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### Exercise 3.1

**Type 1**

**Problems based on definition of terms related to polynomial and identifying polynomials**

WORKING RULE:

1. Write down the given polynomial in such a way that the variable is in the numerator in each term

2. If each term of an algebraic expression contains only non-negative integrals powers of x , then given algebraic expression is a polynomial in x otherwise it is not a polynomial in x

#### Question 1

**Among the following expressions, which are polynomials of a single variable and which are not ? Give reasons for your answer :**

**(i) 4x ^{2}-3x+7**

Sol:

⇒4x^{2}-3x^{1}+7x^{0}

Exponents(power) of given equation are 2,1 and 0 [all are in whole numbers] which shows it is a polynomial in one variable

**(ii) **

Sol :

⇒

Exponents(power) of given equation are 2,1 and 0 [all are in whole numbers] which shows it is a polynomial in one variable

**(iii) 3√t + t√2**

Sol :

Rewritten as

⇒

Here, (exponent)power is in fraction that’s why it is not a polynomial

**(iv) **

Sol :

⇒y^{1}+y^{-2}+3y^{0}

Here in the second term exponent(power) is negative that’s why it’s not a polynomial

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

Sol :

⇒Here , powers(exponents) of variables x,y,z in given algebraic expression are 10,3,50 which are whole numbers.

Hence , given expression is a polynomial in variables x,y,z

**(vi) 2x ^{10}+y^{5}+z**

Sol :

⇒Here , powers(exponents) of variables x,y,z in given algebraic expression are 10,5,1 which are whole numbers.

Hence , given expression is a polynomial in variables x,y,z or we can say that

Polynomial in three variables

#### Question 2

**Find the coefficient of x ^{2} in each of the following :**

**(i) 2x ^{3}+x^{2}+x**

Sol : 1

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

Sol : -7

**(iii) **

Sol : 0

**(iv) **

Sol : 0

**(v) (x-1)(x+1)**

Sol :

Using identity:

(a+b)(a-b)=a^{2}-b^{2}

⇒x^{2}-1^{2}

coefficient of x^{2}=1

**Type 2**

**Problems based on degree of polynomials**

WORKING RULE:

1. Degree of a polynomial in x=highest power of x in the polynomial.

2. Degree of a constant non-zero polynomial is 0.

3. Degree of zero polynomial is undefined.

4. A polynomial of degree one is called a linear polynomial.

General form of a linear polynomial in x is ax+b.

5. A polynomial of degree two is called a quadratic polynomial.

General form of a quadratic polynomial in x is ax^{2}+bx+c.

6. A polynomial of degree three is called a cubic polynomial.

General form of a cubic polynomial in x is ax^{3}+bx^{2}+cx+d

#### Question 3

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

**(i) 5x ^{4}+4x^{3}+10**

Sol:

Degree of polynomial=Highest power of variable x=4

**(ii) 4-4y ^{2}+5y+2**

Sol:

Degree of polynomial=Highest power of variable x=2

**(iii) t ^{3}-5**

Sol:

Degree of polynomial=Highest power of variable t=2

**(iv) 20**

Sol:

20= a constant non-zero polynomial = 20x^{0}

Therefore, degree of this polynomial=0

**(v) z ^{5}-2z^{7}+5**

Sol:

Degree of polynomial=Highest power of variable z=7

**(vii) x ^{7}-2x+1**

Sol:

Degree of polynomial=Highest power of variable x=7

**Type 3**

**Problems based on classification of polynomials on the basis of the number of terms of the polynomials.**

WORKING RULE:

1. A polynomial having only one term is called a monomial.

2. A polynomial having only one term is called a binomial.

3. A polynomial having only one term is called a trinomial.

4. A polynomial having each coefficient zero (0) is called zero polynomial

#### Question 4

**Which of the following polynomials are monomial, binomial and trinomial ? Give reasons for your answer :**

**(i) x ^{2}-x**

Sol: Polynomial p(x) has two terms , therefore , it is a binomial.

**(ii) 3**

Sol: Polynomial p(x) has one terms , therefore , it is a monomial.

**(iii) 3x ^{2}-5**

Sol: Polynomial p(x) has two terms , therefore , it is a binomial.

**(iv) 5x ^{2}+6x+2**

Sol: Polynomial p(x) has three terms , therefore , it is a trinomial.

**(v) 2x**

Sol: Polynomial p(x) has one terms , therefore , it is a monomial.

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**Type 4**

**Problems based on values of a polynomial**

WORKING RULE :

If p(x) is a polynomial in x. then in order to find p(a) , put a in place of x.

1. The value thus obtained will be p(a) i.e. value of p(x) at x=a.

2. A number a is a zero of the polynomial p(x) if p(a)=0

3. If it is to be determined whether a number, a is a zero of the polynomial p(x) or not , then find p(a) [value of p(x) on putting a in place of x]

(i) If p(a)=0 , then a is a zero of the polynomial p(x)

(ii) If p(a)≠0 , then a is not a zero of the polynomial p(x).

4. Zero of a linear polynomial ax+b is -b/a

ax+b=0⇒ax=-b⇒x=-b/a

#### Question 5

**Write the values of the polynomial 5x ^{2} – 2x + 2 at**

**(i) x=0**

Sol :

p(x)=5x^{2} -2x+2

p(0)=5(0)^{2} -2(0)+2

p(0)=+2

**(ii) x=1**

Sol :

p(x)=5x^{2} -2x+2

p(1)=5(1)^{2} -2(1)+2

p(1)=5-2+2

p(1)=5

**(iii) x=-3**

Sol :

p(x)=5x^{2} -2x+2

p(-3)=5(-3)^{2} -2(-3)+2

p(-3)=5×9 +6 +2

p(-3)=45+6 +2

p(-3)=53

#### Question 6

**For each of the following polynomial , find p(0) and p(1) :**

**(i) p(y)=y ^{2}+y+2**

Sol :

p(0)=0^{2}+0+2

p(0)=+2

p(1)=1^{2}+1+2

p(1)=1+1+2

p(1)=4

**(ii) p(t)=5+t+2t ^{3}-t^{4}**

Sol :

p(0)=5+0+2(0)^{3}-(0)^{4}

p(0)=5+0+2-0

p(0)=7

p(1)=5+1+2(1)^{3}-(1)^{4}

p(1)=5+1+2-1

p(1)=7

**(iii) p(x)=x ^{5}**

Sol :

p(0)=0^{5}

p(0)=0

p(1)=1^{5}

p(1)=1

**(iv) p(x)=(x-2)(x+2)**

p(0)=(0-2)(0+2)

p(0)=-4

p(1)=(1-2)(1+2)

p(1)=(-1)(3)

p(1)=-3

**(v) p(x)=2x ^{3}+3x^{2}-1**

p(0)=2(0)^{3}+3(0)^{2}-1

p(0)=-1

p(1)=2(1)^{3}+3(1)^{2}-1

p(1)=2+3-1

p(1)=4

**(vi) p(t)=t ^{4}-t^{2}+3**

p(0)=(0)^{4}-(0)^{2}+3

p(0)=0-0+3

p(0)=3

p(1)=(1)^{4}-(1)^{2}+3

p(1)=1-1+3

p(1)=+3

#### Question 7

**Find the value of the polynomial p(x) at x = a when :**

**(i) p(x)=3x ^{2}+8x+4 and a=-2**

Sol :

⇒p(x)=3x^{2}+8x+4

p(x)=p(a)

⇒p(a)=3a^{2}+8a+4

⇒p(-2)=3(-2)^{2}+8(-2)+4

⇒p(-2)=(3×4)-16+4

⇒p(-2)=12-16+4

⇒p(-2)=0

**(ii) p(x)=x ^{2}+x-6 and a=-3**

Sol :

⇒p(x)=x^{2}+x-6

p(x)=p(a)

⇒p(a)=a^{2}+a-6

⇒p(-3)=(-3)^{2}+(-3)-6

⇒p(-3)=9-3-6

⇒p(-3)=0

**(iii) p(x)=x ^{3}-2x+2 and a=-1**

Sol :

⇒p(x)=x^{2}+x-6

p(x)=p(a)

⇒p(a)=a^{3}-2a+2

⇒p(-1)=(-1)^{3}-2(-1)+2

⇒p(-1)=-1+2+2

⇒p(-1)=3

**(iv) p(x)=x ^{3}-3x^{2}+x and a=-1**

Sol :

⇒p(x)=x^{3}-3x^{2}+x

p(x)=p(a)

⇒p(a)=a^{3}-3a^{2}+a

⇒p(-1)=(-1)^{3}-3(-1)^{2}+(-1)

⇒p(-1)=-1+3-1

⇒p(-1)=1

#### Question 8

**If p(x)=x ^{2}-5x+4 and q(x)=x^{3}+1 . Find the values of the following :**

**(i) p(1)×q(1)**

Sol :

⇒p(1)=1^{2}-5(1)+4

⇒p(1)=1-5+4

⇒p(1)=0..(i)

⇒q(1)=(1)^{3}+1

⇒q(1)=1+1

⇒q(1)=2..(ii)

⇒p(1)×q(1)

From (i) and (ii) , we get

⇒0×2

⇒0

**(ii) **

Sol :

From above p(1)=0 and q(1)=2

⇒

[zero by integer always equal to zero]

⇒0

**(iii) p(2)+q(2)**

Sol :

⇒p(x)=x^{2}-5x+4

⇒p(2)=2^{2}-5(2)+4

⇒p(2)=4-10+4

⇒p(2)=-2..(i)

⇒q(x)=x^{3}+1

⇒q(2)=2^{3}+1

⇒q(2)=8+1

⇒q(2)=9..(ii)

A.T.Q

⇒p(2)+q(2)

putting values of (ii) and (i)

⇒-2+9

⇒7

#### Question 9

**Verify that given values are zeroes of the corresponding polynomials :**

**(i) p(x)=3x+1 ; **

Sol :

Given polynomial p(x)=3x+1

⇒

⇒

⇒

Hence , it is a zero of the given polynomial

**(ii) p(x)=x ^{2}-1 ; x=1 , -1**

Sol :

Given polynomial p(x)=x^{2}-1

When x=1

⇒p(1)=1^{2}-1

⇒p(1)=0

When x=-1

⇒p(-1)=(-1)^{2}-1

⇒p(-1)=1-1=0

Hence , 1 and -1 both are zeros of polynomial

**(iii) p(x)=x ^{2} ; x=0**

Sol :

Given polynomial p(x)=x^{2}

⇒p(0)=0^{2}=0

Hence , 0 is a zero of the given polynomial

**(iv) p(x)=px+q ; **

Sol :

Given polynomial p(x)=px+q

⇒

⇒

Hence , it is the zero of given polynomial

#### Question 10

**Find the zeroes of the given polynomial p(x) under given conditions :**

**(i) p(x)=x+5**

Sol :

Given polynomial p(x)=x+5

⇒p(x)=0

⇒p(x)=x+5=0

⇒p(x)=x=-5

∴ Zero of polynomial p(x) is -5

**(ii) p(x)=2x-5**

Sol :

Given polynomial p(x)=2x-5

⇒p(x)=0

⇒p(x)=2x-5=0

⇒p(x)=2x=+5

⇒p(x)=x=+5/2

∴ Zero of polynomial p(x) is 5/2

**(iii) p(x)=3x-6**

Sol :

Given polynomial p(x)=3x-6

⇒p(x)=0

⇒p(x)=3x-6=0

⇒p(x)=3x=+6

⇒p(x)=x=+6/3

⇒p(x)=x=2

∴ Zero of polynomial p(x) is 2

**(iv) p(x)=5x**

Sol :

Given polynomial p(x)=5x

⇒p(x)=0

⇒p(x)=5x=0

⇒p(x)=x=0/5

[Zero by some integer i s 0]

∴ Zero of polynomial p(x) is 0

**(v) p(x)=(x+1)**

Sol :

Given polynomial p(x)=x+1

⇒p(x)=0

⇒p(x)=x+1=0

⇒p(x)=x=-1

∴ Zero of polynomial p(x) is -1

**(vi) p(x)=ax+b ; a≠0 , a , b are real numbers**

Sol :

Given polynomial p(x)=ax+b

⇒p(x)=0

⇒p(x)=ax+b=0

⇒p(x)=ax=-b

⇒p(x)=x=-b/a

∴ Zero of polynomial p(x) is -b/a