CBSE Class 9 Maths Chapter 2 Polynomials NCERT Solutions

To determine if an expression is a polynomial in one variable, we check two conditions: (1) it must contain only one variable, and (2) the exponents of that variable must be non-negative integers (0, 1, 2, ...).

(i) $4x^2\; -\; 3x\; +\; 7$: This expression contains only the variable 'x'. The exponents of x are 2, 1, and 0 (for the constant term 7). All exponents are non-negative integers. Therefore, it is a polynomial in one variable, x.

(ii) $y^2\; +\; \backslash sqrt\{2\}$: This expression contains only the variable 'y'. The exponents of y are 2 and 0. All exponents are non-negative integers. Therefore, it is a polynomial in one variable, y.

(iii) $3\backslash sqrt\{t\}\; +\; t\backslash sqrt\{2\}$: This expression contains the variable 't'. However, the term $3\backslash sqrt\{t\}$ can be written as $3t^\{1/2\}$. Since the exponent 1/2 is not a whole number, this expression is not a polynomial.

(iv) $y\; +\; \backslash frac\{2\}\{y\}$: This expression contains the variable 'y'. The term $\backslash frac\{2\}\{y\}$ can be written as $2y^\{-1\}$. Since the exponent -1 is a negative integer, this expression is not a polynomial.

(v) $x^\{10\}\; +\; y^3\; +\; t^\{50\}$: This expression contains three different variables: x, y, and t. Therefore, it is not a polynomial in *one* variable, although it is a polynomial in three variables.

The coefficient of a term is the numerical factor that multiplies the variable part of the term. We need to identify the coefficient of the $x^2$ term in each given polynomial.

(i) In the polynomial $2\; +\; x^2\; +\; x$, the term containing $x^2$ is $x^2$. This can be written as $1\; \backslash cdot\; x^2$. Thus, the coefficient of $x^2$ is 1.

(ii) In the polynomial $2\; -\; x^2\; +\; x^3$, the term containing $x^2$ is $-x^2$. This can be written as $-1\; \backslash cdot\; x^2$. Thus, the coefficient of $x^2$ is -1.

(iii) In the polynomial $\backslash frac\{\backslash pi\}\{2\}x^2\; +\; x$, the term containing $x^2$ is $\backslash frac\{\backslash pi\}\{2\}x^2$. Thus, the coefficient of $x^2$ is $\backslash frac\{\backslash pi\}\{2\}$.

(iv) In the polynomial $\backslash sqrt\{2\}x\; -\; 1$, there is no term containing $x^2$. This means the coefficient of $x^2$ is 0. We can think of the polynomial as $0\; \backslash cdot\; x^2\; +\; \backslash sqrt\{2\}x\; -\; 1$.

A binomial is a polynomial with two terms. The degree of a polynomial is the highest power of the variable. A monomial is a polynomial with one term.

Example of a binomial of degree 35:

A binomial has two terms. To have a degree of 35, the highest power of the variable must be 35. An example is x^{35} + 7. Here, the terms are x^{35} and 7, and the highest power is 35.

Example of a monomial of degree 100:

A monomial has one term. To have a degree of 100, the power of the variable must be 100. An example is y^{100}. Another example could be 5z^{100}.

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

(i) In the polynomial $5x^3\; +\; 4x^2\; +\; 7x$, the powers of x are 3, 2, and 1. The highest power is 3. Therefore, the degree of this polynomial is 3.

(ii) In the polynomial $4\; -\; y^2$, the powers of y are 0 (for the constant term 4, which is $4y^0$) and 2. The highest power is 2. Therefore, the degree of this polynomial is 2.

(iii) In the polynomial $5t\; -\; \backslash sqrt\{7\}$, the powers of t are 1 and 0 (for the constant term $-\backslash sqrt\{7\}$). The highest power is 1. Therefore, the degree of this polynomial is 1.

(iv) The polynomial is 3. This is a constant polynomial. A constant non-zero polynomial can be written as $3x^0$ (or $3y^0$, etc.). The power of the variable is 0. Therefore, the degree of this polynomial is 0.

Polynomials are classified based on their degree:

• A polynomial with degree 1 is called a linear polynomial.
• A polynomial with degree 2 is called a quadratic polynomial.
• A polynomial with degree 3 is called a cubic polynomial.

Let's classify each polynomial:

(i) $x^2\; +\; x$: The highest power of x is 2. So, it is a quadratic polynomial.

(ii) $x\; -\; x^3$: The highest power of x is 3. So, it is a cubic polynomial.

(iii) $y\; +\; y^2\; +\; 4$: The highest power of y is 2. So, it is a quadratic polynomial.

(iv) $1\; +\; x$: The highest power of x is 1. So, it is a linear polynomial.

(v) 3t: The highest power of t is 1. So, it is a linear polynomial.

(vi) $r^2$: The highest power of r is 2. So, it is a quadratic polynomial.

(vii) $7x^3$: The highest power of x is 3. So, it is a cubic polynomial.

Let the given polynomial be $p(x)\; =\; 5x\; -\; 4x^2\; +\; 3$. We need to find the value of $p(x)$ for the given values of x.

(i) To find the value at $x\; =\; 0$, substitute 0 for x in the polynomial:

$p(0)\; =\; 5(0)\; -\; 4(0)^2\; +\; 3$

$p(0)\; =\; 0\; -\; 4(0)\; +\; 3$

$p(0)\; =\; 0\; -\; 0\; +\; 3$

$p(0)\; =\; 3$

The value of the polynomial at $x\; =\; 0$ is 3.

(ii) To find the value at $x\; =\; -1$, substitute -1 for x in the polynomial:

$p(-1)\; =\; 5(-1)\; -\; 4(-1)^2\; +\; 3$

$p(-1)\; =\; -5\; -\; 4(1)\; +\; 3$

$p(-1)\; =\; -5\; -\; 4\; +\; 3$

$p(-1)\; =\; -9\; +\; 3$

$p(-1)\; =\; -6$

The value of the polynomial at $x\; =\; -1$ is -6.

(iii) To find the value at $x\; =\; 2$, substitute 2 for x in the polynomial:

$p(2)\; =\; 5(2)\; -\; 4(2)^2\; +\; 3$

$p(2)\; =\; 10\; -\; 4(4)\; +\; 3$

$p(2)\; =\; 10\; -\; 16\; +\; 3$

$p(2)\; =\; -6\; +\; 3$

$p(2)\; =\; -3$

The value of the polynomial at $x\; =\; 2$ is -3.

(i) For the polynomial $p(y)\; =\; y^2\; -\; y\; +\; 1$:

At $y\; =\; 0$: $p(0)\; =\; (0)^2\; -\; (0)\; +\; 1\; =\; 0\; -\; 0\; +\; 1\; =\; 1$

At $y\; =\; 1$: $p(1)\; =\; (1)^2\; -\; (1)\; +\; 1\; =\; 1\; -\; 1\; +\; 1\; =\; 1$

At $y\; =\; 2$: $p(2)\; =\; (2)^2\; -\; (2)\; +\; 1\; =\; 4\; -\; 2\; +\; 1\; =\; 3$

(ii) For the polynomial $p(t)\; =\; 2\; +\; t\; +\; 2t^2\; -\; t^3$:

At $t\; =\; 0$: $p(0)\; =\; 2\; +\; (0)\; +\; 2(0)^2\; -\; (0)^3\; =\; 2\; +\; 0\; +\; 0\; -\; 0\; =\; 2$

At $t\; =\; 1$: $p(1)\; =\; 2\; +\; (1)\; +\; 2(1)^2\; -\; (1)^3\; =\; 2\; +\; 1\; +\; 2(1)\; -\; 1\; =\; 2\; +\; 1\; +\; 2\; -\; 1\; =\; 4$

At $t\; =\; 2$: $p(2)\; =\; 2\; +\; (2)\; +\; 2(2)^2\; -\; (2)^3\; =\; 2\; +\; 2\; +\; 2(4)\; -\; 8\; =\; 4\; +\; 8\; -\; 8\; =\; 4$

(iii) For the polynomial $p(x)\; =\; x^3$:

At $x\; =\; 0$: $p(0)\; =\; (0)^3\; =\; 0$

At $x\; =\; 1$: $p(1)\; =\; (1)^3\; =\; 1$

At $x\; =\; 2$: $p(2)\; =\; (2)^3\; =\; 8$

(iv) For the polynomial $p(x)\; =\; (x\; -\; 1)(x\; +\; 1)$:

At $x\; =\; 0$: $p(0)\; =\; (0\; -\; 1)(0\; +\; 1)\; =\; (-1)(1)\; =\; -1$

At $x\; =\; 1$: $p(1)\; =\; (1\; -\; 1)(1\; +\; 1)\; =\; (0)(2)\; =\; 0$

At $x\; =\; 2$: $p(2)\; =\; (2\; -\; 1)(2\; +\; 1)\; =\; (1)(3)\; =\; 3$

To verify if a given value is a zero of a polynomial, we substitute that value into the polynomial. If the result is 0, then the value is a zero of the polynomial.

(i) For p(x) = 3x + 1, check x = -\frac{1}{3}:

p(-\frac{1}{3}) = 3(-\frac{1}{3}) + 1 = -1 + 1 = 0. Since p(-\frac{1}{3}) = 0, x = -\frac{1}{3} is a zero.

(ii) For p(x) = 5x - \pi, check x = \frac{4}{5}:

p(\frac{4}{5}) = 5(\frac{4}{5}) - \pi = 4 - \pi. Since 4 - \pi

eq 0, x = \frac{4}{5} is not a zero.

(iii) For p(x) = x^2 - 1, check x = 1 and x = -1:

For x = 1: p(1) = (1)^2 - 1 = 1 - 1 = 0. So, x = 1 is a zero.

For x = -1: p(-1) = (-1)^2 - 1 = 1 - 1 = 0. So, x = -1 is a zero.

(iv) For p(x) = (x + 1)(x - 2), check x = -1 and x = 2:

For x = -1: p(-1) = (-1 + 1)(-1 - 2) = (0)(-3) = 0. So, x = -1 is a zero.

For x = 2: p(2) = (2 + 1)(2 - 2) = (3)(0) = 0. So, x = 2 is a zero.

(v) For p(x) = x^2, check x = 0:

p(0) = (0)^2 = 0. Since p(0) = 0, x = 0 is a zero.

(vi) For p(x) = lx + m, check x = -\frac{m}{l}:

p(-\frac{m}{l}) = l(-\frac{m}{l}) + m = -m + m = 0. Since p(-\frac{m}{l}) = 0, x = -\frac{m}{l} is a zero.

(vii) For p(x) = 3x^2 - 1, check x = -\frac{1}{\sqrt{3}} and x = \frac{2}{\sqrt{3}}:

For x = -\frac{1}{\sqrt{3}}: p(-\frac{1}{\sqrt{3}}) = 3(-\frac{1}{\sqrt{3}})^2 - 1 = 3(\frac{1}{3}) - 1 = 1 - 1 = 0. So, x = -\frac{1}{\sqrt{3}} is a zero.

For x = \frac{2}{\sqrt{3}}: p(\frac{2}{\sqrt{3}}) = 3(\frac{2}{\sqrt{3}})^2 - 1 = 3(\frac{4}{3}) - 1 = 4 - 1 = 3. Since p(\frac{2}{\sqrt{3}})

eq 0, x = \frac{2}{\sqrt{3}} is not a zero.

(viii) For p(x) = 2x + 1, check x = \frac{1}{2}:

p(\frac{1}{2}) = 2(\frac{1}{2}) + 1 = 1 + 1 = 2. Since p(\frac{1}{2})

eq 0, x = \frac{1}{2} is not a zero.