NCERT Solutions Class 9 Maths Chapter 2 Polynomial
Chapter 2 of Class 9 Mathematics, "Polynomials," is an essential topic that helps students build a strong math foundation. This chapter covers the basics of polynomials, including their definitions, types, and operations. Understanding polynomials is important because they are the basis for many advanced math concepts studied in higher classes.
The NCERT Solutions for Class 9 Maths Chapter 2 provides clear explanations and step-by-step solutions to textbook exercises, making it easier for students to understand and solve problems related to polynomial expressions, factorisation, and identities. This blog is designed to help students master this chapter, boost their analytical skills, and gain confidence in math as they prepare for exams and future studies.
1.0 Download Maths Class 9 Chapter 2 Polynomials PDF
Downloading the Polynomials Class 9 PDF will help you understand the curriculum's key concepts. Below is the link to download the NCERT solutions for Class 9 Maths Chapter 2 PDF
Class 9 Polynomials NCERT Solution
2.0NCERT Solutions for Class 9 Maths Chapter 2 Polynomials: Overview
Chapter 2 Polynomials is crucial in the Class 9 Maths syllabus. Students are introduced to the basic ideas behind algebraic expressions and their operations. The NCERT Solutions for this chapter make it simpler for students to comprehend and grasp the concepts by offering thorough explanations, examples, and step-by-step solutions to the exercises. Following are the key topics covered in the Polynomial chapter.
Chapter 2 Polynomials Subtopics of NCERT Class 9 Maths
The following is a list of the subjects addressed in CBSE Class 9 Maths Chapter 2:
Polynomials in One Variable
Zeroes of a Polynomial
Factorisation of Polynomials
Remainder Theorem
Factor Theorem
Algebraic Identities
3.0Polynomials in NCERT Solutions for Class 9 Maths, Chapter 2: All Exercises
NCERT Solutions for Class 9 Maths, Chapter 2: Polynomials, covers essential topics such as polynomial definitions, types, and operations, providing a solid foundation for understanding more complex mathematical ideas. The solutions to all exercises are designed to guide students step-by-step. By working through these exercises, learners can enhance their problem-solving skills and gain confidence in their mathematical abilities. You must practice the following number of questions under the NCERT solution for class 9, chapter 2.
Class 9 Polynomials NCERT Solutions Exercise 2.1
1 Question
Class 9 Maths Chapter 2 Exercise 2.2 Solutions
2 Questions
NCERT Class 9 Maths Chapter 2 Exercise 2.3
5 Questions
4.0Advantages of NCERT Solutions for Class 9 Maths Chapter 2
Clear Explanations: The solutions offer comprehensive explanations for each concept, making it easier for students to grasp the underlying principles.
Step-by-Step Solutions: Detailed solutions are provided for all the exercises, guiding students through problem-solving.
Conceptual Clarity: The NCERT Solutions help build a strong foundation in polynomials, which is essential for further studies in mathematics.
Exam Preparation: Practicing with NCERT Solutions can be highly beneficial for preparing for exams as it helps students understand the types of questions that may be asked.
5.0NCERT QUESTIONS WITH SOLUTIONS
6.0Exercise : 2.1
Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2−3x+7
(ii) y2+2
(iii) 3t+t2
(iv) y+y2
(v) x10+y3+t50
Sol. (i) 4x2−3x+7
This expression is a polynomial in one variable x because there is only one variable ( x ) in the expression.
(ii) y2+2
This expression is a polynomial in one variable y because there is only one variable (y) in the expression.
(iii) 3t+t2
The expression is not a polynomial because in the term 3t, the exponent of t is 21, which is not a whole number.
(iv) y+y2=y+2y−1
The expression is not a polynomial because exponent of y is (−1) in term y2 which in not a whole number.
(v) x10+y3+t50
The expression is not a polynomial in one variable, it is a polynomial in 3 variables x,y and t .
Write the coefficients of x2 in each of the following :
(i) 2+x2+x
(ii) 2−x2+x3
(iii) 2πx2+x
(iv) 2x−1
Sol. (i) 2+x2+x
Coefficient of x2=1
(ii) 2−x2+x3
Coefficient of x2=−1
(iii) 2πx2+x
Coefficient of x2=π/2
(iv) 2x−1
Coefficient of x2=0
Give one example each of a binomial of degree 35 , and of a monomial of degree 100.
Sol. One example of a binomial of degree 35 is 3x35−4.
One example of monomial of degree 100 is 5x100.
Write the degree of each of the following polynomials :
(i) 5x3+4x2+7x
(ii) 4−y2
(iii) 5t−7
(iv) 3
Sol. (i) 5x3+4x2+7x
Term with the highest power of x=5x3
Exponent of x in this term =3∴ Degree of this polynomial =3.
(ii) 4−y2
Term with the highest power of y=−y2
Exponent of y in this term =2∴ Degree of this polynomial =2
(iii) 5t−7
Term with highest power of t=5t.
Exponent of t in this term =1∴ Degree of this polynomial =1
(iv) 3
This is a constant which is non-zero
∴ Degree of this polynomial =0
Classify the following as linear, quadratic and cubic polynomials :
(i) x2+x
(ii) x−x3
(iii) y+y2+4
(iv) 1+x
(v) 3t
(vi) r2
(vii) 7x3
Sol.
(i) Quadratic
(ii) Cubic
(iii) Quadratic
(iv) Linear
(v) Linear
(vi) Quadratic
(vii) Cubic
7.0Exercise : 2.2
Find the value of the polynomial 5x−4x2 +3 at
(i) x=0
(ii) x=−1
(iii) x=2
Sol. Let f(x)=5x−4x2+3
(i) Value of f(x) at x=0=f(0)=5(0)−4(0)2+3=3
(ii) Value of f(x) at x=−1=f(−1)=5(−1)−4(−1)2+3=−5−4+3=−6
(iii) Value of f(x) at x=2=f(2)=5(2)−4(2)2+3=10−16+3=−3
Find p(0),p(1) and p(2) for each of the following polynomials :
(i) p(y)=y2−y+1
(ii) p(t)=2+t+2t2−t3
(iii) p(x)=x3
(iv) p(x)=(x−1)(x+1)
Sol. (i) p(y)=y2−y+1∴p(0)=(0)2−(0)+1=1,
p(1)=(1)2−(1)+1=1,
p(2)=(2)2−(2)+1=4−2+1=3.
(ii) p(t)=2+t+2t2−t3p(0)=2+0+2(0)2−(0)3=2p(1)=2+1+2(1)2−(1)3=2+1+2−1=4p(2)=2+2+2(2)2−(2)3=2+2+8−8=4
(iii) p(x)=x3p(0)=(0)3=0p(1)=(1)3=1p(2)=(2)3=8
(iv) p(x)=(x−1)(x+1)p(0)=(0−1)(0+1)=(−1)(1)=−1p(1)=(1−1)(1+1)=0(2)=0p(2)=(2−1)(2+1)=(1)(3)=3
Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1,x=−31
(ii) p(x)=5x−π,x=54
(iii) p(x)=x2−1,x=1,−1
(iv) p(x)=(x+1)(x−2),x=−1,2
(v) p(x)=x2,x=0
(vi) p(x)=ℓx+m,x=−ℓm
(vii) p(x)=3x2−1,x=−31,32
(viii) p(x)=2x+1,x=21
Sol. (i) p(x)=3x+1,x=−31p(−31)=3(−31)+1=−1+1=0−31 is a zero of p(x)
(ii) p(x)=5x−π,x=54p(54)=5(54)−π=4−π=0∴54 is not a zero of p(x)
(iii) p(x)=x2−1,x=1,−1p(1)=(1)2−1=1−1=0p(−1)=(−1)2−1=1−1=0∴1,−1 are zeroes of p(x)
(iv) p(x)=(x+1)(x−2),x=−1,2p(−1)=(−1+1)(−1−2)=(0)(−3)=0p(2)=(2+1)(2−2)=(3)(0)=0∴−1,2 are zeroes of p(x)
(v) p(x)=x2,x=0p(0)=0∴0 is a zero of p(x)
(vi) p(x)=ℓx=m,x=ℓ−mp(ℓ−m)=ℓ(ℓ−m)+m=−m+m=0∴ℓ−m is a zero of p(x).
(vii) p(x)=3x2−1,x=−31,32p(−31)=3(−31)2−1=3(31)−1=1−1=0p(32)=3(32)2−1=3(34)−1=4−1=3=0
So, −31 is a zero of p(x) and 32 is not a zero of p(x).
(viii) p(x)=2x+1,x=21p(21)=2(21)+1=1+1=2=0∴21 is not a zero of p(x).
Find the zero of the polynomial in each of the following cases :
(i) p(x)=x+5
(ii) p(x)=x−5
(iii) p(x)=2x+5
(iv) p(x)=3x−2
(v) p(x)=3x
(vi) p(x)=ax,a=0
(vii) p(x)=cx+d,c=0,c,d are real numbers.
Sol. (i) p(x)=x+5p(x)=0⇒x+5=0⇒x=−5∴−5 is zero of the polynomial p(x).
(ii) p(x)=x−5p(x)=0x−5=0
or x=5∴5 is zero of polynomial p(x).
(iii) p(x)=2x+5p(x)=02x+5=02x=−5⇒x=−25∴−25 is zero of polynomial p(x).
(iv) p(x)=3x−2p(x)=0⇒3x−2=0
or x=32∴32 is zero of polynomial p(x).
(v) p(x)=3xp(x)=0⇒3x=0
or x=0∴0 is zero of polynomial p(x).
(vi) p(x)=ax,a=0⇒ax=0 or x=0∴0 is zero of p(x)
(vii) p(x)=cx+d,c=0,c,d are real numbers
cx+d=0⇒cx=−dx=c−d∴c−d is zero of polynomial p(x).
8.0Exercise : 2.3
Find the remainder when x3+3x2+3x+1 is divided by :
(i) x+1
(ii) x−21
(iii) x
(iv) x+π
(v) 5+2x
Sol. (i) x+1x+1=0⇒x=−1∴ Remainder =p(−1)=(−1)3+3(−1)2+3(− 1) +1=−1+3−3+1=0
(ii) x−21x−21=0⇒x=21∴ Remainder =p(21)=(21)3+3(21)2+3(21)+1=81+43+23+1=827
(iii) x
x=0
Remainder =p(0)=(0)3+3(0)2+3(0)+1=1
(iv) x+πx+π=0⇒x=−π∴ Remainder =p(−π)=(−π)3+3(−π)2+3(−π)+1=−π3+3π2−3π+1
(v) 5+2x5+2x=0⇒x=−5/2∴ Remainder =p(−5/2)=(2−5)3+3(2−5)2+3(2−5)+1=8−125+475−215+1=−827
Find the remainder when x3−ax2+6x−a is divided by x−a.
Sol. Let p(x)=x3−ax2+6x−ax−a=0⇒x=a∴ Remainder =(a)3−a(a)2+6(a)−a=a3−a3+6a−a=5a
Check whether 7+3x is a factor of 3x3+ 7 x .
Sol. 7+3x will be a factor of 3x3+7x only if 7 +3 x divides 3x3+7x leaving 0 as remainder.
Let p(x)=3x3+7x7+3x=0⇒3x=−7⇒x=−7/3∴ Remainder
3(−37)3+7(−37)=9−343−349=9−490=0
so, 7+3x is not a factor of 3x3+7x.
9.0Exercise : 2.4
Determine which of the following polynomials has (x+1) a factor :
(i) x3+x2+x+1
(ii) x4+x3+x2+x+1
(iii) x4+3x3+3x2+x+1
(iv) x3−x2−(2+2)x+2
Sol. (i) x3+x2+x+1
Let p(x)=x3+x2+x+1
The zero of x+1 is -1
p(−1)=(−1)3+(−1)2+(−1)+1=−1+1−1+1=0
By Factor theorem x+1 is a factor of p(x).
(ii) x4+x3+x2+x+1
Let p(x)=x4+x3+x2+x+1
The zero of x+1 is -1
p(−1)=(−1)4+(−1)3+(−1)2+(−1)+1=1=0
By Factor theorem x+1 is not a factor of p(x)
(iii) x4+3x3+3x2+x+1
Let p(x)=x4+3x3+3x2+x+1
Zero of x+1 is -1
p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1=1−3+3−1+1=1=0
By Factor theorem x+1 is not a factor of p(x)
(iv) Let p(x)=x3−x2−(2+2)x+2
zero of x+1 is -1
p(−1)=(−1)3−(−1)2−(2+2)(−1)+2=−1−1+2+2+2=22=0
By Factor theorem, x+1 is not a factor of p(x).
Use the factor theorem to determine whether g(x) is a factor of p(x) in each of the following cases :
(i) p(x)=2x3+x2−2x−1,g(x)=x+1.
(ii) p(x)=x3+3x2+3x+1,g(x)=x+2.
(iii) p(x)=x3−4x2+x+6; g(x)=x−3
Sol. (i) p(x)=2x3+x2−2x−1,g(x)=x+1.
g(x)=0⇒x+1=0⇒x=−1∴ Zero of g(x) is -1
Now, p(−1)=2(−1)3+(−1)2−2(−1)−1=−2+1+2−1=0∴ By Factor theorem, g(x) is a factor of p(x).
(ii) Let p(x)=x3+3x2+3x+1,
g(x)=x+2g(x)=0⇒x+2=0⇒x=−2∴ Zero of g(x) is -2
Now, p(−2)=(−2)3+3(−2)2+3(−2)+1=−8+12−6+1=−1∴ By Factor theorem, g(x) is not a factor of p(x)
(iii) p(x)=x3−4x2+x+6,g(x)=x−3g(x)=0⇒x−3=0⇒x=3∴ Zero of g(x)=3
Now p(3)=33−4(3)2+3+6=27−36+3+6=0∴ By Factor theorem, g(x) is a factor of p(x).
Find the value of k, if x−1 is a factor of p(x) in each of the following cases :
(i) p(x)=x2+x+k
(ii) p(x)=2x2+kx+2
(iii) p(x)=kx2−2x+1
(iv) p(x)=kx2−3x+k
Sol. (i) p(x)=x2+x+k
If x−1 is a factor of p(x), then p(1)=0⇒(1)2+(1)+k=0⇒1+1+k=0⇒2+k=0⇒k=−2
(ii) p(x)=2x2+kx+2
If (x−1) is a factor of p(x) then p(1)=0⇒2(1)2+k(1)+2=0⇒2+k+2=0⇒k=−(2+2)
(iii) p(x)=kx2−2x+1
If (x−1) is a factor of p(x) then p(1)=0k(1)2−2(1)+1=0⇒k−2+1=0k=2−1
(iv) p(x)=kx2−3x+k
If (x−1) is a factor of p(x) then p(1)=0⇒k(1)2−3(1)+k=02k=3k=3/2
Factorise :
(i) 12x2−7x+1
(ii) 2x2+7x+3
(iii) 6x2+5x−6
(iv) 3x2−x−4
Sol. (i) 12x2−7x+1=12x2−4x−3x+1=4x(3x−1)−1(3x−1)=(3x−1)(4x−1)
(ii) 2x2+7x+3=2x2+6x+x+3=2x(x+3)+1(x+3)=(x+3)(2x+1)
(iii) 6x2+5x−6=6x2+9x−4x−6=3x(2x+3)−2(2x+3)=(3x−2)(2x+3)
(iv) 3x2−x−4=3x2−4x+3x−4=x(3x−4)+1(3x−4)=(x+1)(3x−4)
Factorise :
(i) x3−2x2−x+2
(ii) x3−3x2−9x−5
(iii) x3+13x2+32x+20
(iv) 2y3+y2−2y−1
Sol. (i) x3−2x2−x+2
Let p(x)=x3−2x2−x+2
By trial, we find that
p(1)=(1)3−2(1)2−(1)+2=1−2−1+2=0∴ By Factor Theorem, (x−1) is a factor of p(x).
Now, x3−2x2−x+2=x2(x−1)−x(x−1)−2(x−1)=(x−1)(x2−x−2)=(x−1)(x2−2x+x−2)=(x−1){x(x−2)+1(x−2)}=(x−1)(x−2)(x+1)
(ii) x3−3x2−9x−5
Let p(x)=x3−3x2−9x−5
By trial, we find that
p(−1)=(−1)3−3(−1)2−9(−1)−5=−1−3+9−5=0∴ By Factor Theorem, x=−1 or x+1 is factor of p(x).
Now, x3−3x2−9x−5=x2(x+1)−4x(x+1)−5(x+1)=(x+1)(x2−4x−5)=(x+1)(x2−5x+x−5)=(x+1){x(x−5)+1(x−5)}=(x+1)2(x−5)
(iii) x3+13x2+32x+20
Let p(x)=x3+13x2+32x+20
By trial, we find that
p(−1)=(−1)3+13(−1)2+32(−1)+20=−1+13−32+20=0∴ By Factor Theorem, x=−1 or x+1 is a factor of p ( x )
x3+13x2+32x+20=x2(x+1)+12(x)(x+1)+20(x+1)=(x+1)(x2+12x+20)=(x+1)(x2+2x+10x+20)=(x+1){x(x+2)+10(x+2)}=(x+1)(x+2)(x+10)
(iv) 2y3+y2−2y−1p(y)=2y3+y2−2y−1
By trial, we find that
p(1)=2(1)3+(1)2−2(1)−1=0∴ By Factor Theorem, (y−1) is a factor of p(y)=2y3+y2−2y−1=2y2(y−1)+3y(y−1)+1(y−1)=(y−1)(2y2+3y+1)=(y−1)(2y2+2y+y+1)=(y−1){2y(y+1)+1(y+1)}=(y−1)(2y+1)(y+1)
10.0EXERCISE : 2.5
Use suitable identities to find the following products :
(i) (x+4)(x+10)
(ii) (x+8)(x−10)
(iii) (3x+4)(3x−5)
(iv) (y2+23)(y2−23)
(v) (3−2x)(3+2x)
Sol. (i) (x+4)(x+10)=x2+(4+10)x+(4)(10)=x2+14x+40
(ii) (x+8)(x−10)=(x+8){x+(−10)}=x2+{8+(−10)}x+8(−10)=x2−2x−80
(iii) (3x+4)(3x−5)=(3x+4)(3x−5)=(3x+4)(3x+(−5))=(3x)2+{4+(−5)}(3x)+4(−5)=9x2−3x−20
(iv) (y2+23)(y2−23)
Let, y2=x⇒(y2+23)(y2−23)=(x+23)(x−23)=x2−49
[using identity (a+b)(a−b)=a2−b2 ]
Substituting x=y2, we get
=(y2)2−49=y4−49
(v) (3−2x)(3+2x)(3)2−(2x)2=9−4x2
[using identity (a+b)(a−b)=a2−b2 ]
Evaluate the following products without multiplying directly:
(i) 103×107
(ii) 95×96
(iii) 104×96
Sol. (i) 103×107=(100+3)×(100+7)=(100)2+(3+7)(100)+(3)(7)=10000+1000+21=11021
Alternate solution:
103×107=(105−2)×(105+2)=(105)2−(2)2=(100+5)2−4=(100)2+2(100)(5)+(5)2−4=10000+1000+25−4=11021.
(ii) 95×96=(90+5)×(90+6)=(90)2+(5+6)90+(5)(6)=8100+990+30=9120
(iii) 104×96=(100+4)×(100−4)
[using identity (a+b)(a−b)=a2−b2 ]
=(100)2−(4)2=10000−16=9984
Factorise the following using appropriate identities:
(i) 9x2+6xy+y2
(ii) 4y2−4y+1
(iii) x2−100y2
Sol. (i) 9x2+6xy+y2=(3x)2+2(3x)(y)+(y)2=(3x+y)2=(3x+y)(3x+y)
(ii) 4y2−4y+1=(2y)2−2(2y)(1)+(1)2=(2y−1)2=(2y−1)(2y−1)
(iii) x2−100y2
[using identity (a+b)(a−b)=a2−b2 ]
x2−(10y)2=(x+10y)(x−10y)
Expand each of the following using suitable identities :
(i) (x+2y+4z)2
(ii) (2x−y+z)2
(iii) (−2x+3y+2z)2
(iv) (3a−7b−c)2
(v) (−2x+5y−3z)2
(vi) (41a−21b+1)2
Sol. (i) (x+2y+4z)2=(x)2+(2y)2+(4z)2+2(x)(2y)+2(2y)(4z)+2(4z)(x)=x2+4y2+16z2+4xy+16yz+8zx
(ii) (2x−y+z)2=(2x−y+z)(2x−y+z)=(2x)2+(−y)2+(z)2+2(2x)(−y)+2(−
y) (z)+2(z)(2x)=4x2+y2+z2−4xy−2yz+4zx
(iii) (−2x+3y+2z)2=(−2x)2+(3y)2+(2z)2+2(−2x)(3y)+2(−2x)(2z)+2(3y)(2z)=4x2+9y2+4z2−12xy−8xz+12yz
(iv) (3a−7b−c)2=(3a−7b−c)(3a−7b−c)=(3a)2+(−7b)2+(−c)2+2(3a)(−7b)+2(3a)(−c)+2(−7b)(−c)=9a2+49b2+c2−42ab−6ac+14bc
(v) (−2x+5y−3z)2=(−2x+5y−3z)(−2x+5y−3z)=(−2x)2+(5y)2+(−3z)2+2(−2x)(5y)+2(−2x)(−3z)+2(−3z)(5y)=4x2+25y2+9z2−20xy+12xz−30yz
(vi) (41a−21b+1)2=(41a−21b+1)(41a−21b+1)=(41a)2+(−21b)2+(1)2+2(41a)(−21b)+2(41a)(1)2+2(−21b)(1)=161a2+41b2+1−41ab−b+21a
Factorise :
(i) 4x2+9y2+16z2+12xy−24yz−16xz
(ii) 2x2+y2+8z2−22xy+42yz−8xz
Sol. (i) 4x2+9y2+16z2+12xy−24yz−16xz=(2x)2+(3y)2+(−4z)2+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)={2x+3y+(−4z)}2=(2x+3y−4z)2=(2x+3y−4z)(2x+3y−4z)
(ii) 2x2+y2+8z2−22xy+42yz−8xz=(−2x)2+y2+(22z)2+2(−2x)y+2y(22z)+2(22z)(−2x)=(−2x+y+22z)2
Write the following cubes in expanded form :
(i) (2x+1)3
(ii) (2a−3b)3
(iii) (23x+1)3
(iv) (x−32y)3
Sol. (i) (2x+1)3=(2x)3+(1)3+3(2x)(1)(2x+1)=8x3+1+6x(2x+1)=8x3+1+12x2+6x=8x3+12x2+6x+1
(ii) (2a−3b)3=(2a)3−(3b)3−3(2a)(3b)(2a−3b)=8a3−27b3−18ab(2a−3b)=8a3−27b3−36a2b+54ab2
(iii) (23x+1)3=(23x)3+(1)3+3(23x)(1)(23x+1)=827x3+1+427x2+29x=827x3+427x2+29x+1
(iv) (x−32y)3=x3−(32y)3−3x(32y)(x−32y)=x3−278y3−2xy(x−32y)=x3−278y3−2x2y+34x2
Evaluate the following using suitable identities :
(i) (99)3
(ii) (102)3
(iii) (998)3
Sol. (i) (99)3=(100−1)3=(100)3−(1)3−3(100)(1)(100−1)=1000000−1−300(100−1)=1000000−1−30000+300=970299
(ii) (102)3=(100+2)3=(100)3+(2)3+3(100)(2)(100+2)=1000000+8+600(100+2)=1000000+8+60000+1200=1061208
(iii) (998)3=(1000−2)3=(1000)3−(2)3−3(1000)(2)(1000−2)=1000000000−8−6000(1000−2)=994011992
Factorise each of the following :
(i) 8a3+b3+12a2b+6ab2
(ii) 8a3−b3−12a2b+6ab2
(iii) 27−125a3−135a+225a2
(iv) 64a3−27b3−144a2b+108a2
(v) 27p3−2161−29p2+41p
Sol. (i) 8a3+b3+12a2b+6ab2=(2a)3+(b)3+3(2a)(b)(2a+b)=(2a+b)3=(2a+b)(2a+b)(2a+b)
(ii) 8a3−b3−12a2b+6ab2=(2a)3+(−b)3+3(2a)2(−b)+3(2a)(−b)2=(2a−b)3
(iii) 27−125a3−135a+225a2=33−(5a)3−3(3)(5a)(3−5a)=(3−5a)3
(iv) 64a3−27b3−144a2b+180ab2=(4a)3−(3b)3−3(4a)(3b)(4a−3b)=(4a−3b)3
(v) 27p3−2161−29p2+4p1=(3p)3−(61)3−3(3p)(61)(3p−61)=(3p−61)3=(3p−61)(3p−61)(3p−61)
Verify:
(i) x3+y3=(x+y)(x2−xy+y2)
(ii) x3−y3=(x−y)(x2+xy+y2)
Sol. (i) (x+y)3=x3+y3+3xy(x+y)⇒x3+y3=(x+y)3−3xy(x+y)⇒x3+y3=(x+y){(x+y)2−3xy}⇒x3+y3=(x+y)(x2+2xy+y2−3xy)⇒x3+y3=(x+y)(x2−xy+y2)
(ii) (x−y)3=x3−y3−3xy(x−y)⇒x3−y3=(x−y)3+3xy(x−y)⇒x3−y3=(x−y)[(x−y)2+3xy]⇒x3−y3=(x−y)(x2+y2−2xy+3xy)⇒x3−y3=(x−y)(x2+y2+xy)
Factorise each of the following :
(i) 27y3+125z3
(ii) 64m3−343n3
Sol. (i) 27y3+125z3=(3y)3+(5z)3=(3y+5z){(3y)2−(3y)(5z)+(5z)2}=(3y+5z)(9y2−15yz+25z2)
(ii) 64m3−343n3=(4m)3−(7n)3=(4m−7n){16m2+4m.7n+(7n)2}=(4m−7n)(16m2+28mn+49n2)
Verify that x3+y3+z3−3xyz=21(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
Sol. 21(x+y+z)[(x−y)2+(y−z)2+(z−x)2]=21(x+y+z)[(x2−2xy+y2)+(y2−2yz+z2)+(z2−2zx+x2)]=21(x+y+z)(2x2+2y2+2z2−2xy−2yz−2zx)=21(x+y+z)2(x2+y2+z2−xy−yz−zx)=21(x+y+z)(x2+y2+z2−xy−yz−zx)=x3+y3+z3−3xyz
If x+y+z=0, show that x3+y3+z3=3xyz
Sol. We know
x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)
Given: x+y+z=0=(0)(x2+y2+z2−xy−yz−zx)=0
or x3+y3+z3=3xyz
Without actually calculating the cubes, find the value of each of the following :
(i) (−12)3+(7)3+(5)3
(ii) (28)3+(−15)3+(−13)3
Sol. (i) (−12)3+(7)3+(5)3−12+7+5=0(−12)3+(7)3+(5)3=3(−12)(7)(5)=−1260
[using identity]
if a+b+c=0⇒a3+b3+c3=3abc
(ii) (28)3+(−15)3+(−13)328−15−13=0(28)3+(−15)3+(−13)3=3(28)(−15)(−13)=16380
[using identity]
if a+b+c=0⇒a3+b3+c3=3abc
Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given :
(i) Area =25a2−35a+12
(ii) Area =35y2+13y−12
Sol. (i) Area =25a2−35a+12=25a2−20a−15a+12=5a(5a−4)−3(5a−4)=(5a−3)(5a−4) Here, Length =5a−3, Breadth =5a−4
(ii) 35y2+13y−12=35y2+28y−15y−12=7y(5y+4)−3(5y+4)=(5y+4)(7y−3)
Here, Length =5y+4, Breadth =7y−3.
What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume: 3x2−12x
(ii) Volume: 12ky2+8xy−20k
Sol. (i) Volume =3x2−12x=3x(x−4)=3×x×(x−4)∴ Dimensions are 3 units, x -units and ( x−4 ) units
(ii) 12ky2+8ky−20k=4k(3y2+2y−5)=4k(3y2+5y−3y−5)=4k{y(3y+5)−1(3y+5)}=4k(3y+5)(y−1)∴ Dimensions of cuboid are 4k,3y+5,y−1
NCERT Solutions for Class 9 Maths Other Chapters:-
Expressions with one or more terms having non-zero coefficients are called polynomials. Polynomial terms may consist of constants, variables, or both. The exponents of a polynomial should always be whole numbers. Polynomials are another way to represent real numbers.
Yes, practising all the questions is beneficial as they cover various algebraic concepts related to polynomials. Start by solving the examples to understand the steps involved in solving polynomial problems.
Using ALLEN's NCERT solutions for Class 9 Maths Chapter 2 on a regular basis will help you achieve high exam scores. You can learn how to format your exam answers from the solutions. For students to easily understand the concepts, every little detail is covered in an easy-to-understand manner.
The chapter includes important formulas such as the factor theorem, remainder theorem, and key algebraic identities. These are essential for understanding and solving polynomial problems effectively. These are essential for understanding and solving polynomial problems effectively. These are essential for understanding and solving polynomial problems effectively.