KC Sinha: Exercise 3.1 - Mathematics Solution Class 9 Chapter 3 Polynomials

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Exercise 3.1

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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

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Q1 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(ii) 3–√x2+5x−2
Sol :
⇒3–√x2+5x1−2x0
Exponents(power) of given equation are 2,1 and 0 [all are in whole numbers] which shows it is a polynomial in one variable

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(iii) 3√t + t√2
Sol :
Rewritten as
⇒3t12+t12–√
Here, (exponent)power is in fraction that's why it is not a polynomial

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(iv) y+1y2+3
Sol :
⇒y¹+y^-2+3y⁰
Here in the second term exponent(power) is negative that's why it's not a polynomial

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(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

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(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

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Q2 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

Question 2

Find the coeff‌icient of x² in each of the following :
(i) 2x³+x²+x
Sol : 1

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(ii) 5-7x²+x³+2
Sol : -7

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(iii) π2x3+x−1
Sol : 0

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(iv) 2–√−1
Sol : 0

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(v) (x-1)(x+1)
Sol :
Using identity:
(a+b)(a-b)=a²-b²
⇒x²-1²
coeff‌icient of x²=1

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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

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Q3 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(ii) 4-4y²+5y+2
Sol:
Degree of polynomial=Highest power of variable y=2

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(iii) t³-5
Sol:
Degree of polynomial=Highest power of variable t=3

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(iv) 20
Sol:
20= a constant non-zero polynomial = 20x⁰
Therefore, degree of this polynomial=0

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(v) z⁵-2z⁷+5
Sol:
Degree of polynomial=Highest power of variable z=7

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(vii) x⁷-2x+1
Sol:
Degree of polynomial=Highest power of variable x=7

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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

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Q4 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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.

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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

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Q5 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(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

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(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

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Q6 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(ii) p(t)=5+t+2t³-t⁴
Sol :
p(0)=5+0+2(0)³-(0)⁴
p(0)=5+0+0-0
p(0)=5

p(1)=5+1+2(1)³-(1)⁴
p(1)=5+1+2-1
p(1)=7

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(iii) p(x)=x⁵
Sol :
p(0)=0⁵
p(0)=0

p(1)=1⁵
p(1)=1

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(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

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(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

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(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

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Q7 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(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

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(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

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(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)=-5

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Q8 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(ii) p(1)q(1)
Sol :
From above p(1)=0 and q(1)=2
⇒p(1)q(1)=02
[zero by integer always equal to zero]
⇒0

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(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

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Q9 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

Question 9

Verify that given values are zeroes of the corresponding polynomials :
(i) p(x)=3x+1 ; x=−13
Sol :
Given polynomial p(x)=3x+1
⇒p(−13)=3(−13)+1
⇒p(−13)=(−33)+1
⇒p(−13)=−1+1=0
Hence , it is a zero of the given polynomial

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(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

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(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

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(iv) p(x)=px+q ; x=−qp
Sol :
Given polynomial p(x)=px+q
⇒p(−qp)=p(−qp)+q
⇒p(−qp)=−q+q=0
Hence , it is the zero of given polynomial

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Q10 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper

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

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(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

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(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

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(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

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(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

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(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