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 y=2
(iii) t3-5
Sol:
Degree of polynomial=Highest power of variable t=3
(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
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.
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
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
(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
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
(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
(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
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
(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)=-5
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
(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
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 ; $x=-\dfrac{1}{3}$
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
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
(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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