Polynomials — NCERT Solutions | Class 9 Mathematics | Academia Aeternum
Ch 2  ·  Q–
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Class 9 Mathematics Exercise-2.3 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.

5 Questions
10–15 min Ideal time
Q1 Now at
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Q1
NUMERIC2 marks
Determine which of the following polynomials has (x + 1) a factor
(i) \(x^3 + x^2 + x + 1\)
(ii) \(x^4 + x^3 + x^2 + x + 1\)
(iii) \( x^4 + 3x^3 + 3x^2 + x + 1\)
(iv) \( x^3 – x^2 – (2+\sqrt{2})x+\sqrt{2}\)
📘 Concept & Theory In this exercise, we use the Factor Theorem. ›

According to the Factor Theorem:

\[\small \text{If } p(a)=0,\text{ then } (x-a) \text{ is a factor of } p(x) \]

Here, we have to check whether \((x+1)\) is a factor of the given polynomial.

Since

\[\small x+1=0 \] \[\small x=-1 \]

Therefore, we substitute \(x=-1\) in each polynomial.

  • If \(p(-1)=0\), then \((x+1)\) is a factor.
  • If \(p(-1)\neq0\), then \((x+1)\) is not a factor.
🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the zero of \((x+1)\)

  2. Substitute that value in each polynomial.
  3. Evaluate carefully step-by-step.

  4. If the result is 0, then \((x+1)\) is a factor. Otherwise, it is not a factor.

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. Given: \[\small \begin{aligned}x+1&=0\\x&=-1\end{aligned}\]

    Therefore, we evaluate \(p(-1)\) for each polynomial.

  2. \[\small p(x)=x^{3}+x^{2}+x+1\]
  3. \[\small p(-1)=(-1)^{3}+(-1)^{2}+(-1)+1\]
  4. \[\small p(-1)=-1+1-1+1=0\]
    Since \(p(-1)=0\), \((x+1)\) is a factor of \(p(x)\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. \[\small p(x)=x^{4}+x^{3}+x^{2}+x+1\]
  2. Substitute \(x=-1\)
  3. \[\small \begin{aligned} p(-1) &=(-1)^4+(-1)^3+(-1)^2+(-1)+1 \\ &=1-1+1-1+1 \\ &=(1-1)+(1-1)+1 \\ &=0+0+1 \\ &=1 \end{aligned} \]
    Since \(p(-1)\neq0\), \((x+1)\) is not a factor of \(p(x)\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. \[\small p(x)=x^{4}+3x^{3}+3x^{2}+x+1\]
  2. Substitute \(x=-1\)
  3. \[\small \begin{aligned} p(-1) &=(-1)^4+3(-1)^3+3(-1)^2+(-1)+1 \\ &=1-3+3-1+1 \\ &=(1-3)+(3-1)+1 \\ &=-2+2+1 \\ &=1 \end{aligned} \]
    Since \(p(-1)\neq0\), \((x+1)\) is not a factor of \(p(x)\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  3 steps
  1. \[\small p(x)=x^{3}-x^{2}-(2+\sqrt{2})x+\sqrt{2}\]
  2. Substitute \(x=-1\)
  3. \[\small \begin{aligned} p(-1) &=(-1)^3-(-1)^2-(2+\sqrt{2})(-1)+\sqrt{2} \\ &=-1-1+(2+\sqrt{2})+\sqrt{2} \\ &=-2+2+2\sqrt{2} \\ &=2\sqrt{2} \end{aligned} \]
    Since \(p(-1)\neq0\), \((x+1)\) is not a factor of \(p(x)\).
💡 Answer Final Answer ›
Answer: (i) is the only polynomial for which \((x+1)\) is a factor.
🎯 Exam Significance Exam Significance ›
  • Factor Theorem is one of the most important concepts in algebra.
  • It is frequently asked in school examinations and board examinations.
  • This concept is also useful in competitive examinations such as NTSE, Olympiads, Polytechnic Entrance Exams, SSC and other aptitude tests.
  • It helps in factorisation of higher degree polynomials quickly.
  • Understanding this theorem builds a strong foundation for higher algebra.
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1 / 5  ·  20%
Q2 →
Q2
NUMERIC3 marks
Use the Factor Theorem to determine whether g(x) is a factor of p(x) in each of the following cases:
(i) \(p(x) = 2x^3 + x^2 – 2x – 1, g(x) = x + 1\)
(ii) \(p(x) = x^3 + 3x^2 + 3x + 1, g(x) = x + 2\)
(iii) \( p(x) = x^3 – 4x^2 + x + 6, g(x) = x – 3\)
📘 Concept & Theory Application of Factor Theorem ›

In these questions, we use the Factor Theorem to check whether the given polynomial \(g(x)\) is a factor of \(p(x)\).

\[\small \begin{aligned} \text{If } g(x)&=x-a \text{ and }\\ p(a)&=0, \text{ then } g(x)\text{ is a factor of }p(x) \end{aligned} \]

Therefore:

  • Find the zero of \(g(x)\)
  • Substitute that value in \(p(x)\)
  • If the result becomes \(0\), then \(g(x)\) is a factor

Use the Factor Theorem to determine whether \(g(x)\) is a factor of \(p(x)\) in each of the following cases:

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Find the value of \(x\) for which \(g(x)=0\).

  2. Substitute that value into \(p(x)\).

  3. Simplify step-by-step carefully.

  4. If the result is \(0\), then \(g(x)\) is a factor of \(p(x)\). Otherwise, it is not a factor.

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (i) Given: \[\small \begin{aligned}p(x)&=2x^{3}+x^{2}-2x-1\\g(x)&=x+1\end{aligned}\]
  2. Find the zero of \(g(x)\)
  3. \[\small \begin{aligned} g(x)&=0\\ x+1&=0\\ x&=-1 \end{aligned} \]
  4. Substitute \(x=-1\) in \(p(x)\)
  5. \[\small \begin{aligned} p(-1) &=2(-1)^{3}+(-1)^{2}-2(-1)-1 \\ &=-2+1+2-1 \\ &=(1-2)+(2-1) \\ &=0 \end{aligned} \]
    Since \(p(-1)=0\), \(g(x)\) is a factor of \(p(x)\).
💡 Answer Final Answer ›
Answer: (i) \(g(x)\) is a factor of \(p(x)\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (ii) Given: \[\small \begin{aligned}p(x)&=x^{3}+3x^{2}+3x+1\\g(x)&=x+2\end{aligned}\]
  2. Find the zero of \(g(x)\)
  3. \[\small \begin{aligned} g(x)&=0\\ x+2&=0\\ x&=-2 \end{aligned} \]
  4. Substitute \(x=-2\) in \(p(x)\)
  5. \[\small \begin{aligned} p(-2) &=(-2)^{3}+3(-2)^{2}+3(-2)+1 \\ &=-8+12-6+1 \\ &=(12-8)+(1-6) \\ &=4-5 \\ &=-1 \end{aligned} \]
    Since \(p(-2)\neq0\), \(g(x)\) is not a factor of \(p(x)\).
💡 Answer Final Answer ›
Answer: (ii) \(g(x)\) is not a factor of \(p(x)\).
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (iii) Given: \[\small \begin{aligned}p(x)&=x^{3}-4x^{2}+x+6\\g(x)&=x-3\end{aligned}\]
  2. Find the zero of \(g(x)\)
  3. \[\small \begin{aligned} g(x)&=0\\ x-3&=0\\ x&=3 \end{aligned} \]
  4. Substitute \(x=3\) in \(p(x)\)
  5. \[\small \begin{aligned} p(3) &=(3)^{3}-4(3)^{2}+(3)+6 \\ &=27-36+3+6 \\ &=(27-36)+(3+6) \\ &=-9+9 \\ &=0 \end{aligned} \]
    Since \(p(3)=0\), \(g(x)\) is a factor of \(p(x)\).
💡 Answer Final Answer ›
Answer: (iii) \(g(x)\) is a factor of \(p(x)\).
🎯 Exam Significance Exam Significance ›
  • Questions based on Factor Theorem are frequently asked in CBSE and state board examinations.
  • This theorem is also useful in NTSE, Olympiads, SSC, Polytechnic and other competitive examinations.
  • It helps students solve factorisation problems quickly without long division.
  • This concept is the foundation for higher algebra and polynomial equations.
← Q1
2 / 5  ·  40%
Q3 →
Q3
NUMERIC3 marks
Find the value of k, if x – 1 is a factor of p(x) in each of the following cases:
(i) \(p(x) = x^2 + x + k\)
(ii) \(p(x) = 2x^2 + kx + \sqrt{2}\)
(iii) \(p(x) = kx^2 – \sqrt{2}x + 1\)
(iv) \( p(x) = kx^2 – 3x + k\)
📘 Concept & Theory Finding the Value of \(k\) Using Factor Theorem ›

In these questions, we are given that \((x-1)\) is a factor of the polynomial.

Using the Factor Theorem:

\[\small \text{If } (x-a)\text{ is a factor of }p(x), \text{ then }p(a)=0 \]

Since \((x-1)\) is a factor:

\[\small x-1=0 \] \[\small x=1 \]

Therefore, we substitute \(x=1\) into each polynomial and solve for \(k\).

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Since \((x-1)\) is a factor, first find the zero of \((x-1)\).

  2. Put \(x=1\) into the polynomial.

  3. Use \(p(1)=0\).

  4. Solve the resulting equation to find the value of \(k\).

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  6 steps
  1. (i) Given: \[\small \begin{aligned}p(x)&=x^{2}+x+k\\x-1&=0\end{aligned}\]
  2. Find the zero of \(x-1\)
  3. \[\small \begin{aligned} x-1&=0\\ x&=1 \end{aligned} \]
  4. Substitute \(x=1\) in \(p(x)\)
  5. \[\small \begin{aligned} p(1) &=(1)^{2}+(1)+k \\ &=1+1+k \\ &=2+k \end{aligned} \]
    Since \(p(1)=0\), we have:
  6. \[\small \begin{aligned} 2+k&=0\\ k&=-2 \end{aligned} \]
💡 Answer Final Answer ›
Answer: (i) \(k=-2\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. (ii) Given: \[\small p(x)=2x^{2}+kx+\sqrt{2} \]
  2. Substitute \(x=1\) in \(p(x)\)
  3. \[\small \begin{aligned} p(1) &=2(1)^{2}+k(1)+\sqrt{2} \\ &=2+k+\sqrt{2} \end{aligned} \]
    Since \(p(1)=0\), we have:
  4. \[\small \begin{aligned} 2+k+\sqrt{2}&=0\\ k&=-2-\sqrt{2} \end{aligned} \]
💡 Answer Final Answer ›
Answer: (ii) \(k=-2-\sqrt{2}\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. (iii )Given:\[\small p(x)=kx^{2}-\sqrt{2}x+1\]
  2. Substitute \(x=1\) in \(p(x)\)
  3. \[\small \begin{aligned} p(1) &=k(1)^{2}-\sqrt{2}(1)+1 \\ &=k-\sqrt{2}+1 \end{aligned} \]
    Since \(p(1)=0\), we have:
  4. \[\small \begin{aligned} k-\sqrt{2}+1&=0\\ k&=\sqrt{2}-1 \end{aligned} \]
💡 Answer Final Answer ›
Answer: (iii) \(k=\sqrt{2}-1\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  4 steps
  1. (iv) Given:\[\small p(x)=kx^{2}-3x+k\]
  2. Substitute \(x=1\) in \(p(x)\)
  3. \[\small \begin{aligned} p(1) &=k(1)^{2}-3(1)+k \\ &=k-3+k \\ &=2k-3 \end{aligned} \]
    Since \(p(1)=0\), we have:
  4. \[\small \begin{aligned} 2k-3&=0\\ 2k&=3\\ k&=\frac{3}{2} \end{aligned} \]
💡 Answer Final Answer ›
Answer: (iv) \(k=\frac{3}{2}\)
🎯 Exam Significance Exam Significance ›
  • Questions based on unknown coefficients are very common in board examinations.
  • These problems strengthen understanding of Factor Theorem and polynomial roots.
  • Such concepts are useful in competitive examinations like NTSE, Olympiads, SSC, Polytechnic and other entrance tests.
  • Learning these methods improves algebraic manipulation skills.
← Q2
3 / 5  ·  60%
Q4 →
Q4
NUMERIC3 marks
Factorise:
(i) \(12x^2 – 7x + 1\)
(ii) \(2x^2 + 7x + 3\)
(iii) \(6x^2 + 5x – 6\)
(iv) \(3x^2 – x – 4\)
📘 Concept & Theory Factorisation of Polynomials ›

In this exercise, we factorise quadratic polynomials using:

  • Factor Theorem
  • Splitting the Middle Term Method
  • Product–Sum Method

A quadratic polynomial of the form:

\[\small ax^{2}+bx+c \]

can often be factorised into:

\[\small (mx+n)(px+q) \]

We use suitable algebraic techniques to find these factors.

🗺️ Solution Roadmap Step-by-step Plan ›

  1. Rewrite the polynomial carefully.

  2. Find suitable numbers whose:

  3. Product gives the constant term

  4. Sum gives the coefficient of \(x\)

  5. Use Factor Theorem or splitting method.

  6. Write the polynomial as product of factors.

✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (i) Given: \[\small p(x)=12x^{2}-7x+1\]
  2. Find two numbers whose product is \(12 \times 1 = 12\) and sum is \(-7\).
  3. The numbers are \(-3\) and \(-4\).
  4. Rewrite the middle term:
  5. \[\small \begin{aligned} p(x) &=12x^{2}-3x-4x+1 \\ &=(12x^{2}-3x)- (4x-1) \\ &=3x(4x-1)-1(4x-1) \\ &=(4x-1)(3x-1) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (i) \(p(x)=(4x-1)(3x-1)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (ii) Given: \[\small p(x)=2x^{2}+7x+3\]
  2. Find two numbers whose product is \(2 \times 3 = 6\) and sum is \(7\).
  3. The numbers are \(6\) and \(1\).
  4. Rewrite the middle term:
  5. \[\small \begin{aligned} p(x) &=2x^{2}+6x+x+3 \\ &=(2x^{2}+6x)+(x+3) \\ &=2x(x+3)+1(x+3) \\ &=(x+3)(2x+1) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (ii) \(p(x)=(x+3)(2x+1)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (iii) Given: \[\small p(x)=6x^{2}+5x-6\]
  2. Find two numbers whose product is \(6 \times (-6) = -36\) and sum is \(5\).
  3. The numbers are \(9\) and \(-4\).
  4. Rewrite the middle term:
  5. \[\small \begin{aligned} p(x) &=6x^{2}+9x-4x-6 \\ &=(6x^{2}+9x)- (4x+6) \\ &=3x(2x+3)-2(2x+3) \\ &=(2x+3)(3x-2) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (iii) \(p(x)=(2x+3)(3x-2)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  5 steps
  1. (iv) Given: \[\small p(x)=3x^{2}-x-4\]
  2. Find two numbers whose product is \(3 \times (-4) = -12\) and sum is \(-1\).
  3. The numbers are \(3\) and \(-4\).
  4. Rewrite the middle term:
  5. \[\small \begin{aligned} p(x) &=3x^{2}+3x-4x-4 \\ &=(3x^{2}+3x)- (4x+4) \\ &=3x(x+1)-4(x+1) \\ &=(x+1)(3x-4) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (iv) \(p(x)=(x+1)(3x-4)\)
🎯 Exam Significance Exam Significance ›
  • Factorisation is one of the most important algebraic skills in mathematics.
  • It is heavily used in CBSE board examinations and school assessments.
  • Questions based on factorisation frequently appear in Olympiads, NTSE, SSC, Polytechnic and other competitive examinations.
  • It helps in solving polynomial equations quickly.
  • This concept forms the foundation of higher algebra and calculus.
← Q3
4 / 5  ·  80%
Q5 →
Q5
NUMERIC3 marks
Factorise:
(i) \(x^3 – 2x^2 – x + 2\)
(ii) \(x^3 – 3x^2 – 9x – 5\)
(iii) \(x^3 + 13x^2 + 32x + 20\)
(iv) \(2y^3 + y^2 – 2y – 1\)
📘 Concept & Theory Factorisation of Cubic Polynomials ›

In this exercise, we factorise cubic polynomials using:

  • Factor Theorem
  • Grouping Method
  • Division Method
  • Identity Method

For a cubic polynomial:

\[\small p(x)=ax^{3}+bx^{2}+cx+d \]

we first try to find one linear factor using the Factor Theorem. Then the remaining quadratic expression is factorised further.

🗺️ Solution Roadmap Step-by-step Plan ›
  1. Try grouping terms if possible.
  2. Otherwise, use the Factor Theorem.
  3. Find possible roots using factors of the constant term.
  4. Once one factor is found, divide the polynomial to obtain the remaining factors.
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  9 steps
  1. (i) Given: \[\small p(x)=x^{3}-2x^{2}-x+2\]
  2. Try grouping:
  3. \[\small \begin{aligned} x^{3}-2x^{2}-x+2 &=x^{2}(x-2)-1(x-2) \\ &=(x-2)(x^{2}-1) \end{aligned} \]
  4. Now use identity:
  5. \[\small a^{2}-b^{2}=(a-b)(a+b) \]
  6. Therefore
  7. \[\small \begin{aligned} x^{2}-1 &=(x-1)(x+1) \end{aligned} \]
  8. Final factorisation:
  9. \[\small \begin{aligned} x^{3}-2x^{2}-x+2 &=(x-2)(x-1)(x+1) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (i) \(p(x)=(x-2)(x-1)(x+1)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  13 steps
  1. (ii) Given: \[\small p(x)=x^{3}-3x^{2}-9x-5\]
  2. Possible factors of constant term \(-5\) are:
  3. \[\small \pm1,\ \pm5 \]
  4. Try \(x=1\):
  5. \[\small \begin{aligned} p(1) &=(1)^{3}-3(1)^{2}-9(1)-5 \\ &=1-3-9-5 \\ &=(1-3)+(-9-5) \\ &=-2-14 \\ &=-16 \end{aligned} \]
    Since \(p(1)\neq0\), \((x-1)\) is not a factor of \(p(x)\).
  6. Try \(x=-1\):
  7. \[\small \begin{aligned} p(-1) &=(-1)^{3}-3(-1)^{2}-9(-1)-5 \\ &=-1-3+9-5 \\ &=(-1-3)+(9-5) \\ &=-4+4 \\ &=0 \end{aligned} \]
    Since \(p(-1)=0\), \((x+1)\) is a factor of \(p(x)\).
  8. Divide \(p(x)\) by \((x+1)\) to find the remaining factors.
  9. \[\small \begin{aligned} p(x) &=(x+1)(x^{2}-4x-5) \end{aligned} \]
  10. Factorise the quadratic:
  11. \[\small \begin{aligned} x^{2}-4x-5 &=(x-5)(x+1) \end{aligned} \]
  12. Final factorisation:
  13. \[\small \begin{aligned} p(x) &=(x+1)(x-5)(x+1) \\ &=(x+1)^{2}(x-5) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (ii) \(p(x)=(x+1)^{2}(x-5)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  15 steps
  1. (iii) Given: \[\small p(x)=x^{3}+13x^{2}+32x+20\]
  2. Try possible factors of constant term \(20\):
  3. \[\small \pm1,\ \pm2,\ \pm4,\ \pm5,\ \pm10,\ \pm20 \]
  4. Try \(x=-1\):
  5. \[\small \begin{aligned} p(-1) &=(-1)^{3}+13(-1)^{2}+32(-1)+20 \\ &=-1+13-32+20 \\ &=(13-1)+(20-32) \\ &=12-12 \\ &=0 \end{aligned} \]
    Since \(p(-1)=0\), \((x+1)\) is a factor of \(p(x)\).
  6. Try \(x=-2\):
  7. \[\small \begin{aligned} p(-2) &=(-2)^{3}+13(-2)^{2}+32(-2)+20 \\ &=-8+52-64+20 \\ &=(52-8)+(20-64) \\ &=44-44 \\ &=0 \end{aligned} \]
    Since \(p(-2)=0\), \((x+2)\) is a factor of \(p(x)\).
  8. Try \(x=-4\):
  9. \[\small \begin{aligned} p(-4) &=(-4)^{3}+13(-4)^{2}+32(-4)+20 \\ &=-64+208-128+20 \\ &=(208-64)+(20-128) \\ &=144-108 \\ &=36 \end{aligned} \]
    Since \(p(-4)\neq0\), \((x+4)\) is not a factor of \(p(x)\).
  10. Try \(x=-5\):
  11. \[\small \begin{aligned} p(-5) &=(-5)^{3}+13(-5)^{2}+32(-5)+20 \\ &=-125+325-160+20 \\ &=(325-125)+(20-160) \\ &=200-140 \\ &=60 \end{aligned} \]
    Since \(p(-5)\neq0\), \((x+5)\) is not a factor of \(p(x)\).
  12. Try \(x=-10\):
  13. \[\small \begin{aligned} p(-10) &=(-10)^{3}+13(-10)^{2}+32(-10)+20 \\ &=-1000+1300-320+20 \\ &=(1300-1000)+(20-320) \\ &=300-300 \\ &=0 \end{aligned} \]
    Since \(p(-10)=0\), \((x+10)\) is a factor of \(p(x)\).
  14. Try \(x=-20\):
  15. \[\small \begin{aligned} p(-20) &=(-20)^{3}+13(-20)^{2}+32(-20)+20 \\ &=-8000+5200-640+20 \\ &=(5200-8000)+(20-640) \\ &=-2800-620 \\ &=-3420 \end{aligned} \]
    Since \(p(-20)\neq0\), \((x+20)\) is not a factor of \(p(x)\).
💡 Answer Final Answer ›
Answer: (iii) \(p(x)=(x+1)(x+2)(x+10)\)
✏️ Solution Complete Solution ›
Step-by-step Solution  ·  9 steps
  1. (iv) Given: \[\small p(y)=2y^{3}+y^{2}-2y-1\]
  2. Try grouping:
  3. \[\small \begin{aligned} 2y^{3}+y^{2}-2y-1 &=(2y^{3}+y^{2})- (2y+1) \\ &=y^{2}(2y+1)-1(2y+1) \\ &=(2y+1)(y^{2}-1) \end{aligned} \]
  4. Now use identity:
  5. \[\small a^{2}-b^{2}=(a-b)(a+b) \]
  6. Therefore
  7. \[\small \begin{aligned} y^{2}-1 &=(y-1)(y+1) \end{aligned} \]
  8. Final factorisation:
  9. \[\small \begin{aligned} 2y^{3}+y^{2}-2y-1 &=(2y+1)(y-1)(y+1) \end{aligned} \]
💡 Answer Final Answer ›
Answer: (iv) \(p(y)=(2y+1)(y-1)(y+1)\)
🎯 Exam Significance Exam Significance ›
  • Cubic polynomial factorisation is an important topic in algebra.
  • Questions from this concept are frequently asked in CBSE board examinations and school tests.
  • Factor Theorem is also important for Olympiads, NTSE, SSC, Polytechnic and other entrance examinations.
  • It helps in solving polynomial equations efficiently.
  • This topic develops logical thinking and algebraic manipulation skills.
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5 / 5  ·  100%
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NCERT Class 9 Maths Exercise 2.3 Step-by-step Solutions
NCERT Class 9 Maths Exercise 2.3 Step-by-step Solutions — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 9 Maths Polynomials Exercise 2.3 solutions focus on factorisation methods that expand upon the concepts from previous exercises. This exercise explains the Factor Theorem with examples and shows how to apply it in solving polynomial equations. Students also learn factorisation by the Factor Theorem and the split middle term method of factorisation for quadratic expressions. With clear explanations, solved examples, and step-by-step solutions, this exercise helps students master…
🎓 Class 9 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Polynomials — Learning Resources

    📄 Detailed Notes
    🧠 Practice MCQs
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