{"id":623826,"date":"2023-08-31T10:12:37","date_gmt":"2023-08-31T10:12:37","guid":{"rendered":"https:\/\/www.indcareer.com\/schools\/?p=623826"},"modified":"2023-09-04T07:03:24","modified_gmt":"2023-09-04T07:03:24","slug":"kc-sinha-exercise-3-1-mathematics-solution-class-9-chapter-3-polynomials","status":"publish","type":"post","link":"https:\/\/www.indcareer.com\/schools\/kc-sinha-exercise-3-1-mathematics-solution-class-9-chapter-3-polynomials\/","title":{"rendered":"KC Sinha: Exercise 3.1 – Mathematics Solution Class 9 Chapter 3 Polynomials"},"content":{"rendered":"\n\n\n\n\n

Exercise 3.1<\/h3>\n\n\n\n
\n\n\n\n

Type 1<\/strong>
Problems based on definition of terms related to polynomial and identifying polynomials<\/strong>
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<\/p>\n\n\n\n

\n\n\n\n

Q1 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/p>\n\n\n\n

Question 1<\/h4>\n\n\n\n

Among the following expressions, which are polynomials of a single variable and which are not ? Give reasons for your answer :<\/strong>
(i) 4x2<\/sup>-3x+7<\/strong>
Sol:
\u21d24x2<\/sup>-3x1<\/sup>+7x0<\/sup>
Exponents(power) of given equation are 2,1 and 0 [all are in whole numbers] which shows it is a polynomial in one variable<\/p>\n\n\n\n

\n\n\n\n

(ii) 3\u2013\u221ax2+5x\u22122<\/strong>
Sol :
\u21d23\u2013\u221ax2+5×1\u22122×0
Exponents(power) of given equation are 2,1 and 0 [all are in whole numbers] which shows it is a polynomial in one variable<\/p>\n\n\n\n

\n\n\n\n

(iii) 3\u221at + t\u221a2<\/strong>
Sol :
Rewritten as
\u21d23t12+t12\u2013\u221a
Here, (exponent)power is in fraction that’s why it is not a polynomial<\/p>\n\n\n\n

\n\n\n\n

(iv) y+1y2+3<\/strong>
Sol :
\u21d2y1<\/sup>+y-2<\/sup>+3y0<\/sup>
Here in the second term exponent(power) is negative that’s why it’s not a polynomial<\/p>\n\n\n\n

\n\n\n\n

(v) x10<\/sup>+y3<\/sup>+t50<\/sup><\/strong>
Sol :
\u21d2Here , 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<\/p>\n\n\n\n

\n\n\n\n

(vi) 2x10<\/sup>+y5<\/sup>+z<\/strong>
Sol :
\u21d2Here , 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<\/p>\n\n\n\n

\n\n\n\n

Q2 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/p>\n\n\n\n

Question 2<\/h4>\n\n\n\n

Find the coeff\u200cicient of x2<\/sup> in each of the following :<\/strong>
(i) 2x3<\/sup>+x2<\/sup>+x<\/strong>
Sol : 1<\/p>\n\n\n\n

\n\n\n\n

(ii) 5-7x2<\/sup>+x3<\/sup>+2<\/strong>
Sol : -7<\/p>\n\n\n\n

\n\n\n\n

(iii) \u03c02×3+x\u22121<\/strong>
Sol : 0<\/p>\n\n\n\n

\n\n\n\n

(iv) 2\u2013\u221a\u22121<\/strong>
Sol : 0<\/p>\n\n\n\n

\n\n\n\n

(v) (x-1)(x+1)<\/strong>
Sol :
Using identity:
(a+b)(a-b)=a2<\/sup>-b2<\/sup>
\u21d2x2<\/sup>-12<\/sup>
coeff\u200cicient of x2<\/sup>=1<\/p>\n\n\n\n

\n\n\n\n

Type 2<\/strong>
Problems based on degree of polynomials<\/strong>
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 undef\u200cined.
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<\/sup>+bx+c.<\/p>\n\n\n\n

6. A polynomial of degree three is called a cubic polynomial.<\/p>\n\n\n\n

General form of a cubic polynomial in x is ax3<\/sup>+bx2<\/sup>+cx+d<\/p>\n\n\n\n

\n\n\n\n

Q3 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/p>\n\n\n\n

Question 3<\/h4>\n\n\n\n

Find the degree of each of the following polynomials :<\/strong>
(i) 5x4<\/sup>+4x3<\/sup>+10<\/strong>
Sol:
Degree of polynomial=Highest power of variable x=4<\/p>\n\n\n\n

\n\n\n\n

(ii) 4-4y2<\/sup>+5y+2<\/strong>
Sol:
Degree of polynomial=Highest power of variable y=2<\/p>\n\n\n\n

\n\n\n\n

(iii) t3<\/sup>-5<\/strong>
Sol:
Degree of polynomial=Highest power of variable t=3<\/p>\n\n\n\n

\n\n\n\n

(iv) 20<\/strong>
Sol:
20= a constant non-zero polynomial = 20x0<\/sup>
Therefore, degree of this polynomial=0<\/p>\n\n\n\n

\n\n\n\n

(v) z5<\/sup>-2z7<\/sup>+5<\/strong>
Sol:
Degree of polynomial=Highest power of variable z=7<\/p>\n\n\n\n

\n\n\n\n

(vii) x7<\/sup>-2x+1<\/strong>
Sol:
Degree of polynomial=Highest power of variable x=7<\/p>\n\n\n\n

\n\n\n\n

Type 3<\/strong>
Problems based on classif\u200cication of polynomials on the basis of the number of terms of the polynomials.<\/strong>
WORKING RULE:
1. A polynomial having only one term is called a monomial.
2. A polynomial having only one term is called a binomial.<\/p>\n\n\n\n

3. A polynomial having only one term is called a trinomial.<\/p>\n\n\n\n

4. A polynomial having each coefficient zero (0) is called zero polynomial<\/p>\n\n\n\n

\n\n\n\n

Q4 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 4<\/h4>\n\n\n\n

Which of the following polynomials are monomial, binomial and trinomial ? Give reasons for your answer :<\/strong>
(i) x2<\/sup>-x<\/strong>
Sol: Polynomial p(x) has two terms , therefore , it is a binomial.
(ii) 3<\/strong>
Sol: Polynomial p(x) has one terms , therefore , it is a monomial.
(iii) 3x2<\/sup>-5<\/strong>
Sol: Polynomial p(x) has two terms , therefore , it is a binomial.
(iv) 5x2<\/sup>+6x+2<\/strong>
Sol: Polynomial p(x) has three terms , therefore , it is a trinomial.
(v) 2x<\/strong>
Sol: Polynomial p(x) has one terms , therefore , it is a monomial.<\/p>\n\n\n\n

\n\n\n\n

Page 3.11
Type 4<\/strong>
Problems based on values of a polynomial<\/strong>
WORKING RULE :
If p(x) is a polynomial in x. then in order to f\u200cind 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.<\/p>\n\n\n\n

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)<\/p>\n\n\n\n

(ii) If p(a)\u22600 , 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\u21d2ax=-b\u21d2x=-b\/a<\/p>\n\n\n\n

\n\n\n\n

Q5 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 5<\/h4>\n\n\n\n

Write the values of the polynomial 5x2<\/sup> – 2x + 2 at<\/strong>
(i) x=0<\/strong>
Sol :
p(x)=5x2<\/sup> -2x+2
p(0)=5(0)2<\/sup> -2(0)+2
p(0)=+2<\/p>\n\n\n\n

\n\n\n\n

(ii) x=1<\/strong>
Sol :
p(x)=5x2<\/sup> -2x+2
p(1)=5(1)2<\/sup> -2(1)+2
p(1)=5-2+2
p(1)=5<\/p>\n\n\n\n

\n\n\n\n

(iii) x=-3<\/strong>
Sol :
p(x)=5x2<\/sup> -2x+2
p(-3)=5(-3)2<\/sup> -2(-3)+2
p(-3)=5\u00d79 +6 +2
p(-3)=45+6 +2
p(-3)=53<\/p>\n\n\n\n

\n\n\n\n

Q6 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 6<\/h4>\n\n\n\n

For each of the following polynomial , find p(0) and p(1) :<\/strong>
(i) p(y)=y2<\/sup>+y+2<\/strong>
Sol :
p(0)=02<\/sup>+0+2
p(0)=+2<\/p>\n\n\n\n

p(1)=12<\/sup>+1+2
p(1)=1+1+2
p(1)=4<\/p>\n\n\n\n

\n\n\n\n

(ii) p(t)=5+t+2t3<\/sup>-t4<\/sup><\/strong>
Sol :
p(0)=5+0+2(0)3<\/sup>-(0)4<\/sup>
p(0)=5+0+0-0
p(0)=5<\/p>\n\n\n\n

p(1)=5+1+2(1)3<\/sup>-(1)4<\/sup>
p(1)=5+1+2-1
p(1)=7<\/p>\n\n\n\n

\n\n\n\n

(iii) p(x)=x5<\/sup><\/strong>
Sol :
p(0)=05<\/sup>
p(0)=0<\/p>\n\n\n\n

p(1)=15<\/sup>
p(1)=1<\/p>\n\n\n\n

\n\n\n\n

(iv) p(x)=(x-2)(x+2)<\/strong>
p(0)=(0-2)(0+2)
p(0)=-4<\/p>\n\n\n\n

p(1)=(1-2)(1+2)
p(1)=(-1)(3)
p(1)=-3<\/p>\n\n\n\n

\n\n\n\n

(v) p(x)=2x3<\/sup>+3x2<\/sup>-1<\/strong>
p(0)=2(0)3<\/sup>+3(0)2<\/sup>-1
p(0)=-1<\/p>\n\n\n\n

p(1)=2(1)3<\/sup>+3(1)2<\/sup>-1
p(1)=2+3-1
p(1)=4<\/p>\n\n\n\n

\n\n\n\n

(vi) p(t)=t4<\/sup>-t2<\/sup>+3<\/strong>
p(0)=(0)4<\/sup>-(0)2<\/sup>+3
p(0)=0-0+3
p(0)=3<\/p>\n\n\n\n

p(1)=(1)4<\/sup>-(1)2<\/sup>+3
p(1)=1-1+3
p(1)=+3<\/p>\n\n\n\n

\n\n\n\n

Q7 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 7<\/h4>\n\n\n\n

Find the value of the polynomial p(x) at x = a when :<\/strong>
(i) p(x)=3x2<\/sup>+8x+4 and a=-2<\/strong>
Sol :
\u21d2p(x)=3x2<\/sup>+8x+4
p(x)=p(a)
\u21d2p(a)=3a2<\/sup>+8a+4
\u21d2p(-2)=3(-2)2<\/sup>+8(-2)+4
\u21d2p(-2)=(3\u00d74)-16+4
\u21d2p(-2)=12-16+4
\u21d2p(-2)=0<\/p>\n\n\n\n

\n\n\n\n

(ii) p(x)=x2<\/sup>+x-6 and a=-3<\/strong>
Sol :
\u21d2p(x)=x2<\/sup>+x-6
p(x)=p(a)
\u21d2p(a)=a2<\/sup>+a-6
\u21d2p(-3)=(-3)2<\/sup>+(-3)-6
\u21d2p(-3)=9-3-6
\u21d2p(-3)=0<\/p>\n\n\n\n

\n\n\n\n

(iii) p(x)=x3<\/sup>-2x+2 and a=-1<\/strong>
Sol :
\u21d2p(x)=x2<\/sup>+x-6
p(x)=p(a)
\u21d2p(a)=a3<\/sup>-2a+2
\u21d2p(-1)=(-1)3<\/sup>-2(-1)+2
\u21d2p(-1)=-1+2+2
\u21d2p(-1)=3<\/p>\n\n\n\n

\n\n\n\n

(iv) p(x)=x3<\/sup>-3x2<\/sup>+x and a=-1<\/strong>
Sol :
\u21d2p(x)=x3<\/sup>-3x2<\/sup>+x
p(x)=p(a)
\u21d2p(a)=a3<\/sup>-3a2<\/sup>+a
\u21d2p(-1)=(-1)3<\/sup>-3(-1)2<\/sup>+(-1)
\u21d2p(-1)=-1-3-1
\u21d2p(-1)=-5<\/p>\n\n\n\n

\n\n\n\n

Q8 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 8<\/h4>\n\n\n\n

If p(x)=x2<\/sup>-5x+4 and q(x)=x3<\/sup>+1 . Find the values of the following :<\/strong>
(i) p(1)\u00d7q(1)<\/strong>
Sol :
\u21d2p(1)=12<\/sup>-5(1)+4
\u21d2p(1)=1-5+4
\u21d2p(1)=0..(i)<\/p>\n\n\n\n

\u21d2q(1)=(1)3<\/sup>+1
\u21d2q(1)=1+1
\u21d2q(1)=2..(ii)<\/p>\n\n\n\n

\u21d2p(1)\u00d7q(1)
From (i) and (ii) , we get
\u21d20\u00d72
\u21d20<\/p>\n\n\n\n

\n\n\n\n

(ii) p(1)q(1)<\/strong>
Sol :
From above p(1)=0 and q(1)=2
\u21d2p(1)q(1)=02
[zero by integer always equal to zero]
\u21d20<\/p>\n\n\n\n

\n\n\n\n

(iii) p(2)+q(2)<\/strong>
Sol :
\u21d2p(x)=x2<\/sup>-5x+4
\u21d2p(2)=22<\/sup>-5(2)+4
\u21d2p(2)=4-10+4
\u21d2p(2)=-2..(i)<\/p>\n\n\n\n

\u21d2q(x)=x3<\/sup>+1
\u21d2q(2)=23<\/sup>+1
\u21d2q(2)=8+1
\u21d2q(2)=9..(ii)<\/p>\n\n\n\n

A.T.Q
\u21d2p(2)+q(2)
putting values of (ii) and (i)
\u21d2-2+9
\u21d27<\/p>\n\n\n\n

\n\n\n\n

Q9 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 9<\/h4>\n\n\n\n

Verify that given values are zeroes of the corresponding polynomials :<\/strong>
(i) p(x)=3x+1 ; x=\u221213<\/strong>
Sol :
Given polynomial p(x)=3x+1
\u21d2p(\u221213)=3(\u221213)+1
\u21d2p(\u221213)=(\u221233)+1
\u21d2p(\u221213)=\u22121+1=0
Hence , it is a zero of the given polynomial<\/p>\n\n\n\n

\n\n\n\n

(ii) p(x)=x2<\/sup>-1 ; x=1 , -1<\/strong>
Sol :
Given polynomial p(x)=x2<\/sup>-1
When x=1
\u21d2p(1)=12<\/sup>-1
\u21d2p(1)=0<\/p>\n\n\n\n

When x=-1
\u21d2p(-1)=(-1)2<\/sup>-1
\u21d2p(-1)=1-1=0
Hence , 1 and -1 both are zeros of polynomial<\/p>\n\n\n\n

\n\n\n\n

(iii) p(x)=x2<\/sup> ; x=0<\/strong>
Sol :
Given polynomial p(x)=x2<\/sup>
\u21d2p(0)=02<\/sup>=0
Hence , 0 is a zero of the given polynomial<\/p>\n\n\n\n

\n\n\n\n

(iv) p(x)=px+q ; x=\u2212qp<\/strong>
Sol :
Given polynomial p(x)=px+q
\u21d2p(\u2212qp)=p(\u2212qp)+q
\u21d2p(\u2212qp)=\u2212q+q=0
Hence , it is the zero of given polynomial<\/p>\n\n\n\n

\n\n\n\n

Q10 | Ex-3.1 | Class 9 | Polynomials | KC SINHA Mathematics | myhelper<\/strong><\/h4>\n\n\n\n

Question 10<\/h4>\n\n\n\n

Find the zeroes of the given polynomial p(x) under given conditions :<\/strong>
(i) p(x)=x+5<\/strong>
Sol :
Given polynomial p(x)=x+5
\u21d2p(x)=0
\u21d2p(x)=x+5=0
\u21d2p(x)=x=-5
\u2234 Zero of polynomial p(x) is -5<\/p>\n\n\n\n

\n\n\n\n

(ii) p(x)=2x-5<\/strong>
Sol :
Given polynomial p(x)=2x-5
\u21d2p(x)=0
\u21d2p(x)=2x-5=0
\u21d2p(x)=2x=+5
\u21d2p(x)=x=+5\/2
\u2234 Zero of polynomial p(x) is 5\/2<\/p>\n\n\n\n

\n\n\n\n

(iii) p(x)=3x-6<\/strong>
Sol :
Given polynomial p(x)=3x-6
\u21d2p(x)=0
\u21d2p(x)=3x-6=0
\u21d2p(x)=3x=+6
\u21d2p(x)=x=+6\/3
\u21d2p(x)=x=2
\u2234 Zero of polynomial p(x) is 2<\/p>\n\n\n\n

\n\n\n\n

(iv) p(x)=5x<\/strong>
Sol :
Given polynomial p(x)=5x
\u21d2p(x)=0
\u21d2p(x)=5x=0
\u21d2p(x)=x=0\/5
[Zero by some integer i s 0]
\u2234 Zero of polynomial p(x) is 0<\/p>\n\n\n\n

\n\n\n\n

(v) p(x)=(x+1)<\/strong>
Sol :
Given polynomial p(x)=x+1
\u21d2p(x)=0
\u21d2p(x)=x+1=0
\u21d2p(x)=x=-1
\u2234 Zero of polynomial p(x) is -1<\/p>\n\n\n\n

\n\n\n\n

(vi) p(x)=ax+b ; a\u22600 , a , b are real numbers<\/strong>
Sol :
Given polynomial p(x)=ax+b
\u21d2p(x)=0
\u21d2p(x)=ax+b=0
\u21d2p(x)=ax=-b
\u21d2p(x)=x=-b\/a
\u2234 Zero of polynomial p(x) is -b\/a<\/p>\n\n\n\n

\n
KC Sinha Solutions<\/a><\/div>\n\n\n\n
KC Sinha Class 9 Solutions<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

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If each term of an algebraic expression contains only non-negative integrals powers of x , then given algebraic expression is […]<\/p>\n","protected":false},"author":302,"featured_media":623828,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"newspack_featured_image_position":"","newspack_post_subtitle":"","newspack_article_summary_title":"Overview:","newspack_article_summary":"","newspack_hide_updated_date":false,"newspack_show_updated_date":false,"footnotes":""},"categories":[921],"tags":[],"boards":[],"class_list":["post-623826","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-class-9","entry"],"yoast_head":"\nKC Sinha: Exercise 3.1 - Mathematics Solution Class 9 Chapter 3 Polynomials - IndCareer Schools<\/title>\n<meta name=\"description\" content=\"Exercise 3.1 Type 1Problems based on definition of terms related to polynomial and identifying polynomialsWORKING RULE:1. 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