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KC Sinha Mathematics Solution Class 9 Chapter 3 Polynomials exercise 3.1


Page 3.10

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) 4x2-3x+7

Sol:

⇒4x2-3x1+7x0

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) \sqrt{3}x^2+5x-2

Sol :

\sqrt{3}x^2+5x^1-2x^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

 


(iii) 3√t + t√2

Sol :

Rewritten as

3t^{\frac{1}{2}}+t^1\sqrt{2}

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

 


(iv) y+\dfrac{1}{y^2}+3

Sol :

⇒y1+y-2+3y0

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


(v) x10+y3+t50

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) 2x10+y5+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 x2 in each of the following :

(i) 2x3+x2+x

Sol : 1


(ii) 5-7x2+x3+2

Sol : -7


(iii) \dfrac{\pi}{2}x^3+x-1

Sol : 0


(iv) \sqrt{2}-1

Sol : 0


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

Sol :

Using identity:

(a+b)(a-b)=a2-b2

⇒x2-12

coeff‌icient of x2=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 ax2+bx+c.

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

General form of a cubic polynomial in x is ax3+bx2+cx+d


Question 3

Find the degree of each of the following polynomials :

(i) 5x4+4x3+10

Sol:

Degree of polynomial=Highest power of variable x=4


(ii) 4-4y2+5y+2

Sol:

Degree of polynomial=Highest power of variable x=2


(iii) t3-5

Sol:

Degree of polynomial=Highest power of variable t=2


(iv) 20

Sol:

20= a constant non-zero polynomial = 20x0

Therefore, degree of this polynomial=0


(v) z5-2z7+5

Sol:

Degree of polynomial=Highest power of variable z=7


(vii) x7-2x+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) x2-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) 3x2-5

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

(iv) 5x2+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.


Page 3.11

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 5x2 – 2x + 2 at

(i) x=0

Sol :

p(x)=5x2 -2x+2

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

p(0)=+2

 


(ii) x=1

Sol :

p(x)=5x2 -2x+2

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

p(1)=5-2+2

p(1)=5

 


(iii) x=-3

Sol :

p(x)=5x2 -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)=y2+y+2

Sol :

p(0)=02+0+2

p(0)=+2

 

p(1)=12+1+2

p(1)=1+1+2

p(1)=4


(ii) p(t)=5+t+2t3-t4

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)=x5

Sol :

p(0)=05

p(0)=0

 

p(1)=15

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)=2x3+3x2-1

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)=t4-t2+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)=3x2+8x+4 and a=-2

Sol :

⇒p(x)=3x2+8x+4

p(x)=p(a)

⇒p(a)=3a2+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)=x2+x-6 and a=-3

Sol :

⇒p(x)=x2+x-6

p(x)=p(a)

⇒p(a)=a2+a-6

⇒p(-3)=(-3)2+(-3)-6

⇒p(-3)=9-3-6

⇒p(-3)=0


(iii) p(x)=x3-2x+2 and a=-1

Sol :

⇒p(x)=x2+x-6

p(x)=p(a)

⇒p(a)=a3-2a+2

⇒p(-1)=(-1)3-2(-1)+2

⇒p(-1)=-1+2+2

⇒p(-1)=3


(iv) p(x)=x3-3x2+x and a=-1

Sol :

⇒p(x)=x3-3x2+x

p(x)=p(a)

⇒p(a)=a3-3a2+a

⇒p(-1)=(-1)3-3(-1)2+(-1)

⇒p(-1)=-1+3-1

⇒p(-1)=1


Question 8

If p(x)=x2-5x+4 and q(x)=x3+1 . Find the values of the following :

(i) p(1)×q(1)

Sol :

⇒p(1)=12-5(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) \dfrac{p(1)}{q(1)}

Sol :

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

\dfrac{p(1)}{q(1)}=\dfrac{0}{2}

[zero by integer always equal to zero]

⇒0


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

Sol :

⇒p(x)=x2-5x+4

⇒p(2)=22-5(2)+4

⇒p(2)=4-10+4

⇒p(2)=-2..(i)

 

⇒q(x)=x3+1

⇒q(2)=23+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 ; x=-\dfrac{1}{3}

Sol :

Given polynomial p(x)=3x+1

p\left(-\dfrac{1}{3}\right)=3\left(-\dfrac{1}{3}\right)+1

p\left(-\dfrac{1}{3}\right)=\left(-\dfrac{3}{3}\right)+1

p\left(-\dfrac{1}{3}\right)=-1+1=0

Hence , it is a zero of the given polynomial

 


(ii) p(x)=x2-1 ; x=1 , -1

Sol :

Given polynomial p(x)=x2-1

When x=1

⇒p(1)=12-1

⇒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)=x2 ; x=0

Sol :

Given polynomial p(x)=x2

⇒p(0)=02=0

Hence , 0 is a zero of the given polynomial


(iv) p(x)=px+q ; x=-\dfrac{q}{p}

Sol :

Given polynomial p(x)=px+q

p\left(-\dfrac{q}{p}\right)=p\left(-\dfrac{q}{p}\right)+q

p\left(-\dfrac{q}{p}\right)=-q+q=0

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


 

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