**Question 1: **

**Which of the following expressions are polynomials?**

**(i) x**^{5 }**-2x**^{3}** + x+7**

It is a polynomial, Degree = 5.

(ii**) y**^{3 }**– \(\sqrt{3}y\)**

It is polynomial, Degree = 3.

**(iii) t**^{2}**–\(\frac{2}{5}t+\sqrt{2}\)**

It is polynomial, Degree = 2.

**(iv)** **\(5\sqrt{z}-6\)**

It is not a polynomial.

**(v) \(x-\frac{1}{x}\)**

It is not a polynomial.

**(vi)** **\(x^{108}-1\)**

It is polynomial, Degree = 108.

**(vii) \(\sqrt[3]{x}-27\)**

It is not a polynomial.

**(viii) \(\frac{1}{\sqrt{2}}x^{2}-\sqrt{2}x+2\)**

It is a polynomial, Degree = 2.

**(ix)** **\(x^{-2}+2x^{-1}+3\)**

It is not a polynomial.

**(x) 1**

It is a polynomial, Degree = 0.

**(xi) –\(\frac{3}{5}\)**

It is a polynomial, Degree = 0.

**(xii)** **\(\sqrt[3]{2}y^{2}-8\)**

It is a polynomial, Degree = 2.

**Question 2:**

**Write the degree of each of the following polynomials:**

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

**(i)2x-\(\sqrt{5}\)**

Degree of 2x – \(\sqrt{5}\) is 1.

**(ii) 3 – x + x**^{2}**-6x**^{3}

Degree of 3 – x + x^{2}– 6x^{3} is 3.

**(iii)9**

Degree of 9 is 0.

**(iv) 8x**^{4}**-36x+5x**^{7}

Degree of 8x^{4} – 36x + 5x^{7} is 7.

**(v)x**^{9}**-x**^{5}**+3x**^{10}**+8**

Degree of x^{9} – x^{5} 3x^{10} + 8 is 10.

**(vi) 2-3x**^{2}

Degree of 2 – 3x^{2} is 2.

**Question 3:**

**Write :**

**(i) Coefficient of x**^{3}** in 2x + x**^{2}** – 5x**^{3}**+ x**^{4}

-5

**(ii)** **Coefficient of x in \(\sqrt{3}\) – 2\(\sqrt{2x}\) + 4x**^{2}

— 2\(\sqrt{2}\)

**(iii) Coefficient of x**^{2}** in \(\prod \div 3\) x**^{2}** + 7x-3**

**(iv)** ** Coefficient of x**^{2}** in 3x – 5**

0.

**Question 4: **

**(i)** **Give an example of a binomial of degree 27.**

x^{27} — 36

**(ii) Give an example of a monomial of degree 16.**

y1^{16}

**(iii) Give an example of a trinomial of degree 3.**

5x^{3} — 8x + 7

**Question 5: **

**Classify the following as linear, quadratic and cubic polynomials :**

**(i)** **2x**^{2 }**+ 4x**

It is a quadratic polynomial.

**(ii)** **x – x**^{3}

It is a cubic polynomial.

**(iii)** **2 – y – y**^{2}

It is a quadratic polynomial.

**(iv)** **-7 + z**

It is a linear polynomial.

**(v) 5t**

It is a linear polynomial.

**(vi**) ^{ } **p**^{3 }** **

It is a cubic polynomial.

** **

**Question 6**:

**If p(x) = 5 – 4x + 2x**^{2}**, find**

**(i)** **p(0)**

= 5 – 4(0) + 2(0)^{2}

= 5

**(ii)** **p(3)**

= 5 – 4(3) + 2(3)^{2}

= 5 – 12 + 18

= 23 – 12

= 11

**(iii) p(-2)**

= 5 – 4(-2) + 2(-2)^{2}

= 5 + 8 + 8

= 21

**Question 7:**

**If P(Y) = 4 + 3Y – Y**^{2}**+ 5y**^{3 }**, find**

**(i) p(0)**

= 4 + 3(0) – 0^{2} + 5(0)^{3}

= 4 + 0 – 0 + 0

= 4

**(ii) p(2)**

= 4 + 3(2) – 2^{2} + 5(2)^{3}

= 4 + 6 – 4 + 40

= 10 – 4 + 40

= 46

**(III) p(-1) **

= 4 + 3(-1) – (-1)2 + 5(-1)^{3}

= 4 – 3 – 1 – 5

= -5

**Question 8**:

**If f(t) = 4t**^{2}** – 3t + 6, find **

**(i) f(0)**

= 4(0)^{2} – 3(0) + 6

= 0 – 0 + 6

= 6

**(ii) f(4)**

= 4(4)^{2} – 3(4) + 6

= 64 – 12 + 6

= 58

**(iii) f(-5)**

= 4(-5)^{2} – 3(-5) + 6

= 100 + 15 + 6

= 121

**Question 9: **

**Find the zero of the polynomial :**

**(i)** **p(x) = x – 5**

p(x) = 0

x – 5 = 0

x = 5

5 is the zero of the polynomial p(x).

**(ii) q(x) = x + 4**

q(x) = 0

x + 4 = 0

x = -4

-4 is the zero of the polynomial q(x).

**(iii)** **p(t) = 2t – 3**

p(t) = 0

2t – 3 = 0

2t =3

\(t=\frac{3}{2}\)\(t=\frac{3}{2}\) is the zero of the polynomial p(t).

**(iv)** **f(x) = 3x + 1**

f(x) = 0

3x + 1= 0

3x = -1

\(x=\frac{-1}{3}\)\(x=\frac{-1}{3}\) is the zero of the polynomial f(x).

**(v)** **g(x) = 5 – 4x**

g(x) = 0

5 – 4x = 0

-4x = -5

\(x=\frac{5}{4}\)\(x=\frac{5}{4}\) is the zero of the polynomial g(x).

**(vi)** **h(x) = 6x – 1**

h(x) = 0

6x – 1 = 0

6x = 1

\(x=\frac{1}{6}\)\(x=\frac{1}{6}\) is the zero of the polynomial h(x).

**(vii)** **p(x) = ax + b , a** **\(\not\equiv\) 0**

p(x) = 0

ax + b = 0

ax = -b

\(x=\frac{-b}{a}\)\(x=\frac{-b}{a}\) is the zero of the polynomial p(x)

**(viii)** **q(x) = 4x**

q(x) = 0

4x = 0

x = 0

0 is the zero of the polynomial q(x).

**(ix) p(x) = ax, a** **\(\not\equiv\) 0**

p(x) = 0

ax = 0

x = 0

0 is the zero of the polynomial p(x).

**Question 10:**

**Verify that :**

**(i) 4 is a zero of the polynomial p(x) = x – 4 **

Then, p(4) = 4 – 4 = 0

4 is a zero of the polynomial p(x).

**(ii) -3 is a zero of the polynomial p(x) = x – 3 **

Then, p(-3) = -3 – 3

= -6

-3 is not a zero of the polynomial p(x).

**(iii) \(\frac{-1}{2}\) is a zero of the polynomial p(y) = 2y + 1**

Then,

\(\frac{-1}{2}\) is a zero of the polynomial p(y).

**(iv) \(\frac{2}{5}\) is a zero of the polynomial p(x) = 2 – 5x**

Then,

\(\frac{2}{5}\) is a zero of the polynomial p(x).

**(v) 1 and 2 are the zeros of the polynomial p(x) = (x – 1) (x – 2) **

Then, p(1) = (1 – 1) (1 – 2)

= 0 -1

= 0

1 is a zero of the polynomial p(x).

Also, p(2) = (2 – 1)(2 – 2)

= 1- 0

= 0

2 is a zero of the polynomial p(x).

Hence, 1 and 2 are the zeroes of the polynomial p(x).

**(vi) 0 and 3 are the zeros of the polynomial p(x) = x**^{2}** – 3x**.

Then, p(0) = 0^{2} – 3(0) = 0

p(3) = (3^{2}) – 3(3)

= 9 – 9

= 0

0 and 3 are the zeroes of the polynomial p(x).

**(vii) 2 and -3 are the zeros of the polynomial p(x) = x**^{2}** + x – 6 **

Then, p(2) = 2^{2} + 2 – 6

= 4 + 2 – 6

= 6 – 6

= 0

2 is a zero of the polynomial p(x).

Also, p(-3) = (-3)2 – 3 – 6

= 9 – 3 – 6

= 0

-3 is a zero of the polynomial p(x).

Hence, 2 and -3 are the zeroes of the polynomial p(x).

**Using remainder theorem, find the remainder when :**

**Question 11:**

**(x**^{3}** – 6x**^{2}** + 9x + 3) is divided by (x – 1)**

f(x) = x^{3} – 6x^{2} + 9x + 3

Now, x – 1 = 0 x = 1

By the remainder theorem, we know that when f(x) is divided by (x – 1) the remainder is f(1).

Now, f(1) = 1^{3}-6 x 1^{2}+9 x 1 + 3

= 1 – 6 + 9 + 3

= 13 – 6

= 7

The required remainder is 7.

**Question 12: **

**(2x**^{3}** – 5x**^{2}** + 9x – 8) is divided by (x-3)**

f(x) = (2x^{3} – 5x^{2} + 9x – 8)

Now, x – 3 = 0

x = 3

By the remainder theorem, we know that when f(x) is divided by (x – 3) the remainder is f(3).

Now, f(3) = 2 x 3^{3} – 5 x 3^{2} + 9 x 3 – 8

= 54 – 45 + 27 – 8

= 81 – 53

= 28

Therefore, The required remainder is 28.

**Question 13: **

**(3x**^{4}** – 6x**^{2}** – 8x + 2) is divided by (x-2)**

f(x) = (3x^{4} – 6x^{2} – 8x + 2)

Now, x – 2 = 0

x = 2

By the remainder theorem, we know that when f(x) is divided by (x – 2) the remainder is f(2).

Now, f(2) = 3 x 2^{4} – 6 x 2^{2}– 8 x 2 + 2

= 48 – 24 – 16 + 2

= 50 – 40 = 10

Therefore, The required remainder is 10.

Related Links |
||
---|---|---|

NCERT Books | NCERT Solutions | RS Aggarwal |

Lakhmir Singh | RD Sharma Solutions | NCERT Solutions Class 6 to 12 |