Polynomial Class 9 NCERT Study Guide

Welcome, Class 9 champions! The chapter on polynomials is one of the most critical foundational blocks in CBSE Class 9 Mathematics. In this guide on polynomial class 9 ncert, we will break down algebraic expressions into easy-to-digest concepts. You will master how to identify polynomials, determine their degrees, find their zeroes, and apply powerful algebraic techniques like the Remainder Theorem and Factor Theorem. We will also dive deep into algebraic identities that will help you solve complex simplifications instantly. Whether you are preparing for your class tests or aiming for a perfect score in your school board exams, mastering this chapter is essential as it lays the groundwork for Class 10 Quadratic Equations and beyond. Let's pick up our virtual sketchpad, follow the worked steps patiently, and make polynomials your strongest topic!

What is a Polynomial?

An algebraic expression is called a polynomial if the exponents of the variables in all terms are non-negative integers (i.e., whole numbers like 0, 1, 2, ...). For example, $x^2 + 5x + 6$ is a polynomial because the powers of $x$ (2 and 1) are non-negative integers. Conversely, expressions like $x^{1/2} + 3$ or $1/x$ are not polynomials because their powers are fractions or negative numbers.

Key Classifications:

  1. Based on Terms:
  • Monomial: Consists of only one term (e.g., $5x$, $3$).
  • Binomial: Consists of exactly two terms (e.g., $x^2 - 4$, $2y + 1$).
  • Trinomial: Consists of exactly three terms (e.g., $x^2 + 3x + 2$).
  1. Based on Degree:

The highest power of the variable in a polynomial is called its degree.

  • Linear Polynomial: Degree is 1 (e.g., $ax + b$).
  • Quadratic Polynomial: Degree is 2 (e.g., $ax^2 + bx + c$).
  • Cubic Polynomial: Degree is 3 (e.g., $ax^3 + bx^2 + cx + d$).
  • Constant Polynomial: Degree is 0 (e.g., $5$ can be written as $5x^0$).
  • Zero Polynomial: The constant number 0. Its degree is not defined.

Understanding the Remainder and Factor Theorems

  1. Step 1: Finding Zeroes of a Polynomial — A real number '$c
    is a zero of a polynomial $p(x)$ if and only if $p(c) = 0$. To find the zero of a linear polynomial like $ax + b$, equate it to zero: $ax + b = 0 \implies x = -b/a$.
  2. Step 2: Applying the Remainder Theorem — If a polynomial $p(x)$ of degree greater than or equal to 1 is divided by a linear polynomial $(x - a)$, then the remainder of this division is exactly $p(a)$. You do not need to perform long division to find the remainder!
  3. Step 3: Utilizing the Factor Theorem — The Factor Theorem is a special case of the Remainder Theorem. It states that if $p(x)$ is a polynomial of degree $n \ge 1$ and '$a
    is any real number, then: 1. $(x - a)$ is a factor of $p(x)$ if $p(a) = 0$. 2. Conversely, if $(x - a)$ is a factor of $p(x)$, then $p(a) = 0$.

Worked Examples of Factorisation

  • Example 1: Factorising a Quadratic Polynomial ($6x^2 + 17x + 5$) by splitting the middle term. Step 1: Identify $a=6, b=17, c=5$. We need to find two numbers whose sum is $b = 17$ and whose product is $a \times c = 6 \times 5 = 30$. Step 2: The numbers are $15$ and $2$ because $15 + 2 = 17$ and $15 \times 2 = 30$. Step 3: Rewrite the middle term: $6x^2 + 15x + 2x + 5$. Step 4: Group and factor out common terms: $3x(2x + 5) + 1(2x + 5) = (3x + 1)(2x + 5)$. * Final Answer: $(3x + 1)(2x + 5)$
  • Example 2: Verify if $(x + 2)$ is a factor of $p(x) = x^3 + 3x^2 + 5x + 6$ using the Factor Theorem. Step 1: To check if $(x + 2)$ is a factor, we equate $x + 2 = 0 \implies x = -2$. Step 2: Substitute $x = -2$ in $p(x)$: $p(-2) = (-2)^3 + 3(-2)^2 + 5(-2) + 6$ Step 3: Calculate the value: $p(-2) = -8 + 3(4) - 10 + 6 = -8 + 12 - 10 + 6 = 0$. Conclusion: Since $p(-2) = 0$, by the Factor Theorem, $(x + 2)$ is indeed a factor of the polynomial.

Algebraic Identities & Board Exam Tips

To score full marks on cbse class 9 polynomial questions, you must memorise these 8 standard identities:

  1. $(x+y)^2 = x^2 + 2xy + y^2$
  2. $(x-y)^2 = x^2 - 2xy + y^2$
  3. $x^2 - y^2 = (x+y)(x-y)$
  4. $(x+a)(x+b) = x^2 + (a+b)x + ab$
  5. $(x+y+z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx$
  6. $(x+y)^3 = x^3 + y^3 + 3xy(x+y)$
  7. $(x-y)^3 = x^3 - y^3 - 3xy(x-y)$
  8. $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$

Special Case for Identity 8: If $x + y + z = 0$, then $x^3 + y^3 + z^3 = 3xyz$. This is a highly tested concept in exams!

Common Mistake to Avoid: Students often write $(x-y)^3$ as $x^3 - y^3 - 3x^2y - 3xy^2$. Remember the sign rules carefully: $(x-y)^3 = x^3 - y^3 - 3x^2y + 3xy^2$.

Practice Questions with Solutions

  • Q: Find the value of $k$ if $x - 1$ is a factor of $p(x) = 2x^2 + kx + \sqrt{2}$. A: Step 1: Since $(x - 1)$ is a factor, by the Factor Theorem, $p(1) = 0$. Step 2: Substitute $x = 1$ into $p(x)$: $p(1) = 2(1)^2 + k(1) + \sqrt{2} = 0$ $2 + k + \sqrt{2} = 0$ Step 3: Solve for $k$: $k = -(2 + \sqrt{2})$ Final answer: $k = -2 - \sqrt{2}$
  • Q: Factorise the cubic polynomial: $p(x) = x^3 - 23x^2 + 142x - 120$ using the Factor Theorem. A: Step 1: Look at the constant term $-120$. Factors of 120 include $\pm 1, \pm 2, \pm 3, \dots$ Step 2: Test $x = 1$: $p(1) = 1^3 - 23(1)^2 + 142(1) - 120 = 1 - 23 + 142 - 120 = 143 - 143 = 0$. Since $p(1) = 0$, $(x - 1)$ is a factor. Step 3: Divide $p(x)$ by $(x-1)$ using long division or synthetic division to get the quadratic quotient: $x^3 - 23x^2 + 142x - 120 = (x-1)(x^2 - 22x + 120)$. Step 4: Factorise the quadratic part $x^2 - 22x + 120$ by splitting the middle term: $x^2 - 12x - 10x + 120 = x(x - 12) - 10(x - 12) = (x-12)(x-10)$. Final answer: $(x-1)(x-10)(x-12)$
  • Q: Evaluate $(104)^3$ using a suitable algebraic identity. A: Step 1: Write $104$ as $(100 + 4)$. Step 2: Use the identity $(a + b)^3 = a^3 + b^3 + 3ab(a + b)$. Here, $a = 100$ and $b = 4$. Step 3: Substitute the values: $(100 + 4)^3 = (100)^3 + 4^3 + 3(100)(4)(100 + 4)$ $= 1000000 + 64 + 1200(104)$ $= 1000000 + 64 + 124800$ Step 4: Add the terms: $= 1124864$ Final answer: $1,124,864$
  • Q: If $x + y + z = 0$, show that $x^3 + y^3 + z^3 = 3xyz$. Also find the value of $(-12)^3 + (7)^3 + (5)^3$. A: Step 1: We know the identity: $x^3 + y^3 + z^3 - 3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$. Step 2: Substitute $x + y + z = 0$ in the identity: $x^3 + y^3 + z^3 - 3xyz = (0)(x^2+y^2+z^2-xy-yz-zx) = 0$. Therefore, $x^3 + y^3 + z^3 = 3xyz$. Step 3: Let $x = -12, y = 7, z = 5$. Check their sum: $x + y + z = -12 + 7 + 5 = 0$. Step 4: Since their sum is 0, we can apply the identity: $(-12)^3 + (7)^3 + (5)^3 = 3(-12)(7)(5) = 3 \times (-420) = -1260$. Final answer: $-1260$

Frequently Asked Questions

What is the degree of a non-zero constant polynomial?

The degree of a non-zero constant polynomial (like 5 or -7) is 0. This is because any non-zero constant can be written as constant times $x^0$, where the power of the variable is zero.

What is the difference between the Remainder Theorem and the Factor Theorem?

The Remainder Theorem calculates the remainder $p(a)$ when $p(x)$ is divided by $(x-a)$. The Factor Theorem is a special case of this theorem which states that if the remainder $p(a)$ is exactly 0, then $(x-a)$ is a factor of $p(x)$.

Is 1/x a polynomial?

No, $1/x$ is not a polynomial. When written in exponent form, $1/x$ is $x^{-1}$, and the exponent $-1$ is a negative integer. Polynomial exponents must always be non-negative integers.