CBSE Class 10 Maths: Polynomials Exercise 2.4 (NCERT)

Understanding the Division Algorithm for Polynomials

Just like we divide numbers, we can also divide polynomials! The Division Algorithm for Polynomials is a fundamental concept that states: Given any polynomial $p(x)$ and any non-zero polynomial $g(x)$, there exist polynomials $q(x)$ (quotient) and $r(x)$ (remainder) such that:

$p(x) = g(x) \times q(x) + r(x)$

where $r(x) = 0$ or degree of $r(x) <$ degree of $g(x)$.

This algorithm is incredibly useful. If the remainder $r(x)$ is $0$, it means that $g(x)$ is a factor of $p(x)$. This is particularly helpful when you need to find factors of a polynomial or determine if a given polynomial is a factor of another. For example, if you know one zero of a polynomial, you can use it to form a linear factor, divide the polynomial by that factor, and then find the remaining zeros from the quotient. This systematic approach simplifies the process of factorizing higher-degree polynomials and is a cornerstone of this exercise.

Key Concepts and Definitions

Zero of a Polynomial

A real number 'a' is a zero of a polynomial $p(x)$ if $p(a) = 0$. Graphically, these are the x-intercepts where the polynomial's graph crosses the x-axis.

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. $p(a) = 0$ if $(x-a)$ is a factor of $p(x)$.

Cubic Polynomial

A polynomial of degree 3, generally written as $ax^3 + bx^2 + cx + d$, where $a \neq 0$. A cubic polynomial can have at most three zeros.

Biquadratic Polynomial

A polynomial of degree 4, generally written as $ax^4 + bx^3 + cx^2 + dx + e$, where $a \neq 0$. A biquadratic polynomial can have at most four zeros.

Finding All Zeros Using the Division Algorithm

1. Problem Statement — Obtain all other zeros of $p(x) = 3x^4 + 6x^3 - 2x^2 - 10x - 5$, if two of its zeros are $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$.
2. Step 1: Form a quadratic factor from the given zeros — If $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$ are zeros, then $(x - \sqrt{\frac{5}{3}})$ and $(x + \sqrt{\frac{5}{3}})$ are factors. Their product will also be a factor: $(x - \sqrt{\frac{5}{3}})(x + \sqrt{\frac{5}{3}}) = x^2 - (\sqrt{\frac{5}{3}})^2 = x^2 - \frac{5}{3}$. To simplify, we can multiply by 3, making it $3x^2 - 5$. (Note: $k(x^2 - 5/3)$ is also a factor for any non-zero k. Choosing $k=3$ makes division easier).
3. Step 2: Divide the given polynomial by the quadratic factor — Perform polynomial long division of $p(x) = 3x^4 + 6x^3 - 2x^2 - 10x - 5$ by $g(x) = 3x^2 - 5$. `` x^2 + 2x + 1 _________________ 3x^2-5 | 3x^4 + 6x^3 - 2x^2 - 10x - 5 - (3x^4 - 5x^2) _________________ 6x^3 + 3x^2 - 10x - (6x^3 - 10x) _________________ 3x^2 - 5 - (3x^2 - 5) _________________ 0 `` The quotient is $q(x) = x^2 + 2x + 1$ and the remainder is $r(x) = 0$.
4. Step 3: Factorize the quotient to find the remaining zeros — Since the remainder is 0, $3x^2 - 5$ is indeed a factor, and the original polynomial can be written as: $p(x) = (3x^2 - 5)(x^2 + 2x + 1)$ Now, we need to find the zeros of the quotient $x^2 + 2x + 1$. This is a perfect square trinomial: $x^2 + 2x + 1 = (x+1)^2$. Setting $(x+1)^2 = 0$, we get $x+1 = 0$, which means $x = -1$. Since it's $(x+1)^2$, this zero has a multiplicity of 2, meaning $x=-1$ is a zero repeated twice.
5. Step 4: State all zeros of the polynomial — The given zeros were $\sqrt{\frac{5}{3}}$ and $-\sqrt{\frac{5}{3}}$. The zeros found from the quotient are $-1$ and $-1$. Therefore, all the zeros of the polynomial $3x^4 + 6x^3 - 2x^2 - 10x - 5$ are $\sqrt{\frac{5}{3}}$, $-\sqrt{\frac{5}{3}}$, $-1$, and $-1$.

Common Mistakes and Exam Tips

1. Sign Errors in Division: Polynomial long division requires careful handling of signs, especially during subtraction. A common mistake is forgetting to change the signs of all terms in the subtrahend. Double-check your arithmetic and signs at each step.
2. Incorrectly Forming the Factor: When given zeros 'a' and 'b', the corresponding factors are $(x-a)$ and $(x-b)$. The product is $(x-a)(x-b)$. Students sometimes incorrectly form the factor as $(x+a)(x+b)$ or make algebraic errors in expanding the product.
3. **Forgetting to Find All Zeros**: The question usually asks for all other zeros or all zeros. After dividing and finding the quotient, remember to factorize the quotient (which is usually a quadratic) to find the remaining zeros. Don't stop after finding the quotient.
4. Misinterpreting the Factor Theorem: Remember, $p(a)=0$ if and only if $(x-a)$ is a factor. This means if you substitute a value 'a' into $p(x)$ and get 0, you've found a zero, and $(x-a)$ can be used for division.
5. Dealing with Fractional Zeros: If zeros are fractions like $\frac{a}{b}$, the corresponding factor is $(bx-a)$. For example, if $x=2/3$ is a zero, then $3x-2$ is a factor.

Practice Questions with Solutions

• Q: Verify whether 2, -1, -3 are the zeros of the cubic polynomial $p(x) = x^3 + 2x^2 - 5x - 6$. A: Step 1: Check $p(2)$. $p(2) = (2)^3 + 2(2)^2 - 5(2) - 6 = 8 + 2(4) - 10 - 6 = 8 + 8 - 10 - 6 = 16 - 16 = 0$. Since $p(2)=0$, 2 is a zero. Step 2: Check $p(-1)$. $p(-1) = (-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2(1) + 5 - 6 = -1 + 2 + 5 - 6 = 7 - 7 = 0$. Since $p(-1)=0$, -1 is a zero. Step 3: Check $p(-3)$. $p(-3) = (-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 2(9) + 15 - 6 = -27 + 18 + 15 - 6 = 33 - 33 = 0$. Since $p(-3)=0$, -3 is a zero. Final answer: Yes, 2, -1, and -3 are the zeros of the given polynomial.
• Q: Find all the zeros of the polynomial $p(x) = x^3 - 3x^2 - x + 3$, if one of its zeros is 1. A: Step 1: Since 1 is a zero, $(x-1)$ is a factor of $p(x)$. Step 2: Divide $p(x)$ by $(x-1)$ using long division. `` x^2 - 2x - 3 _________________ x - 1 | x^3 - 3x^2 - x + 3 - (x^3 - x^2) _________________ -2x^2 - x - (-2x^2 + 2x) _________________ -3x + 3 - (-3x + 3) _________________ 0 `` The quotient is $x^2 - 2x - 3$. Step 3: Factorize the quotient $x^2 - 2x - 3$. $x^2 - 2x - 3 = x^2 - 3x + x - 3 = x(x-3) + 1(x-3) = (x-3)(x+1)$. Step 4: Set the factors to zero to find the remaining zeros. $x-3=0 \Rightarrow x=3$ $x+1=0 \Rightarrow x=-1$ Final answer: The zeros of the polynomial are 1, 3, and -1.
• Q: If the polynomial $x^4 - 6x^3 + 16x^2 - 25x + 10$ is divided by another polynomial $x^2 - 2x + k$, the remainder comes out to be $x + a$. Find $k$ and $a$. A: Step 1: Perform polynomial long division of $p(x) = x^4 - 6x^3 + 16x^2 - 25x + 10$ by $g(x) = x^2 - 2x + k$. `` x^2 - 4x + (8-k) _________________________ x^2-2x+k | x^4 - 6x^3 + 16x^2 - 25x + 10 - (x^4 - 2x^3 + kx^2) _________________________ -4x^3 + (16-k)x^2 - 25x - (-4x^3 + 8x^2 - 4kx) _________________________ (8-k)x^2 + (4k-25)x + 10 - ((8-k)x^2 - 2(8-k)x + k(8-k)) _________________________ (4k-25 + 16 - 2k)x + (10 - 8k + k^2) (2k-9)x + (k^2-8k+10) `` Step 2: Equate the remainder to the given remainder $x+a$. Comparing coefficients of $x$: $2k - 9 = 1 \Rightarrow 2k = 10 \Rightarrow k = 5$. Comparing constant terms: $k^2 - 8k + 10 = a$. Substitute $k=5$: $(5)^2 - 8(5) + 10 = a \Rightarrow 25 - 40 + 10 = a \Rightarrow -5 = a$. Final answer: $k=5$ and $a=-5$.
• Q: Find all the zeros of $2x^4 - 3x^3 - 3x^2 + 6x - 2$, if you know that two of its zeros are $\sqrt{2}$ and $-\sqrt{2}$. A: Step 1: Since $\sqrt{2}$ and $-\sqrt{2}$ are zeros, then $(x - \sqrt{2})$ and $(x + \sqrt{2})$ are factors. Their product is a factor: $(x - \sqrt{2})(x + \sqrt{2}) = x^2 - 2$. Step 2: Divide $p(x) = 2x^4 - 3x^3 - 3x^2 + 6x - 2$ by $g(x) = x^2 - 2$. `` 2x^2 - 3x + 1 _________________ x^2-2 | 2x^4 - 3x^3 - 3x^2 + 6x - 2 - (2x^4 - 4x^2) _________________ -3x^3 + x^2 + 6x - (-3x^3 + 6x) _________________ x^2 - 2 - (x^2 - 2) _________________ 0 `` The quotient is $2x^2 - 3x + 1$. Step 3: Factorize the quotient $2x^2 - 3x + 1$. $2x^2 - 3x + 1 = 2x^2 - 2x - x + 1 = 2x(x-1) - 1(x-1) = (2x-1)(x-1)$. Step 4: Set the factors to zero to find the remaining zeros. $2x-1=0 \Rightarrow x=1/2$ $x-1=0 \Rightarrow x=1$ Final answer: The zeros of the polynomial are $\sqrt{2}$, $-\sqrt{2}$, $1/2$, and 1.

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