write the given polynomial.
  2. Substitute the given value of the variable.
  3. Simplify powers and multiplication carefully.
  4. Check whether the result equals zero.
  5. Conclude whether the given number is a zero or not
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=3x+1,\quad x=-\dfrac{1}{3}\)
  2. Substitute \(x = -\frac{1}{3}\) into the polynomial:
  3. \[\small \begin{aligned} p\left(-\frac{1}{3}\right) &=3\left(-\frac{1}{3}\right)+1 \end{aligned} \]
  4. Perform multiplication:
  5. \[\small \begin{aligned} p\left(-\frac{1}{3}\right) &=-1+1 \end{aligned} \]
    Since \(p\left(-\frac{1}{3}\right)=0\), the number \(-\frac{1}{3}\) is a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=5x-\pi,\quad x=\dfrac{4}{5}\)
  2. Substitute \(x = \frac{4}{5}\) into the polynomial:
  3. \[\small \begin{aligned} p\left(\frac{4}{5}\right) &=5\left(\frac{4}{5}\right)-\pi \end{aligned} \]
  4. Perform multiplication:
  5. \[\small \begin{aligned} p\left(\frac{4}{5}\right) &=4-\pi \end{aligned} \]
    Since \(p\left(\frac{4}{5}\right)\neq 0\), the number \(\frac{4}{5}\) is not a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=x^2-1,\quad x=1,-1\)
  2. Substitute \(x = 1\) into the polynomial:
  3. \[\small \begin{aligned} p(1) &=1^2-1 \end{aligned} \]
    Since \(p(1)=0\), the number \(1\) is a zero of the polynomial.
  4. Substitute \(x = -1\) into the polynomial:
  5. \[\small \begin{aligned} p(-1) &=(-1)^2-1 \end{aligned} \]
    Since \(p(-1)=0\), the number \(-1\) is also a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=(x+1)(x-2),\quad x=-1,2\)
  2. Substitute \(x = -1\) into the polynomial:
  3. \[\small \begin{aligned} p(-1) &=(-1+1)(-1-2) \end{aligned} \]
    Since \(p(-1)=0\), the number \(-1\) is a zero of the polynomial.
  4. Substitute \(x = 2\) into the polynomial:
  5. \[\small \begin{aligned} p(2) &=(2+1)(2-2) \end{aligned} \]
    Since \(p(2)=0\), the number \(2\) is also a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. \(p(x)=x^2,\quad x=0\)
  2. Substitute \(x = 0\) into the polynomial:
  3. \[\small \begin{aligned} p(0) &=0^2 \end{aligned} \]
    Since \(p(0)=0\), the number \(0\) is a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=lx+m,\quad x=-\dfrac{m}{l}\)
  2. Substitute \(x = -\frac{m}{l}\) into the polynomial:
  3. \[\small \begin{aligned} p\left(-\frac{m}{l}\right) &=l\left(-\frac{m}{l}\right)+m \end{aligned} \]
  4. Perform multiplication:
  5. \[\small \begin{aligned} p\left(-\frac{m}{l}\right) &=-m+m \end{aligned} \]
    Since \(p\left(-\frac{m}{l}\right)=0\), the number \(-\frac{m}{l}\) is a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. \(p(x)=3x^2-1,\quad x=-\dfrac{1}{\sqrt{3}},\dfrac{2}{\sqrt{3}}\)
  2. Substitute \(x = -\frac{1}{\sqrt{3}}\) into the polynomial:
  3. \[\small \begin{aligned} p\left(-\frac{1}{\sqrt{3}}\right) &=3\left(-\frac{1}{\sqrt{3}}\right)^2-1 \end{aligned} \]
    Since \(p\left(-\frac{1}{\sqrt{3}}\right)=0\), the number \(-\frac{1}{\sqrt{3}}\) is a zero of the polynomial.
  4. Substitute \(x = \frac{2}{\sqrt{3}}\) into the polynomial:
  5. \[\small \begin{aligned} p\left(\frac{2}{\sqrt{3}}\right) &=3\left(\frac{2}{\sqrt{3}}\right)^2-1 \end{aligned} \]
    Since \(p\left(\frac{2}{\sqrt{3}}\right)\neq 0\), the number \(\frac{2}{\sqrt{3}}\) is not a zero of the polynomial.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. \(p(x)=2x+1,\quad x=\frac{1}{2}\)
  2. Substitute \(x = \frac{1}{2}\) into the polynomial:
  3. \[\small \begin{aligned} p\left(\frac{1}{2}\right) &=2\left(\frac{1}{2}\right)+1 \end{aligned} \]
    Since \(p\left(\frac{1}{2}\right)=2\), the number \(x=\frac{1}{2}\) is not a zero of the polynomial.
🎯 Exam Significance Exam Significance ›

2-mark standard board question.

← Q2
3 / 4  ·  75%
Q4 →
Q4
NUMERIC3 marks
Find the zero of the polynomial in each of the following cases:
ol type="i">
  • \(p(x) = x + 5\)
  • \( p(x) = x – 5 \)
  • \(p(x) = 2x + 5\)
  • \(p(x) = 3x – 2\)
  • \( p(x) = 3x\)
  • \( p(x) = ax, a ≠ 0\)
  • \( p(x) = cx + d, c ≠ 0, c, d \text{ are real numbers.}\)
    • 📘 Concept & Theory Theory / Concept Used ›

    A zero of a polynomial is the value of the variable for which the polynomial becomes equal to zero.

    If \(p(x)\) is a polynomial, then:

    \[\small p(x)=0 \]

    gives the zeroes of the polynomial.

    For a linear polynomial of the form:

    \[\small ax+b \] where \(a\neq0\), the zero is obtained by solving: \[\small ax+b=0 \]

    This concept is very important because zeroes help us understand where the graph of the polynomial cuts the x-axis.

    🗺️ Solution Roadmap Step-by-step Plan ›
    1. Write the polynomial.
    2. Equate the polynomial to zero.
    3. Solve the resulting linear equation step-by-step.
    4. Find the value of \(x\).
    5. Write the final zero clearly.
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  5 steps
    1. \(p(x)=x+5\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} x+5&=0 \end{aligned} \]
    4. Transpose \(5\) to the right side:
    5. \[\small \begin{aligned} x&=-5 \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{-5} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  5 steps
    1. \(p(x)=x-5\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} x-5&=0 \end{aligned} \]
    4. Transpose \(-5\) to the right side:
    5. \[\small \begin{aligned} x&=5 \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{5} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  7 steps
    1. \(p(x)=2x+5\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} 2x+5&=0 \end{aligned} \]
    4. Transpose \(5\) to the right side:
    5. \[\small \begin{aligned} 2x&=-5 \end{aligned} \]
    6. Divide both sides by \(2\):
    7. \[\small \begin{aligned} x&=-\frac{5}{2} \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{-\frac{5}{2}} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  7 steps
    1. \(p(x)=3x-2\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} 3x-2&=0 \end{aligned} \]
    4. Transpose \(-2\) to the right side:
    5. \[\small \begin{aligned} 3x&=2 \end{aligned} \]
    6. Divide both sides by \(3\):
    7. \[\small \begin{aligned} x&=\frac{2}{3} \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{\frac{2}{3}} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  5 steps
    1. \(p(x)=3x\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} 3x&=0 \end{aligned} \]
    4. Divide both sides by \(3\):
    5. \[\small \begin{aligned} x&=0 \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{0} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  5 steps
    1. \(p(x)=ax,\quad a\neq0\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} ax&=0 \end{aligned} \]
    4. Divide both sides by \(a\):
    5. \[\small \begin{aligned} x&=0 \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{0} \]
    ✏️ Solution Complete Solution ›
    Step-by-step Solution  ·  7 steps
    1. \(p(x)=cx+d,\quad c\neq0\)
    2. To find the zero, equate the polynomial to zero:
    3. \[\small \begin{aligned} cx+d&=0 \end{aligned} \]
    4. Transpose \(d\) to the right side:
    5. \[\small \begin{aligned} cx&=-d \end{aligned} \]
    6. Divide both sides by \(c\):
    7. \[\small \begin{aligned} x&=-\frac{d}{c} \end{aligned} \]
      Hence, the zero of the polynomial is: \[\small \boxed{-\frac{d}{c}} \]
    🎯 Exam Significance Exam Significance ›
    • Finding zeroes of linear polynomials is a fundamental algebraic skill tested frequently in school examinations.
    • These questions help students understand the relation between algebraic equations and graphical representation.
    • The concept is the foundation for solving quadratic and higher degree polynomial equations in higher classes.
    • Competitive examinations often test speed and accuracy in solving linear equations and identifying zeroes.
    • Proper stepwise solving improves presentation quality and helps secure full marks in board examinations.
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    Step by StepSolutions NCERT Class 9 Maths Exercise 2.2 — Complete Notes & Solutions · academia-aeternum.com
    Exercise 2.2 of NCERT Class 9 Mathematics (Chapter: Polynomials) focuses on finding the value of polynomials at different points, understanding and calculating the zeros of polynomials, and verifying them with step-by-step solutions. All textbook questions and examples have been solved clearly to help students strengthen their fundamentals. This post is especially useful for Class 9 students preparing for exams who often face difficulty in solving polynomial problems. At the end of the post, we…
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