Polynomial Exercise 2.4 Class 9 NCERT: Factor Theorem & Factorization

Welcome to YoLearn AI's deep-dive guide to NCERT Class 9 Mathematics Chapter 2, Exercise 2.4. This exercise is the cornerstone of polynomial factorization, focusing heavily on the Factor Theorem, splitting the middle term for quadratic polynomials, and factorizing cubic polynomials. Mastering the topic polynomial ex 2 4 class 9 ncert is critical because these algebraic skills are used constantly throughout high school math and science. In this guide, our AI tutor will break down the Core Factor Theorem, walk you through the trial-and-error method for cubic polynomials, point out tricky areas where students often lose marks, and provide challenging practice questions with step-by-step solutions. Ready to secure full marks? Let's start with the core concepts!

Understanding the Factor Theorem

The Factor Theorem is a crucial deduction from the Remainder Theorem. It states that for any polynomial p(x) of degree n >= 1, where a is any real number:

  1. If p(a) = 0, then (x - a) is a factor of p(x).
  2. Conversely, if (x - a) is a factor of p(x), then p(a) = 0.

This theorem gives us a highly efficient algebraic tool: instead of performing long polynomial division to check if a linear binomial divides a polynomial completely, we can simply evaluate the polynomial at x = a. If the result is zero, the division has a remainder of zero, confirming that (x - a) is indeed a factor. This simplifies factorization immensely, especially when finding unknown coefficients (like k) or when trying to factorize complex cubics.

How to Factorize Cubic Polynomials Step-by-Step

  1. Step 1: Find one factor by trial — Look at the constant term of the cubic polynomial p(x). Find its factors. Substitute these factor values (like +1, -1, +2, -2) into the polynomial until you find a value 'a' such that p(a) = 0. By the Factor Theorem, (x - a) is a factor.
  2. Step 2: Perform polynomial division — Divide the cubic polynomial p(x) by the linear factor (x - a) using polynomial long division to find the quadratic quotient q(x).
  3. Step 3: Factorize the quadratic quotient — Take the quadratic quotient q(x) and factorize it completely using the method of splitting the middle term.
  4. Step 4: Combine all factors — Express the original cubic polynomial as the product of the three linear factors: p(x) = (x - a) (factor_1) (factor_2).

Step-by-Step Solved NCERT-Style Problems

  • Example 1: Determine whether (x + 1) is a factor of x³ + x² + x + 1. Step 1: Let p(x) = x³ + x² + x + 1. To test if (x + 1) is a factor, we find the zero of x + 1, which is x = -1. Step 2: Substitute x = -1 into p(x): p(-1) = (-1)³ + (-1)² + (-1) + 1 = -1 + 1 - 1 + 1 = 0. Step 3: Since p(-1) = 0, by the Factor Theorem, (x + 1) is indeed a factor.
  • Example 2: Find the value of k, if x - 1 is a factor of p(x) = 2x² + kx + √2. Step 1: Since x - 1 is a factor of p(x), by the Factor Theorem, p(1) = 0. Step 2: Substitute x = 1 into p(x) and set it to 0: 2(1)² + k(1) + √2 = 0 2 + k + √2 = 0 Step 3: Solve for k: k = -(2 + √2).

Avoid These Common Exam Mistakes

  1. The Sign Flip Trap: When checking if (x + a) is a factor, remember to substitute x = -a into the polynomial. For example, if checking (x + 3), compute p(-3), not p(3).
  1. Incomplete Cubic Factorization: Many students find the first factor (e.g., x - 1) and stop there. A cubic polynomial of degree 3 can have up to 3 linear factors. You must perform division and factorize the remaining quadratic quotient completely.
  1. Splitting the Middle Term Sign Errors: When splitting bx into px + qx, ensure that both the sum p + q = b and the product p q = a c match perfectly, including their positive or negative signs.

Practice Questions with Solutions

  • Q: Examine whether x + 2 is a factor of x³ + 3x² + 5x + 6. A: Step 1: Identify the divisor. The divisor is x + 2, so its zero is x = -2. Step 2: Let p(x) = x³ + 3x² + 5x + 6. We need to compute p(-2). Step 3: Substitute x = -2: p(-2) = (-2)³ + 3(-2)² + 5(-2) + 6 p(-2) = -8 + 3(4) - 10 + 6 p(-2) = -8 + 12 - 10 + 6 = 0. Step 4: Since p(-2) = 0, by the Factor Theorem, x + 2 is a factor of the given polynomial. Final answer: Yes, x + 2 is a factor.
  • Q: Find the value of k if x - 1 is a factor of 4x³ + 3x² - 4x + k. A: Step 1: Since x - 1 is a factor of p(x) = 4x³ + 3x² - 4x + k, we know by the Factor Theorem that p(1) = 0. Step 2: Substitute x = 1 into p(x): 4(1)³ + 3(1)² - 4(1) + k = 0 4 + 3 - 4 + k = 0 Step 3: Simplify the equation: 3 + k = 0 k = -3. Final answer: k = -3.
  • Q: Factorize the quadratic polynomial: 6x² + 5x - 6 by splitting the middle term. A: Step 1: Compare the given expression 6x² + 5x - 6 with ax² + bx + c. Here, a = 6, b = 5, and c = -6. Step 2: We need two numbers whose product is a c = 6 (-6) = -36 and whose sum is b = 5. These numbers are 9 and -4. Step 3: Rewrite the middle term and factor by grouping: 6x² + 9x - 4x - 6 = 3x(2x + 3) - 2(2x + 3) = (2x + 3)(3x - 2). Final answer: (2x + 3)(3x - 2).
  • Q: Factorize the cubic polynomial: x³ - 23x² + 142x - 120. A: Step 1: Let p(x) = x³ - 23x² + 142x - 120. Find factors of the constant term 120. Let's test x = 1: p(1) = 1³ - 23(1)² + 142(1) - 120 = 1 - 23 + 142 - 120 = 0. Since p(1) = 0, (x - 1) is a factor. Step 2: Divide p(x) by (x - 1) using long division to get the quadratic quotient: x² - 22x + 120. Step 3: Factorize x² - 22x + 120 by splitting the middle term. We need two numbers whose product is 120 and sum is -22. These numbers are -12 and -10. x² - 12x - 10x + 120 = x(x - 12) - 10(x - 12) = (x - 12)(x - 10). Step 4: Combine all factors: p(x) = (x - 1)(x - 10)(x - 12). Final answer: (x - 1)(x - 10)(x - 12).

Frequently Asked Questions

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

The Remainder Theorem calculates the remainder p(a) when a polynomial is divided by (x - a). The Factor Theorem is a special case where if that remainder p(a) = 0, we conclude that (x - a) is a perfect factor.

How do I choose trial values when factorizing a cubic polynomial?

Look at the constant term of the cubic polynomial. Find its integer factors and substitute them starting from the simplest ones, like +1, -1, +2, or -2, until you get a polynomial value of zero.

Can a quadratic polynomial have three factors?

No, a polynomial can only have as many linear factors as its degree. Since a quadratic polynomial is of degree 2, it can have at most 2 linear factors.