**NEXT**

#### NCERT Class 9 maths chapter 2 Polynomials

#### NCERT Solution Chapter 2: Polynomials

### Important points about Polynomials:

**TEXTBOOK’S EXERCISE – 2.1**

- A polynomial can have many terms but not infinite terms.
- The exponent of a variable of a polynomial cannot be negative. This means a variable with power – 2, -3, -4, etc. are not allowed. If the power of a variable in an algebraic expression is negative, then that cannot be considered a polynomial.
- The exponent of a variable of a polynomial must be a whole number.
- The exponent of a variable of a polynomial cannot be a fraction. This means a variable with power 1/2, 3/2, etc. is not allowed. If the power of a variable in an algebraic expression is an infraction, then that cannot be considered a polynomial.
- A polynomial with the only constant term is called constant polynomial.
- The degree of a non-zero constant polynomial is zero.
- Degree of a zero polynomial is not defined.

**Q 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer**

(i) 4x^{2}−3x+7

**Ans:** Since the given expression has only one variable ‘x’, hence this is a polynomial in one variable.

(ii) y^{2}+√2

**Ans:** Since the given expression has only one variable ‘y’, hence this is a polynomial in one variable.

(iii) 3√t+t√2

**Ans:** Since, variable, ‘t’ in this expression exponent of the variable is not a whole number. (Exponent of a variable in a polynomial must be a whole number. Expression with an exponent of a variable infraction is not considered as a polynomial.)

(iv) y+2/y

**Ans:** Since the exponent of the variable is a negative integer, and not a whole number, hence it cannot be considered a polynomial.

(v) x^{10}+y^{3}+t^{50}

**Ans:** Since, given expression have three variables, i.e. x, y and t, so it is not a polynomial in one variable.

**Q 2: Write the coefficients of x**^{2}** in each of the following:**

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

**Ans:** Given; 2+x2+x

Here, the coefficient of x2=1

(ii) 2–x^{2}+x^{3}

**Ans:** Here, the coefficient of x^{2}=1 (iii) π/2x^{2}+x **Answer:** Here, the coefficient of x^{2}=π/2 (iv) √2x–12x–1

**Ans:** Here, the coefficient of x^{2}=0

**3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.**

Sol. x^{35} + 5 is a binomial of degree 35.

2y^{100} is a monomial of degree 100.

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

*(i) *5*x*^{3} + 4*x*^{2} + 7*x (ii) *4 – *y*^{2} *(iii) *5*t *–`√ 7 (iv) 3

Sol.
(i) The degree is 3 as x^{3} is the highest power.

(ii) The degree is
2 as y^{2} is the highest power.

(iii) The degree is
1 as t is the highest power.

(iv) The degree is
0 as x^{0} is the highest power.

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

**(i) x**^{2}** + x (ii) x – x**^{3}** (iii) y + y**^{2}** + 4 (iv) 1 + x **

**(v) 3t (vi) r**^{2}** (vii) 7x**^{3}

**Sol. **(i) *x*^{2} +
*x *is quadratic.

(ii) *x *– *x*^{3} is
cubic.

(iii) *y *+ *y*^{2} +
4 is quadratic.

(iv) 1 + *x *is linear.

(v) 3*t *is linear.

(vi) *r*^{2} is quadratic.

(vii) 7*x*^{3} is cubic.

**TEXTBOOK’S
EXERCISE – 2.2 **

**1. **Find
the value of the polynomial 5x – 4x^{2} + 3 at

(i) x = 0 (ii) x = –1 (iii) x = 2

**Sol. **p(x) = 5x – 4x^{2} + 3

(i) At x = 0, p(0) = 5 × 0 – 4 × 0^{2} + 3 = 3

(ii) At x = –1,

p(–1) = 5 × (–1) – 4 ×
(–1)^{2} + 3

= –5 – 4 + 3 = – 6

(iii) At x = 2, p(2) = 5 × 2 – 4 × (2)^{2} +
3

= 10 – 16 + 3 = – 3

**2. **Find
p(0), p(1) and p(2) for each of the following polynomials :

(i) p(y) = y^{2} – y +
1 (ii) p(t) = 2 + t + 2t^{2} – t^{3}

(iii) p(x) = x^{3} (iv) p(x)
= (x – 1) (x + 1)

**Sol. **(i) p(y) = y^{2}
– y + 1

p(0) = 0^{2} – 0 +
1 = 1

p(1) = 1^{2} – 1 +
1 = 1

p(2) = 2^{2} – 2 + 1 = 3.

(ii) p(t) = 2 + t + 2t^{2} – t^{3 }

p(0) = 2 + 0 + 2 × 0^{2} – 0^{3} =
2

p(1) = 2 + 1 + 2 × 1^{2} – 1^{3} =
4

p(2) = 2 + 2 + 2 × 2^{2} – 2^{3 }

= 4 + 8 – 8 = 4.

(iii) p(x) = x^{3 }

p(0) = 0

p(1) = 1

p(2) = 8.

(iv) p(x) = (x – 1) (x + 1)

p(0) = (–1) (1) = –1

p(1) = (1 – 1) (1 + 1) = 0

p(2) = (2 – 1) (2 + 1) = 3

**3. **Verify
whether the following are zeroes of the polynomial, indicated against them.

(i) p(x) = 3x + 1, x = – 1/3 (ii) p(x) = 5x – π, x = 4/5

(iii) p(x) = x^{2} – 1, x =
1, –1 (iv) p(x) = (x + 1) (x – 2), x = –1, 2

(v) p(x) = x^{2}, x = 0

(vi) p(x) = lx + m, x = – m/l

(vii) p(x) = 3x^{2} – 1, x = – 1/√3, 2/√3 ,

(viii) p(x) = 2x + 1, x = 1

**Sol. **(i) Yes. 3x +
1 = 0, for x = – 1/3 .

(ii) No. 5x – π = 5 × 4/5 – π = 4 – π ≠ 0

(iii) Yes. x^{2} – 1 = 1^{2} – 1 = 0 for x = 1 and x^{2} – 1 = (–1)^{2} –1 = 0 for x = 1

(iv) Yes. (x + 1) (x – 2) = 0 for x = –1, or, x = 2.

(v) Yes. x^{2} = 0 for x = 0

(vi) Yes. lx + m = 0 for x = – m/l

(vii) 3x^{2} –
1 = 3. 1/3 – 1 = 0 for x = −1/ √3 and 3x^{2} – 1 = 3· 4/3 –1 = 3
≠ 0

Thus, for −1/ √3 is a zero but for – 2/ √3 is not a zero of the polynomial.

(viii) No. 2x + 1 ≠ 0 for x = 1/2.

**4. **Find the zero of the
polynomial in each of the following cases :

(i) p(x) = x + 5

(ii) p(x) = x – 5

(iii) p(x) = 2x + 5

(iv) p(x) = 3x – 2

(v) p(x) = 3x

(vi) p(x) = ax, a ≠ 0

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

**Sol. **(i) x + 5 = 0, x = –5,

so, –5 is the zero of x + 5.

⇒ x = −5/2 ,

so – 5/2 is the zero of 2x + 5.

(iv) 3x – 2 = 0 ⇒ 3x = 2

⇒ x = 2/3 , so 2/3 is the zero of 3x – 2.

(v) 3x = 0 ⇒ x = 0, so 0 is the zero of 3x.

(vi) ax = 0 (a ≠ 0) ⇒ x = 0/a = 0,

so, 0 is the zero of ax.

(vii) cx + d = 0 (c ≠ 0)

⇒ cx = – d

⇒ x = −d/ c ,

so, −d /c is the 0 of cx + d. **PREVIOUS**