CBSE Class 9 Maths: Polynomials Exercise 2.3 - Remainder and Factor Theorems

Understanding the Remainder Theorem

The Remainder Theorem is a fundamental concept in algebra that provides an efficient way to find the remainder of a polynomial division. Instead of going through the lengthy process of polynomial long division, this theorem offers a direct method when the divisor is a linear polynomial.

Statement of the Remainder Theorem:
Let \(p(x)\) be any polynomial of degree greater than or equal to one, and let \(a\) be any real number. If \(p(x)\) is divided by the linear polynomial \((x - a)\), then the remainder is \(p(a)\).

In simpler terms, if you want to find the remainder when a polynomial \(p(x)\) is divided by \((x - a)\), all you need to do is substitute \(x = a\) into the polynomial \(p(x)\) and calculate the value. The result will be your remainder. For example, if you divide \(p(x) = x^2 + 2x + 1\) by \((x - 1)\), the zero of \((x-1)\) is \(x=1\). So, the remainder is \(p(1) = (1)^2 + 2(1) + 1 = 1 + 2 + 1 = 4\).

This theorem is incredibly useful for quickly checking remainders and is a precursor to the Factor Theorem, which helps in identifying factors of polynomials.

Step-by-Step: Applying the Remainder Theorem

1. Identify the Polynomial and Divisor — Clearly state the given polynomial, \(p(x)\), and the linear divisor, such as \((x - a)\) or \((x + a)\).
2. Find the Zero of the Divisor — Set the linear divisor equal to zero and solve for \(x\). This value of \(x\) will be the 'a' that you need to substitute into \(p(x)\). For example, if the divisor is \((x - 2)\), then \(x - 2 = 0 \implies x = 2\). If the divisor is \((x + 3)\), then \(x + 3 = 0 \implies x = -3\). If the divisor is \((2x - 1)\), then \(2x - 1 = 0 \implies x = 1/2\).
3. Substitute the Zero into the Polynomial — Substitute the value of \(x\) found in the previous step (let's call it 'a') into the polynomial \(p(x)\) to find \(p(a)\). Be careful with signs and powers during substitution.
4. State the Remainder — The numerical value obtained after substitution, \(p(a)\), is the remainder when \(p(x)\) is divided by the given linear divisor.

The Factor Theorem: A Special Case of Remainder Theorem

The Factor Theorem is a direct consequence and a special application of the Remainder Theorem. It provides a quick and efficient way to determine if a linear polynomial is a factor of a given polynomial.

Statement of the Factor Theorem:

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

Essentially, the Factor Theorem tells us that a linear expression \((x - a)\) is a factor of a polynomial \(p(x)\) if and only if dividing \(p(x)\) by \((x - a)\) leaves a remainder of zero. This means that \(a\) is a root or a zero of the polynomial \(p(x)\).

This theorem is incredibly powerful, especially when it comes to factorizing higher-degree polynomials. For instance, to factorize a cubic polynomial, you can use the Factor Theorem to find one linear factor by testing simple values like \(x = \pm 1, \pm 2\), etc. Once one factor is found (i.e., \(p(a) = 0\)), you can then use polynomial long division or synthetic division to divide \(p(x)\) by \((x - a)\) to obtain a quadratic polynomial, which can then be factorized further. This systematic approach is key to solving many problems in polynomial ex 2 3 class 9 ncert.

Common Mistakes and Exam Tips for Polynomial Ex 2.3

Here are some crucial tips to help you avoid common pitfalls and excel in questions related to Remainder and Factor Theorems:

1. Incorrect Zero of the Divisor: A very common mistake is finding the wrong 'a' value. If the divisor is \((x + a)\), the zero is \(-a\), not \(a\). If the divisor is \((ax - b)\), the zero is \(b/a\). Always set the divisor to zero and solve for \(x\) to find the correct value for substitution.
2. Sign Errors During Substitution: When substituting negative values into a polynomial, be extremely careful with signs. Remember that \((-x)^2 = x^2\) but \((-x)^3 = -x^3\). A single sign error can lead to an incorrect remainder or factor determination.
3. Misinterpreting the Result: For the Remainder Theorem, the result \(p(a)\) IS the remainder. For the Factor Theorem, only if \(p(a) = 0\) is \((x - a)\) a factor. A non-zero result means it's not a factor.
4. Practice Mental Calculation: While you should always write down steps, try to mentally estimate the result, especially for simple substitutions, to catch obvious errors.
5. Understanding the "If and Only If": The Factor Theorem works both ways. If \(p(a) = 0\), then \((x-a)\) is a factor. And if \((x-a)\) is a factor, then \(p(a) = 0\). Understand this bidirectional relationship.
6. For Factorisation of Cubics: If asked to factorize a cubic polynomial, start by using the Factor Theorem to find one factor. Test simple integer values like \(x = \pm 1, \pm 2\). Once you find an 'a' such that \(p(a) = 0\), you know \((x-a)\) is a factor. Then, use polynomial long division to divide \(p(x)\) by \((x-a)\) to get a quadratic, which you can factorize by splitting the middle term.

Practice Questions with Solutions

• Q: Find the remainder when \(x^3 + 3x^2 + 3x + 1\) is divided by \((x + 1)\). A: Step 1: Identify the polynomial \(p(x) = x^3 + 3x^2 + 3x + 1\) and the divisor \((x + 1)\). Step 2: Find the zero of the divisor. Set \(x + 1 = 0\), which gives \(x = -1\). Step 3: Substitute \(x = -1\) into \(p(x)\). \(p(-1) = (-1)^3 + 3(-1)^2 + 3(-1) + 1\) \(p(-1) = -1 + 3(1) - 3 + 1\) \(p(-1) = -1 + 3 - 3 + 1\) \(p(-1) = 0\) Final answer: The remainder is 0.
• Q: Check whether \((x - 2)\) is a factor of \(p(x) = x^3 - 3x^2 + 4x - 4\). A: Step 1: Identify the polynomial \(p(x) = x^3 - 3x^2 + 4x - 4\) and the potential factor \((x - 2)\). Step 2: Find the zero of the potential factor. Set \(x - 2 = 0\), which gives \(x = 2\). Step 3: Substitute \(x = 2\) into \(p(x)\). \(p(2) = (2)^3 - 3(2)^2 + 4(2) - 4\) \(p(2) = 8 - 3(4) + 8 - 4\) \(p(2) = 8 - 12 + 8 - 4\) \(p(2) = 16 - 16\) \(p(2) = 0\) Final answer: Since \(p(2) = 0\), by the Factor Theorem, \((x - 2)\) is a factor of \(p(x)\).
• Q: Find the value of \(k\) if \((x - 1)\) is a factor of \(4x^3 + 3x^2 - 4x + k\). A: Step 1: Identify the polynomial \(p(x) = 4x^3 + 3x^2 - 4x + k\) and the factor \((x - 1)\). Step 2: Since \((x - 1)\) is a factor, by the Factor Theorem, \(p(1)\) must be equal to 0. Step 3: Find the zero of the factor: \(x - 1 = 0 \implies x = 1\). Step 4: Substitute \(x = 1\) into \(p(x)\) and set the result to 0. \(p(1) = 4(1)^3 + 3(1)^2 - 4(1) + k = 0\) \(4(1) + 3(1) - 4 + k = 0\) \(4 + 3 - 4 + k = 0\) \(3 + k = 0\) \(k = -3\) Final answer: The value of \(k\) is -3.
• Q: What is the remainder when \(x^4 - x^2 + 2\) is divided by \((x - 1)\)? A: Step 1: Identify \(p(x) = x^4 - x^2 + 2\) and the divisor \((x - 1)\). Step 2: Find the zero of the divisor: \(x - 1 = 0 \implies x = 1\). Step 3: Substitute \(x = 1\) into \(p(x)\). \(p(1) = (1)^4 - (1)^2 + 2\) \(p(1) = 1 - 1 + 2\) \(p(1) = 2\) Final answer: The remainder is 2.
• Q: Is \((2x + 1)\) a factor of \(4x^3 + 4x^2 - x - 1\)? A: Step 1: Identify \(p(x) = 4x^3 + 4x^2 - x - 1\) and the potential factor \((2x + 1)\). Step 2: Find the zero of the potential factor: \(2x + 1 = 0 \implies 2x = -1 \implies x = -1/2\). Step 3: Substitute \(x = -1/2\) into \(p(x)\). \(p(-1/2) = 4(-1/2)^3 + 4(-1/2)^2 - (-1/2) - 1\) \(p(-1/2) = 4(-1/8) + 4(1/4) + 1/2 - 1\) \(p(-1/2) = -1/2 + 1 + 1/2 - 1\) \(p(-1/2) = 0\) Final answer: Since \(p(-1/2) = 0\), by the Factor Theorem, \((2x + 1)\) is a factor of \(4x^3 + 4x^2 - x - 1\).

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