NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
NCERT Solutions Class 9 Maths Chapter 2 โ Free PDF Download
NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. These NCERT Solutions are created by BYJUโS expert faculties to help students in the preparation of their board exams. These expert faculties solve and provide the NCERT Solutions for Class 9 so that it would help students to solve the problems comfortably. They give a detailed and stepwise explanation of each answer to the problems given in the exercises in the NCERT textbook for Class 9.
In NCERT Solutions for Class 9, students are introduced to a lot of important topics which will be helpful for those who wish to pursue Mathematics as a subject in further classes. Based on these NCERT Solutions. These solutions help students to prepare for their upcoming Board Exams by covering the whole syllabus which follows NCERT guidelines.Chapter 2- Polynomials
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Exercise 2.1 Page: 32
1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2โ3x+7
Solution:
The equation 4x2โ3x+7 can be written as 4x2โ3x1+7x0
Since x is the only variable in the given equation and the powers of x (i.e., 2, 1 and 0) are whole numbers, we can say that the expression 4x2โ3x+7 is a polynomial in one variable.
(ii) y2+โ2
Solution:
The equation y2+โ2 can be written as y2+โ2y0
Since y is the only variable in the given equation and the powers of y (i.e., 2 and 0) are whole numbers, we can say that the expression y2+โ2 is a polynomial in one variable.
(iii) 3โt+tโ2
Solution:
The equation 3โt+tโ2 can be written as 3t1/2+โ2t
Though, t is the only variable in the given equation, the powers of t (i.e.,1/2) is not a whole number. Hence, we can say that the expression 3โt+tโ2 is not a polynomial in one variable.
(iv) y+2/y
Solution:
The equation y+2/y an be written as y+2y-1
Though, y is the only variable in the given equation, the powers of y (i.e.,-1) is not a whole number. Hence, we can say that the expression y+2/y is not a polynomial in one variable.
(v) x10+y3+t50
Solution:
Here, in the equation x10+y3+t50
Though, the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression
x10+y3+t50. Hence, it is not a polynomial in one variable.
2. Write the coefficients of x2 in each of the following:
(i) 2+x2+x
Solution:
The equation 2+x2+x can be written as 2+(1)x2+x
We know that, coefficient is the number which multiplies the variable.
Here, the number that multiplies the variable x2 is 1
, the coefficients of x2 in 2+x2+x is 1.
(ii) 2โx2+x3
Solution:
The equation 2โx2+x3 can be written as 2+(โ1)x2+x3
We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.
Here, the number that multiplies the variable x2 is -1
the coefficients of x2 in 2โx2+x3 is -1.
(iii) (/2)x2+x
Solution:
The equation (/2)x2 +x can be written as (/2)x2 + x
We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.
Here, the number that multiplies the variable x2 is /2.
the coefficients of x2 in (/2)x2 +x is /2.
(iii)โ2x-1
Solution:
The equation โ2x-1 can be written as 0x2+โ2x-1 [Since 0x2 is 0]
We know that, coefficient is the number (along with its sign, i.e., โ or +) which multiplies the variable.
Here, the number that multiplies the variable x2is 0
, the coefficients of x2 in โ2x-1 is 0.
3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
Binomial of degree 35: A polynomial having two terms and the highest degree 35 is called a binomial of degree 35
Eg., 3x35+5
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100
Eg., 4x100
4. Write the degree of each of the following polynomials:
(i) 5x3+4x2+7x
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, 5x3+4x2+7x = 5x3+4x2+7x1
The powers of the variable x are: 3, 2, 1
the degree of 5x3+4x2+7x is 3 as 3 is the highest power of x in the equation.
(ii) 4โy2
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, in 4โy2,
The power of the variable y is 2
the degree of 4โy2 is 2 as 2 is the highest power of y in the equation.
(iii) 5tโโ7
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, in 5tโโ7 ,
The power of the variable t is: 1
the degree of 5tโโ7 is 1 as 1 is the highest power of y in the equation.
(iv) 3
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, 3 = 3ร1 = 3ร x0
The power of the variable here is: 0
the degree of 3 is 0.
5. Classify the following as linear, quadratic and cubic polynomials:
Solution:
We know that,
Linear polynomial: A polynomial of degree one is called a linear polynomial.
Quadratic polynomial: A polynomial of degree two is called a quadratic polynomial.
Cubic polynomial: A polynomial of degree three is called a cubic polynomial.
(i) x2+x
Solution:
The highest power of x2+x is 2
the degree is 2
Hence, x2+x is a quadratic polynomial
(ii) xโx3
Solution:
The highest power of xโx3 is 3
the degree is 3
Hence, xโx3 is a cubic polynomial
(iii) y+y2+4
Solution:
The highest power of y+y2+4 is 2
the degree is 2
Hence, y+y2+4is a quadratic polynomial
(iv) 1+x
Solution:
The highest power of 1+x is 1
the degree is 1
Hence, 1+x is a linear polynomial.
(v) 3t
Solution:
The highest power of 3t is 1
the degree is 1
Hence, 3t is a linear polynomial.
(vi) r2
Solution:
The highest power of r2 is 2
the degree is 2
Hence, r2is a quadratic polynomial.
(vii) 7x3
Solution:
The highest power of 7x3 is 3
the degree is 3
Hence, 7x3 is a cubic polynomial.
Exercise 2.2 Page: 34
1. Find the value of the polynomial (x)=5xโ4x2+3
(i) x = 0
(ii) x = โ 1
(iii) x = 2
Solution:
Let f(x) = 5xโ4x2+3
(i) When x = 0
f(0) = 5(0)-4(0)2+3
= 3
(ii) When x = -1
f(x) = 5xโ4x2+3
f(โ1) = 5(โ1)โ4(โ1)2+3
= โ5โ4+3
= โ6
(iii) When x = 2
f(x) = 5xโ4x2+3
f(2) = 5(2)โ4(2)2+3
= 10โ16+3
= โ3
2. Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y)=y2โy+1
Solution:
p(y) = y2โy+1
โดp(0) = (0)2โ(0)+1=1
p(1) = (1)2โ(1)+1=1
p(2) = (2)2โ(2)+1=3
(ii) p(t)=2+t+2t2โt3
Solution:
p(t) = 2+t+2t2โt3
โดp(0) = 2+0+2(0)2โ(0)3=2
p(1) = 2+1+2(1)2โ(1)3=2+1+2โ1=4
p(2) = 2+2+2(2)2โ(2)3=2+2+8โ8=4
(iii) p(x)=x3
Solution:
p(x) = x3
โดp(0) = (0)3 = 0
p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) P(x) = (xโ1)(x+1)
Solution:
p(x) = (xโ1)(x+1)
โดp(0) = (0โ1)(0+1) = (โ1)(1) = โ1
p(1) = (1โ1)(1+1) = 0(2) = 0
p(2) = (2โ1)(2+1) = 1(3) = 3
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1, x=โ1/3
Solution:
For, x = -1/3, p(x) = 3x+1
โดp(โ1/3) = 3(-1/3)+1 = โ1+1 = 0
โด -1/3 is a zero of p(x).
(ii) p(x)=5xโฯ, x = 4/5
Solution:
For, x = 4/5, p(x) = 5xโฯ
โด p(4/5) = 5(4/5)- = 4-
โด 4/5 is not a zero of p(x).
(iii) p(x)=x2โ1, x=1, โ1
Solution:
For, x = 1, โ1;
p(x) = x2โ1
โดp(1)=12โ1=1โ1 = 0
p(โ1)=(-1)2โ1 = 1โ1 = 0
โด1, โ1 are zeros of p(x).
(iv) p(x) = (x+1)(xโ2), x =โ1, 2
Solution:
For, x = โ1,2;
p(x) = (x+1)(xโ2)
โดp(โ1) = (โ1+1)(โ1โ2)
= (0)(โ3) = 0
p(2) = (2+1)(2โ2) = (3)(0) = 0
โดโ1,2 are zeros of p(x).
(v) p(x) = x2, x = 0
Solution:
For, x = 0 p(x) = x2
p(0) = 02 = 0
โด 0 is a zero of p(x).
(vi) p(x) = lx+m, x = โm/l
Solution:
For, x = -m/l ; p(x) = lx+m
โด p(-m/l)= l(-m/l)+m = โm+m = 0
โด-m/l is a zero of p(x).
(vii) p(x) = 3x2โ1, x = -1/โ3 , 2/โ3
Solution:
For, x = -1/โ3 , 2/โ3 ; p(x) = 3x2โ1
โดp(-1/โ3) = 3(-1/โ3)2-1 = 3(1/3)-1 = 1-1 = 0
โดp(2/โ3 ) = 3(2/โ3)2-1 = 3(4/3)-1 = 4โ1=3 โ 0
โด-1/โ3 is a zero of p(x) but 2/โ3 is not a zero of p(x).
(viii) p(x) =2x+1, x = 1/2
Solution:
For, x = 1/2 p(x) = 2x+1
โด p(1/2)=2(1/2)+1 = 1+1 = 2โ 0
โด1/2 is not a zero of p(x).
4. Find the zero of the polynomials in each of the following cases:
(i) p(x) = x+5
Solution:
p(x) = x+5
โ x+5 = 0
โ x = โ5
โด -5 is a zero polynomial of the polynomial p(x).
(ii) p(x) = xโ5
Solution:
p(x) = xโ5
โ xโ5 = 0
โ x = 5
โด 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x+5
Solution:
p(x) = 2x+5
โ 2x+5 = 0
โ 2x = โ5
โ x = -5/2
โดx = -5/2 is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3xโ2
Solution:
p(x) = 3xโ2
โ 3xโ2 = 0
โ 3x = 2
โx = 2/3
โดx = 2/3 is a zero polynomial of the polynomial p(x).
(v) p(x) = 3x
Solution:
p(x) = 3x
โ 3x = 0
โ x = 0
โด0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a0
Solution:
p(x) = ax
โ ax = 0
โ x = 0
โดx = 0 is a zero polynomial of the polynomial p(x).
(vii)p(x) = cx+d, c โ 0, c, d are real numbers.
Solution:
p(x) = cx + d
โ cx+d =0
โ x = -d/c
โด x = -d/c is a zero polynomial of the polynomial p(x).
Exercise 2.3 Page: 40
1. Find the remainder when x3+3x2+3x+1 is divided by
(i) x+1
Solution:
x+1= 0
โx = โ1
โดRemainder:
p(โ1) = (โ1)3+3(โ1)2+3(โ1)+1
= โ1+3โ3+1
= 0
(ii) xโ1/2
Solution:
x-1/2 = 0
โ x = 1/2
โดRemainder:
p(1/2) = (1/2)3+3(1/2)2+3(1/2)+1
= (1/8)+(3/4)+(3/2)+1
= 27/8
(iii) x
Solution:
x = 0
โดRemainder:
p(0) = (0)3+3(0)2+3(0)+1
= 1
(iv) x+ฯ
Solution:
x+ฯ = 0
โ x = โฯ
โดRemainder:
p(0) = (โฯ)3 +3(โฯ)2+3(โฯ)+1
= โฯ3+3ฯ2โ3ฯ+1
(v) 5+2x
Solution:
5+2x=0
โ 2x = โ5
โ x = -5/2
โดRemainder:
(-5/2)3+3(-5/2)2+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1
= -27/8
2. Find the remainder when x3โax2+6xโa is divided by x-a.
Solution:
Let p(x) = x3โax2+6xโa
xโa = 0
โดx = a
Remainder:
p(a) = (a)3โa(a2)+6(a)โa
= a3โa3+6aโa = 5a
3. Check whether 7+3x is a factor of 3x3+7x.
Solution:
7+3x = 0
โ 3x = โ7
โ x = -7/3
โดRemainder:
3(-7/3)3+7(-7/3) = -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 โ 0
โด7+3x is not a factor of 3x3+7x
Exercise 2.4 Page: 43
1. Determine which of the following polynomials has (x + 1) a factor:
(i) x3+x2+x+1
Solution:
Let p(x) = x3+x2+x+1
The zero of x+1 is -1. [x+1 = 0 means x = -1]
p(โ1) = (โ1)3+(โ1)2+(โ1)+1
= โ1+1โ1+1
= 0
โดBy factor theorem, x+1 is a factor of x3+x2+x+1
(ii) x4+x3+x2+x+1
Solution:
Let p(x)= x4+x3+x2+x+1
The zero of x+1 is -1. . [x+1= 0 means x = -1]
p(โ1) = (โ1)4+(โ1)3+(โ1)2+(โ1)+1
= 1โ1+1โ1+1
= 1 โ 0
โดBy factor theorem, x+1 is not a factor of x4 + x3 + x2 + x + 1
(iii) x4+3x3+3x2+x+1
Solution:
Let p(x)= x4+3x3+3x2+x+1
The zero of x+1 is -1.
p(โ1)=(โ1)4+3(โ1)3+3(โ1)2+(โ1)+1
=1โ3+3โ1+1
=1 โ 0
โดBy factor theorem, x+1 is not a factor of x4+3x3+3x2+x+1
(iv) x3 โ x2โ (2+โ2)x +โ2
Solution:
Let p(x) = x3โx2โ(2+โ2)x +โ2
The zero of x+1 is -1.
p(โ1) = (-1)3โ(-1)2โ(2+โ2)(-1) + โ2 = โ1โ1+2+โ2+โ2
= 2โ2 โ 0
โดBy factor theorem, x+1 is not a factor of x3โx2โ(2+โ2)x +โ2
2. 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+x2โ2xโ1, g(x) = x+1
Solution:
p(x) = 2x3+x2โ2xโ1, g(x) = x+1
g(x) = 0
โ x+1 = 0
โ x = โ1
โดZero of g(x) is -1.
Now,
p(โ1) = 2(โ1)3+(โ1)2โ2(โ1)โ1
= โ2+1+2โ1
= 0
โดBy factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x3+3x2+3x+1, g(x) = x+2
Solution:
p(x) = x3+3x2+3x+1, g(x) = x+2
g(x) = 0
โ x+2 = 0
โ x = โ2
โด Zero of g(x) is -2.
Now,
p(โ2) = (โ2)3+3(โ2)2+3(โ2)+1
= โ8+12โ6+1
= โ1 โ 0
โดBy factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x3โ4x2+x+6, g(x) = xโ3
Solution:
p(x) = x3โ4x2+x+6, g(x) = x -3
g(x) = 0
โ xโ3 = 0
โ x = 3
โด Zero of g(x) is 3.
Now,
p(3) = (3)3โ4(3)2+(3)+6
= 27โ36+3+6
= 0
โดBy factor theorem, g(x) is a factor of p(x).
3. Find the value of k, if xโ1 is a factor of p(x) in each of the following cases:
(i) p(x) = x2+x+k
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
โ (1)2+(1)+k = 0
โ 1+1+k = 0
โ 2+k = 0
โ k = โ2
(ii) p(x) = 2x2+kx+โ2
Solution:
If x-1 is a factor of p(x), then p(1)=0
โ 2(1)2+k(1)+โ2 = 0
โ 2+k+โ2 = 0
โ k = โ(2+โ2)
(iii) p(x) = kx2โโ2x+1
Solution:
If x-1 is a factor of p(x), then p(1)=0
By Factor Theorem
โ k(1)2-โ2(1)+1=0
โ k = โ2-1
(iv) p(x)=kx2โ3x+k
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
โ k(1)2โ3(1)+k = 0
โ kโ3+k = 0
โ 2kโ3 = 0
โ k= 3/2
4. Factorize:
(i) 12x2โ7x+1
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1ร12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3ร-4 = 12]
12x2โ7x+1= 12x2-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)
(ii) 2x2+7x+3
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2ร3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6ร1 = 6]
2x2+7x+3 = 2x2+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x2+5x-6
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6ร-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4ร9 = -36]
6x2+5x-6 = 6x2+9xโ4xโ6
= 3x(2x+3)โ2(2x+3)
= (2x+3)(3xโ2)
(iv) 3x2โxโ4
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3ร-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4ร3 = -12]
3x2โxโ4 = 3x2โxโ4
= 3x2โ4x+3xโ4
= x(3xโ4)+1(3xโ4)
= (3xโ4)(x+1)
5. Factorize:
(i) x3โ2x2โx+2
Solution:
Let p(x) = x3โ2x2โx+2
Factors of 2 are ยฑ1 and ยฑ 2
Now,
p(x) = x3โ2x2โx+2
p(โ1) = (โ1)3โ2(โ1)2โ(โ1)+2
= โ1โ2+1+2
= 0
Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor ร Quotient + Remainder
(x+1)(x2โ3x+2) = (x+1)(x2โxโ2x+2)
= (x+1)(x(xโ1)โ2(xโ1))
= (x+1)(xโ1)(x-2)
(ii) x3โ3x2โ9xโ5
Solution:
Let p(x) = x3โ3x2โ9xโ5
Factors of 5 are ยฑ1 and ยฑ5
By trial method, we find that
p(5) = 0
So, (x-5) is factor of p(x)
Now,
p(x) = x3โ3x2โ9xโ5
p(5) = (5)3โ3(5)2โ9(5)โ5
= 125โ75โ45โ5
= 0
Therefore, (x-5) is the factor of p(x)
Now, Dividend = Divisor ร Quotient + Remainder
(xโ5)(x2+2x+1) = (xโ5)(x2+x+x+1)
= (xโ5)(x(x+1)+1(x+1))
= (xโ5)(x+1)(x+1)
(iii) x3+13x2+32x+20
Solution:
Let p(x) = x3+13x2+32x+20
Factors of 20 are ยฑ1, ยฑ2, ยฑ4, ยฑ5, ยฑ10 and ยฑ20
By trial method, we find that
p(-1) = 0
So, (x+1) is factor of p(x)
Now,
p(x)= x3+13x2+32x+20
p(-1) = (โ1)3+13(โ1)2+32(โ1)+20
= โ1+13โ32+20
= 0
Therefore, (x+1) is the factor of p(x)
Now, Dividend = Divisor ร Quotient +Remainder
(x+1)(x2+12x+20) = (x+1)(x2+2x+10x+20)
= (xโ5)x(x+2)+10(x+2)
= (xโ5)(x+2)(x+10)
(iv) 2y3+y2โ2yโ1
Solution:
Let p(y) = 2y3+y2โ2yโ1
Factors = 2ร(โ1)= -2 are ยฑ1 and ยฑ2
By trial method, we find that
p(1) = 0
So, (y-1) is factor of p(y)
Now,
p(y) = 2y3+y2โ2yโ1
p(1) = 2(1)3+(1)2โ2(1)โ1
= 2+1โ2
= 0
Therefore, (y-1) is the factor of p(y)
Now, Dividend = Divisor ร Quotient + Remainder
(yโ1)(2y2+3y+1) = (yโ1)(2y2+2y+y+1)
= (yโ1)(2y(y+1)+1(y+1))
= (yโ1)(2y+1)(y+1)
Exercise 2.5 Page: 48
1. Use suitable identities to find the following products:
(i) (x+4)(x +10)
Solution:
Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab[Here, a = 4 and b = 10]
We get,
(x+4)(x+10) = x2+(4+10)x+(4ร10)
= x2+14x+40
(ii) (x+8)(x โ10)
Solution:
Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab[Here, a = 8 and b = โ10]
We get,
(x+8)(xโ10) = x2+(8+(โ10))x+(8ร(โ10))
= x2+(8โ10)xโ80
= x2โ2xโ80
(iii) (3x+4)(3xโ5)
Solution:
Using the identity, (x+a)(x+b) = x 2+(a+b)x+ab[Here, x = 3x, a = 4 and b = โ5]
We get,
(3x+4)(3xโ5) = (3x)2+[4+(โ5)]3x+4ร(โ5)
= 9x2+3x(4โ5)โ20
= 9x2โ3xโ20
(iv) (y2+3/2)(y2-3/2)
Solution:
Using the identity, (x+y)(xโy) = x2โy 2[Here, x = y2and y = 3/2]
We get,
(y2+3/2)(y2โ3/2) = (y2)2โ(3/2)2
= y4โ9/4
2. Evaluate the following products without multiplying directly:
(i) 103ร107
Solution:
103ร107= (100+3)ร(100+7)
Using identity, [(x+a)(x+b) = x2+(a+b)x+ab
Here, x = 100
a = 3
b = 7
We get, 103ร107 = (100+3)ร(100+7)
= (100)2+(3+7)100+(3ร7))
= 10000+1000+21
= 11021
(ii) 95ร96
Solution:
95ร96 = (100-5)ร(100-4)
Using identity, [(x-a)(x-b) = x2-(a+b)x+ab
Here, x = 100
a = -5
b = -4
We get, 95ร96 = (100-5)ร(100-4)
= (100)2+100(-5+(-4))+(-5ร-4)
= 10000-900+20
= 9120
(iii) 104ร96
Solution:
104ร96 = (100+4)ร(100โ4)
Using identity, [(a+b)(a-b)= a2-b2]
Here, a = 100
b = 4
We get, 104ร96 = (100+4)ร(100โ4)
= (100)2โ(4)2
= 10000โ16
= 9984
3. Factorize the following using appropriate identities:
(i) 9x2+6xy+y2
Solution:
9x2+6xy+y2 = (3x)2+(2ร3xรy)+y2
Using identity, x2+2xy+y2 = (x+y)2
Here, x = 3x
y = y
9x2+6xy+y2 = (3x)2+(2ร3xรy)+y2
= (3x+y)2
= (3x+y)(3x+y)
(ii) 4y2โ4y+1
Solution:
4y2โ4y+1 = (2y)2โ(2ร2yร1)+1
Using identity, x2 โ 2xy + y2 = (x โ y)2
Here, x = 2y
y = 1
4y2โ4y+1 = (2y)2โ(2ร2yร1)+12
= (2yโ1)2
= (2yโ1)(2yโ1)
(iii) x2โy2/100
Solution:
x2โy2/100 = x2โ(y/10)2
Using identity, x2-y2 = (x-y)(x+y)
Here, x = x
y = y/10
x2โy2/100 = x2โ(y/10)2
= (xโy/10)(x+y/10)
4. Expand each of the following, using suitable identities:
(i) (x+2y+4z)2
(ii) (2xโy+z)2
(iii) (โ2x+3y+2z)2
(iv) (3a โ7bโc)2
(v) (โ2x+5yโ3z)2
((1/4)a-(1/2)b +1)2
Solution:
(i) (x+2y+4z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)2 = x2+(2y)2+(4z)2+(2รxร2y)+(2ร2yร4z)+(2ร4zรx)
= x2+4y2+16z2+4xy+16yz+8xz
(ii) (2xโy+z)2
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 2x
y = โy
z = z
(2xโy+z)2 = (2x)2+(โy)2+z2+(2ร2xรโy)+(2รโyรz)+(2รzร2x)
= 4x2+y2+z2โ4xyโ2yz+4xz
(iii) (โ2x+3y+2z)2
Solution:
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = โ2x
y = 3y
z = 2z
(โ2x+3y+2z)2 = (โ2x)2+(3y)2+(2z)2+(2รโ2xร3y)+(2ร3yร2z)+(2ร2zรโ2x)
= 4x2+9y2+4z2โ12xy+12yzโ8xz
(iv) (3a โ7bโc)2
Solution:
Using identity (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = 3a
y = โ 7b
z = โ c
(3a โ7bโ c)2 = (3a)2+(โ 7b)2+(โ c)2+(2ร3a รโ 7b)+(2รโ 7b รโ c)+(2รโ c ร3a)
= 9a2 + 49b2 + c2โ 42ab+14bcโ6ca
(v) (โ2x+5yโ3z)2
Solution:
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = โ2x
y = 5y
z = โ 3z
(โ2x+5yโ3z)2 = (โ2x)2+(5y)2+(โ3z)2+(2รโ2x ร 5y)+(2ร 5yรโ 3z)+(2รโ3z รโ2x)
= 4x2+25y2 +9z2โ 20xyโ30yz+12zx
(vi) ((1/4)a-(1/2)b+1)2
Solution:
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z = 1
5. Factorize:
(i) 4x2+9y2+16z2+12xyโ24yzโ16xz
(ii ) 2x2+y2+8z2โ2โ2xy+4โ2yzโ8xz
Solution:
(i) 4x2+9y2+16z2+12xyโ24yzโ16xz
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
4x2+9y2+16z2+12xyโ24yzโ16xz = (2x)2+(3y)2+(โ4z)2+(2ร2xร3y)+(2ร3yรโ4z)+(2รโ4zร2x)
= (2x+3yโ4z)2
= (2x+3yโ4z)(2x+3yโ4z)
(ii) 2x2+y2+8z2โ2โ2xy+4โ2yzโ8xz
Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx = (x+y+z)2
2x2+y2+8z2โ2โ2xy+4โ2yzโ8xz
= (-โ2x)2+(y)2+(2โ2z)2+(2ร-โ2xรy)+(2รyร2โ2z)+(2ร2โ2รโโ2x)
= (โโ2x+y+2โ2z)2
= (โโ2x+y+2โ2z)(โโ2x+y+2โ2z)
6. Write the following cubes in expanded form:
(i) (2x+1)3
(ii) (2aโ3b)3
(iii) ((3/2)x+1)3
(iv) (xโ(2/3)y)3
Solution:
(i) (2x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
(2x+1)3= (2x)3+13+(3ร2xร1)(2x+1)
= 8x3+1+6x(2x+1)
= 8x3+12x2+6x+1
(ii) (2aโ3b)3
Using identity,(xโy)3 = x3โy3โ3xy(xโy)
(2aโ3b)3 = (2a)3โ(3b)3โ(3ร2aร3b)(2aโ3b)
= 8a3โ27b3โ18ab(2aโ3b)
= 8a3โ27b3โ36a2b+54ab2
(iii) ((3/2)x+1)3
Using identity,(x+y)3 = x3+y3+3xy(x+y)
((3/2)x+1)3=((3/2)x)3+13+(3ร(3/2)xร1)((3/2)x +1)
(iv) (xโ(2/3)y)3
Using identity, (x โy)3 = x3โy3โ3xy(xโy)
7. Evaluate the following using suitable identities:
(i) (99)3
(ii) (102)3
(iii) (998)3
Solutions:
(i) (99)3
Solution:
We can write 99 as 100โ1
Using identity, (x โy)3 = x3โy3โ3xy(xโy)
(99)3 = (100โ1)3
= (100)3โ13โ(3ร100ร1)(100โ1)
= 1000000 โ1โ300(100 โ 1)
= 1000000โ1โ30000+300
= 970299
(ii) (102)3
Solution:
We can write 102 as 100+2
Using identity,(x+y)3 = x3+y3+3xy(x+y)
(100+2)3 =(100)3+23+(3ร100ร2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)3
Solution:
We can write 99 as 1000โ2
Using identity,(xโy)3 = x3โy3โ3xy(xโy)
(998)3 =(1000โ2)3
=(1000)3โ23โ(3ร1000ร2)(1000โ2)
= 1000000000โ8โ6000(1000โ 2)
= 1000000000โ8- 6000000+12000
= 994011992
8. Factorise each of the following:
(i) 8a3+b3+12a2b+6ab2
(ii) 8a3โb3โ12a2b+6ab2
(iii) 27โ125a3โ135a +225a2
(iv) 64a3โ27b3โ144a2b+108ab2
(v) 27p3โ(1/216)โ(9/2) p2+(1/4)p
Solutions:
(i) 8a3+b3+12a2b+6ab2
Solution:
The expression, 8a3+b3+12a2b+6ab2 can be written as (2a)3+b3+3(2a)2b+3(2a)(b)2
8a3+b3+12a2b+6ab2 = (2a)3+b3+3(2a)2b+3(2a)(b)2
= (2a+b)3
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)3 = x3+y3+3xy(x+y) is used.
(ii) 8a3โb3โ12a2b+6ab2
Solution:
The expression, 8a3โb3โ12a2b+6ab2 can be written as (2a)3โb3โ3(2a)2b+3(2a)(b)2
8a3โb3โ12a2b+6ab2 = (2a)3โb3โ3(2a)2b+3(2a)(b)2
= (2aโb)3
= (2aโb)(2aโb)(2aโb)
Here, the identity,(xโy)3 = x3โy3โ3xy(xโy) is used.
(iii) 27โ125a3โ135a+225a2
Solution:
The expression, 27โ125a3โ135a +225a2 can be written as 33โ(5a)3โ3(3)2(5a)+3(3)(5a)2
27โ125a3โ135a+225a2 =
33โ(5a)3โ3(3)2(5a)+3(3)(5a)2
= (3โ5a)3
= (3โ5a)(3โ5a)(3โ5a)
Here, the identity, (xโy)3 = x3โy3-3xy(xโy) is used.
(iv) 64a3โ27b3โ144a2b+108ab2
Solution:
The expression, 64a3โ27b3โ144a2b+108ab2can be written as (4a)3โ(3b)3โ3(4a)2(3b)+3(4a)(3b)2
64a3โ27b3โ144a2b+108ab2=
(4a)3โ(3b)3โ3(4a)2(3b)+3(4a)(3b)2
=(4aโ3b)3
=(4aโ3b)(4aโ3b)(4aโ3b)
Here, the identity, (x โ y)3 = x3 โ y3 โ 3xy(x โ y) is used.
(v) 7p3โ (1/216)โ(9/2) p2+(1/4)p
Solution:
The expression, 27p3โ(1/216)โ(9/2) p2+(1/4)p
can be written as (3p)3โ(1/6)3โ3(3p)2(1/6)+3(3p)(1/6)2
27p3โ(1/216)โ(9/2) p2+(1/4)p =
(3p)3โ(1/6)3โ3(3p)2(1/6)+3(3p)(1/6)2
= (3pโ16)3
= (3pโ16)(3pโ16)(3pโ16)
9. Verify:
(i) x3+y3 = (x+y)(x2โxy+y2)
(ii) x3โy3 = (xโy)(x2+xy+y2)
Solutions:
(i) x3+y3 = (x+y)(x2โxy+y2)
We know that, (x+y)3 = x3+y3+3xy(x+y)
โ x3+y3 = (x+y)3โ3xy(x+y)
โ x3+y3 = (x+y)[(x+y)2โ3xy]
Taking (x+y) common โ x3+y3 = (x+y)[(x2+y2+2xy)โ3xy]
โ x3+y3 = (x+y)(x2+y2โxy)
(ii) x3โy3 = (xโy)(x2+xy+y2)
We know that,(xโy)3 = x3โy3โ3xy(xโy)
โ x3โy3 = (xโy)3+3xy(xโy)
โ x3โy3 = (xโy)[(xโy)2+3xy]
Taking (x+y) common โ x3โy3 = (xโy)[(x2+y2โ2xy)+3xy]
โ x3+y3 = (xโy)(x2+y2+xy)
10. Factorize each of the following:
(i) 27y3+125z3
(ii) 64m3โ343n3
Solutions:
(i) 27y3+125z3
The expression, 27y3+125z3 can be written as (3y)3+(5z)3
27y3+125z3 = (3y)3+(5z)3
We know that, x3+y3 = (x+y)(x2โxy+y2)
27y3+125z3 = (3y)3+(5z)3
= (3y+5z)[(3y)2โ(3y)(5z)+(5z)2]
= (3y+5z)(9y2โ15yz+25z2)
(ii) 64m3โ343n3
The expression, 64m3โ343n3can be written as (4m)3โ(7n)3
64m3โ343n3 =
(4m)3โ(7n)3
We know that, x3โy3 = (xโy)(x2+xy+y2)
64m3โ343n3 = (4m)3โ(7n)3
= (4m-7n)[(4m)2+(4m)(7n)+(7n)2]
= (4m-7n)(16m2+28mn+49n2)
11. Factorise: 27x3+y3+z3โ9xyz
Solution:
The expression27x3+y3+z3โ9xyz can be written as (3x)3+y3+z3โ3(3x)(y)(z)
27x3+y3+z3โ9xyz = (3x)3+y3+z3โ3(3x)(y)(z)
We know that, x3+y3+z3โ3xyz = (x+y+z)(x2+y2+z2โxy โyzโzx)
27x3+y3+z3โ9xyz = (3x)3+y3+z3โ3(3x)(y)(z)
= (3x+y+z)[(3x)2+y2+z2โ3xyโyzโ3xz]
= (3x+y+z)(9x2+y2+z2โ3xyโyzโ3xz)
12. Verify that:
x3+y3+z3โ3xyz = (1/2) (x+y+z)[(xโy)2+(yโz)2+(zโx)2]
Solution:
We know that,
x3+y3+z3โ3xyz = (x+y+z)(x2+y2+z2โxyโyzโxz)
โ x3+y3+z3โ3xyz = (1/2)(x+y+z)[2(x2+y2+z2โxyโyzโxz)]
= (1/2)(x+y+z)(2x2+2y2+2z2โ2xyโ2yzโ2xz)
= (1/2)(x+y+z)[(x2+y2โ2xy)+(y2+z2โ2yz)+(x2+z2โ2xz)]
= (1/2)(x+y+z)[(xโy)2+(yโz)2+(zโx)2]
13. If x+y+z = 0, show that x3+y3+z3 = 3xyz.
Solution:
We know that,
x3+y3+z3-3xyz = (x +y+z)(x2+y2+z2โxyโyzโxz)
Now, according to the question, let (x+y+z) = 0,
then, x3+y3+z3 -3xyz = (0)(x2+y2+z2โxyโyzโxz)
โ x3+y3+z3โ3xyz = 0
โ x3+y3+z3 = 3xyz
Hence Proved
14. Without actually calculating the cubes, find the value of each of the following:
(i) (โ12)3+(7)3+(5)3
(ii) (28)3+(โ15)3+(โ13)3
Solution:
(i) (โ12)3+(7)3+(5)3
Let a = โ12
b = 7
c = 5
We know that if x+y+z = 0, then x3+y3+z3=3xyz.
Here, โ12+7+5=0
(โ12)3+(7)3+(5)3 = 3xyz
= 3ร-12ร7ร5
= -1260
(ii) (28)3+(โ15)3+(โ13)3
Solution:
(28)3+(โ15)3+(โ13)3
Let a = 28
b = โ15
c = โ13
We know that if x+y+z = 0, then x3+y3+z3 = 3xyz.
Here, x+y+z = 28โ15โ13 = 0
(28)3+(โ15)3+(โ13)3 = 3xyz
= 0+3(28)(โ15)(โ13)
= 16380
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a2โ35a+12
(ii) Area : 35y2+13yโ12
Solution:
(i) Area : 25a2โ35a+12
Using the splitting the middle term method,
We have to find a number whose sum = -35 and product =25ร12=300
We get -15 and -20 as the numbers [-15+-20=-35 and -15ร-20=300]
25a2โ35a+12 = 25a2โ15aโ20a+12
= 5a(5aโ3)โ4(5aโ3)
= (5aโ4)(5aโ3)
Possible expression for length = 5aโ4
Possible expression for breadth = 5a โ3
(ii) Area : 35y2+13yโ12
Using the splitting the middle term method,
We have to find a number whose sum = 13 and product = 35ร-12 = 420
We get -15 and 28 as the numbers [-15+28 = 13 and -15ร28=420]
35y2+13yโ12 = 35y2โ15y+28yโ12
= 5y(7yโ3)+4(7yโ3)
= (5y+4)(7yโ3)
Possible expression for length = (5y+4)
Possible expression for breadth = (7yโ3)
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume : 3x2โ12x
(ii) Volume : 12ky2+8kyโ20k
Solution:
(i) Volume : 3x2โ12x
3x2โ12x can be written as 3x(xโ4) by taking 3x out of both the terms.
Possible expression for length = 3
Possible expression for breadth = x
Possible expression for height = (xโ4)
(ii) Volume:
12ky2+8kyโ20k
12ky2+8kyโ20k can be written as 4k(3y2+2yโ5) by taking 4k out of both the terms.
12ky2+8kyโ20k = 4k(3y2+2yโ5)[Here, 3y2+2yโ5 can be written as 3y2+5yโ3yโ5 using splitting the middle term method.]
= 4k(3y2+5yโ3yโ5)
= 4k[y(3y+5)โ1(3y+5)]
= 4k(3y+5)(yโ1)
Possible expression for length = 4k
Possible expression for breadth = (3y +5)
Possible expression for height = (y -1)
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Summary
As this is one of the important topics in Maths, it comes under the unit โ Algebra which has a weightage of 20 marks in Class 9 Maths board exams. This chapter talks about:
- Polynomials in One Variable
- Zeroes of a Polynomial
- Remainder Theorem
- Factorization of Polynomials
- Algebraic Identities
Students can refer to the NCERT Solutions for Class 9 Maths while solving exercise problems and preparing for their Class 9 Maths exams.
List of Exercises in Class 9 Maths Chapter 2:
Exercise 2.1 Solutions 5 Questions
Exercise 2.2 Solutions 4 Questions
Exercise 2.3 Solutions 3 Questions
Exercise 2.4 Solutions 5 Questions
Exercise 2.5 Solutions 16 Questions
NCERT Solutions for Class 9 Maths Chapter 2 โ Polynomials
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials is the second chapter of Class 9 Maths. Polynomials are introduced and discussed in detail here. The chapter discusses Polynomials and their applications. The introduction of the chapter includes whole numbers, integers, and rational numbers.
The chapter starts with the introduction of Polynomials in section 2.1 followed by two very important topics in sections 2.2 and 2.3
- Polynomials in one Variable โ Discussion of Linear, Quadratic and Cubic Polynomial.
- Zeroes of a Polynomial โ A zero of a polynomial need not be zero and can have more than one zero.
- Real Numbers and their Decimal Expansions โ Here you study the decimal expansions of real numbers and see whether they can help in distinguishing between rationals and irrationals.
Next, it discusses the following topics:
- Representing Real Numbers on the Number Line โ In this the solutions for 2 problems in Exercise 2.4.
- Operations on Real Numbers โ Here you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
- Laws of Exponents for Real Numbers โ Use these laws of exponents to solve the questions.
Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 โ Polynomials
- These NCERT Solutions for Class 9 Maths helps you solve and revise the whole syllabus of Class 9.
- After going through the stepwise solutions given by our subject expert teachers, you will be able to score more marks.
- It follows NCERT guidelines which help in preparing the students accordingly.
- It contains all the important questions from the examination point of view.
- It helps in scoring well in Maths in board exams.
Frequently Asked Questions on NCERT Solutions for Class 9 Maths Chapter 2
How many exercises are present in NCERT Solutions for Class 9 Maths Chapter 2?
NCERT Solutions for Class 9 Maths Chapter 2 has 5 exercises. The topics discussed in these exercises are polynomials in one Variable, zeros of a polynomial, real numbers and their decimal expansions, representing real numbers on the number line and operations on real numbers laws of exponents for real numbers. Practice is an essential task to learn and score well in Mathematics. Hence the solutions are designed by BYJUโS experts to boost confidence among students in understanding the concepts covered in this chapter.
Why should I opt for NCERT Solutions for Class 9 Maths Chapter 2?
The concepts present in NCERT Solutions for Class 9 Maths Chapter 2 are explained in simple language, which makes it possible even for a student not proficient in Maths to understand the subject better. Solutions are prepared by a set of experts at BYJUโS with the aim of helping students boost their exam preparation.
Is NCERT Solutions for Class 9 Maths Chapter 2 difficult to learn?
No, if you practice regularly with NCERT Solutions for Class 9 Maths Chapter 2 you can achieve your goal by scoring high in finals. These solutions are formulated by a set of Maths experts at BYJUโS. Students can score good marks in the exams by solving all the questions and cross-checking the answers with the NCERT Solutions for Class 9 Maths Chapter 2.