NCERT Solutions Class 9 Maths Chapter 2 â€“ CBSE Term II Free PDF Download
NCERT Solutions Class 9 Maths Chapter 2 Polynomials are provided here. These NCERT Solutions are created by BYJUâ€™S expert faculties to help students in the preparation for their second term 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,Â 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 Term II exams by covering the updated term wise CBSE syllabus for 2021-22 and its guidelines.
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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) 4x^{2}â€“3x+7
Solution:
The equation 4x^{2}â€“3x+7 can be written as 4x^{2}â€“3x^{1}+7x^{0}
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 4x^{2}â€“3x+7 is a polynomial in one variable.
(ii) y^{2}+âˆš2
Solution:
The equation y^{2}+âˆš2 can be written as y^{2}+âˆš2y^{0}
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 y^{2}+âˆš2 is a polynomial in one variable.
(iii) 3âˆšt+tâˆš2
Solution:
The equation 3âˆšt+tâˆš2 can be written as 3t^{1/2}+âˆš2t
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 3âˆšt+tâˆš2 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) x^{10}+y^{3}+t^{50}
Solution:
Here, in the equation x^{10}+y^{3}+t^{50}
Though, the powers, 10, 3, 50, are whole numbers, there are 3 variables used in the expression
x^{10}+y^{3}+t^{50}. Hence, it is not a polynomial in one variable.
2. Write the coefficients of x^{2} in each of the following:
(i) 2+x^{2}+x
Solution:
The equation 2+x^{2}+x can be written as 2+(1)x^{2}+x
We know that, coefficient is the number which multiplies the variable.
Here, the number that multiplies the variable x^{2} is 1
, the coefficients of x^{2 }in 2+x^{2}+x is 1.
(ii) 2â€“x^{2}+x^{3}
Solution:
The equation 2â€“x^{2}+x^{3 }can be written as 2+(â€“1)x^{2}+x^{3}
We know that, coefficient is the number (along with its sign, i.e., â€“ or +) which multiplies the variable.
Here, the number that multiplies the variable x^{2} is -1
the coefficients of x^{2 }in 2â€“x^{2}+x^{3 }is -1.
(iii) (/2)x^{2}+x
Solution:
The equation (/2)x^{2 }+x can be written as (/2)x^{2} + x
We know that, coefficient is the number (along with its sign, i.e., â€“ or +) which multiplies the variable.
Here, the number that multiplies the variable x^{2} is /2.
the coefficients of x^{2 }in (/2)x^{2 }+x is /2.
(iii)âˆš2x-1
Solution:
The equation âˆš2x-1 can be written as 0x^{2}+âˆš2x-1 [Since 0x^{2} is 0]
We know that, coefficient is the number (along with its sign, i.e., â€“ or +) which multiplies the variable.
Here, the number that multiplies the variable x^{2}is 0
, the coefficients of x^{2 }in âˆš2x-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., Â 3x^{35}+5
Monomial of degree 100: A polynomial having one term and the highest degree 100 is called a monomial of degree 100
Eg., Â 4x^{100}
4. Write the degree of each of the following polynomials:
(i) 5x^{3}+4x^{2}+7x
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, 5x^{3}+4x^{2}+7x = 5x^{3}+4x^{2}+7x^{1}
The powers of the variable x are: 3, 2, 1
the degree of 5x^{3}+4x^{2}+7x is 3 as 3 is the highest power of x in the equation.
(ii) 4â€“y^{2}
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, in 4â€“y^{2},
The power of the variable y is 2
the degree of 4â€“y^{2} is 2 as 2 is the highest power of y in the equation.
(iii) 5tâ€“âˆš7
Solution:
The highest power of the variable in a polynomial is the degree of the polynomial.
Here, in 5tâ€“âˆš7 ,
The power of the variable t is: 1
the degree of 5tâ€“âˆš7 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 = 3Ã—1 = 3Ã— x^{0}
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) x^{2}+x
Solution:
The highest power of x^{2}+x is 2
the degree is 2
Hence, x^{2}+x is a quadratic polynomial
(ii) xâ€“x^{3}
Solution:
The highest power of xâ€“x^{3 }is 3
the degree is 3
Hence, xâ€“x^{3} is a cubic polynomial
(iii) y+y^{2}+4
Solution:
The highest power of y+y^{2}+4 is 2
the degree is 2
Hence, y+y^{2}+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) r^{2}
Solution:
The highest power of r^{2 }is 2
the degree is 2
Hence, r^{2}is a quadratic polynomial.
(vii) 7x^{3}
Solution:
The highest power of 7x^{3 }is 3
the degree is 3
Hence, 7x^{3} is a cubic polynomial.
Exercise 2.2 Page: 34
1. Find the value of the polynomial (x)=5xâˆ’4x^{2}+3Â
(i) x = 0
(ii) xÂ = â€“ 1
(iii) xÂ = 2
Solution:
LetÂ f(x) = 5xâˆ’4x^{2}+3
(i) When x = 0
f(0) = 5(0)-4(0)^{2}+3
= 3
(ii)Â When x = -1
f(x) = 5xâˆ’4x^{2}+3
f(âˆ’1) = 5(âˆ’1)âˆ’4(âˆ’1)^{2}+3
= âˆ’5â€“4+3
= âˆ’6
(iii)Â When x = 2
f(x) = 5xâˆ’4x^{2}+3
f(2) = 5(2)âˆ’4(2)^{2}+3
= 10â€“16+3
= âˆ’3
2. FindÂ p(0),Â p(1) andÂ p(2) for each of the following polynomials:
(i) p(y)=y^{2}âˆ’y+1
Solution:
p(y) = y^{2}â€“y+1
âˆ´p(0) = (0)^{2}âˆ’(0)+1=1
p(1) = (1)^{2}â€“(1)+1=1
p(2) = (2)^{2}â€“(2)+1=3
(ii) p(t)=2+t+2t^{2}âˆ’t^{3}
Solution:
p(t) = 2+t+2t^{2}âˆ’t^{3}
âˆ´p(0) = 2+0+2(0)^{2}â€“(0)^{3}=2
p(1) = 2+1+2(1)^{2}â€“(1)^{3}=2+1+2â€“1=4
p(2) = 2+2+2(2)^{2}â€“(2)^{3}=2+2+8â€“8=4
(iii) p(x)=x^{3}
Solution:
p(x) = x^{3}
âˆ´p(0) = (0)^{3 }= 0
p(1) = (1)^{3 }= 1
p(2) = (2)^{3 }= 8
(iv) P(x) = (xâˆ’1)(x+1)
Solution:
p(x) = (xâ€“1)(x+1)
âˆ´p(0) = (0â€“1)(0+1) = (âˆ’1)(1) = â€“1
p(1) = (1â€“1)(1+1) = 0(2) = 0
p(2) = (2â€“1)(2+1) = 1(3) = 3
3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x)=3x+1, x=âˆ’1/3
Solution:
For,Â x = -1/3, p(x) = 3x+1
âˆ´p(âˆ’1/3) = 3(-1/3)+1 = âˆ’1+1 = 0
âˆ´ -1/3 is a zero of p(x).
(ii) p(x)=5xâ€“Ï€, x = 4/5
Solution:
For,Â x = 4/5, p(x) = 5xâ€“Ï€
âˆ´ p(4/5) = 5(4/5)- = 4-
âˆ´ 4/5 is not a zero of p(x).
(iii) p(x)=x^{2}âˆ’1, x=1, âˆ’1
Solution:
For,Â x = 1, âˆ’1;
p(x) = x^{2}âˆ’1
âˆ´p(1)=1^{2}âˆ’1=1âˆ’1 = 0
p(âˆ’1)=(-1)^{2}âˆ’1 = 1âˆ’1 = 0
âˆ´1, âˆ’1Â are zeros of p(x).
(iv) p(x) = (x+1)(xâ€“2), x =âˆ’1, 2
Solution:
For,Â x = âˆ’1,2;
p(x) = (x+1)(xâ€“2)
âˆ´p(âˆ’1) = (âˆ’1+1)(âˆ’1â€“2)
= (0)(âˆ’3) = 0
p(2) = (2+1)(2â€“2) = (3)(0) = 0
âˆ´âˆ’1,2Â are zeros of p(x).
(v) p(x) = x^{2}, x = 0
Solution:
For,Â x = 0Â p(x) = x^{2}
p(0) = 0^{2 }= 0
âˆ´ 0Â is aÂ zero of p(x).
(vi) p(x) = lx+m, x = âˆ’m/l
Solution:
For,Â x = -m/l ;Â p(x) = lx+m
âˆ´ p(-m/l)= l(-m/l)+m = âˆ’m+m = 0
âˆ´-m/l is a zero of p(x).
(vii) p(x) = 3x^{2}âˆ’1, x = -1/âˆš3 , 2/âˆš3
Solution:
For,Â x = -1/âˆš3 , 2/âˆš3 ; p(x) = 3x^{2}âˆ’1
âˆ´p(-1/âˆš3) = 3(-1/âˆš3)^{2}-1 = 3(1/3)-1 = 1-1 = 0
âˆ´p(2/âˆš3 ) = 3(2/âˆš3)^{2}-1 = 3(4/3)-1 = 4âˆ’1=3 â‰ 0
âˆ´-1/âˆš3Â is a zero of p(x) but 2/âˆš3Â Â is not a zero of p(x).
(viii) p(x) =2x+1, x = 1/2
Solution:
For,Â x = 1/2 p(x) = 2x+1
âˆ´ p(1/2)=2(1/2)+1 = 1+1 = 2â‰ 0
âˆ´1/2 is not aÂ zero of p(x).
4. Find the zero of the polynomials in each of the following cases:
(i) p(x) =Â x+5Â
Solution:
p(x) = x+5
â‡’ x+5 = 0
â‡’ x = âˆ’5
âˆ´ -5 is a zero polynomial of the polynomial p(x).
(ii) p(x) =Â xâ€“5
Solution:
p(x) = xâˆ’5
â‡’ xâˆ’5 = 0
â‡’ x = 5
âˆ´ 5 is a zero polynomial of the polynomial p(x).
(iii) p(x) = 2x+5
Solution:
p(x) = 2x+5
â‡’ 2x+5 = 0
â‡’ 2x = âˆ’5
â‡’ x = -5/2
âˆ´x = -5/2Â is a zero polynomial of the polynomial p(x).
(iv) p(x) = 3xâ€“2Â
Solution:
p(x) = 3xâ€“2
â‡’ 3xâˆ’2 = 0
â‡’ 3x = 2
â‡’x = 2/3
âˆ´x = 2/3 Â is a zero polynomial of the polynomial p(x).
(v) p(x) = 3xÂ
Solution:
p(x) = 3x
â‡’ 3x = 0
â‡’ x = 0
âˆ´0 is a zero polynomial of the polynomial p(x).
(vi) p(x) = ax, a0
Solution:
p(x) = ax
â‡’ ax = 0
â‡’ x = 0
âˆ´x = 0 is a zero polynomial of the polynomial p(x).
(vii)p(x) = cx+d, c â‰ 0, c, d are real numbers.
Solution:
p(x) = cx + d
â‡’ cx+d =0
â‡’ x = -d/c
âˆ´ x = -d/c is a zero polynomial of the polynomial p(x).
Exercise 2.3 Page: 40
1. Find the remainder whenÂ x^{3}+3x^{2}+3x+1Â is divided by
(i) x+1
Solution:
x+1= 0
â‡’x = âˆ’1
âˆ´Remainder:
p(âˆ’1) = (âˆ’1)^{3}+3(âˆ’1)^{2}+3(âˆ’1)+1
= âˆ’1+3âˆ’3+1
= 0
(ii) xâˆ’1/2
Solution:
x-1/2 = 0
â‡’ x = 1/2
âˆ´Remainder:
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
âˆ´Remainder:
p(0) = (0)^{3}+3(0)^{2}+3(0)+1
= 1
(iv) x+Ï€
Solution:
x+Ï€ = 0
â‡’ x = âˆ’Ï€
âˆ´Remainder:
p(0) = (âˆ’Ï€)^{3 }+3(âˆ’Ï€)^{2}+3(âˆ’Ï€)+1
= âˆ’Ï€^{3}+3Ï€^{2}âˆ’3Ï€+1
(v) 5+2x
Solution:
5+2x=0
â‡’ 2x = âˆ’5
â‡’ x = -5/2
âˆ´Remainder:
(-5/2)^{3}+3(-5/2)^{2}+3(-5/2)+1 = (-125/8)+(75/4)-(15/2)+1
= -27/8
2. Find the remainder whenÂ x^{3}âˆ’ax^{2}+6xâˆ’aÂ is divided by x-a.
Solution:
LetÂ p(x) = x^{3}âˆ’ax^{2}+6xâˆ’a
xâˆ’a = 0
âˆ´x = a
Remainder:
p(a) = (a)^{3}âˆ’a(a^{2})+6(a)âˆ’a
= a^{3}âˆ’a^{3}+6aâˆ’a = 5a
3. Check whether 7+3x is a factor ofÂ 3x^{3}+7x.
Solution:
7+3x = 0
â‡’ 3x = âˆ’7
â‡’ x = -7/3
âˆ´Remainder:
3(-7/3)^{3}+7(-7/3) = -(343/9)+(-49/3)
= (-343-(49)3)/9
= (-343-147)/9
= -490/9 â‰ 0
âˆ´7+3x is not a factor ofÂ 3x^{3}+7x
Exercise 2.4 Page: 43
1. Determine which of the following polynomials has (xÂ + 1) a factor:
(i) x^{3}+x^{2}+x+1
Solution:
Let p(x) =Â x^{3}+x^{2}+x+1
The zero of x+1 is -1. [x+1 = 0 means x = -1]
p(âˆ’1) = (âˆ’1)^{3}+(âˆ’1)^{2}+(âˆ’1)+1
= âˆ’1+1âˆ’1+1
= 0
âˆ´By factor theorem, x+1 is a factor ofÂ x^{3}+x^{2}+x+1
(ii) x^{4}+x^{3}+x^{2}+x+1
Solution:
Let p(x)=Â x^{4}+x^{3}+x^{2}+x+1
The zero of x+1 is -1. . [x+1= 0 means x = -1]
p(âˆ’1) = (âˆ’1)^{4}+(âˆ’1)^{3}+(âˆ’1)^{2}+(âˆ’1)+1
= 1âˆ’1+1âˆ’1+1
= 1 â‰ 0
âˆ´By factor theorem, x+1 is not a factor ofÂ x^{4}Â +Â x^{3}Â +Â x^{2}Â +Â xÂ + 1
(iii) x^{4}+3x^{3}+3x^{2}+x+1Â
Solution:
Let p(x)=Â x^{4}+3x^{3}+3x^{2}+x+1
The zero of x+1 is -1.
p(âˆ’1)=(âˆ’1)^{4}+3(âˆ’1)^{3}+3(âˆ’1)^{2}+(âˆ’1)+1
=1âˆ’3+3âˆ’1+1
=1 â‰ 0
âˆ´By factor theorem, x+1 is not a factor of x^{4}+3x^{3}+3x^{2}+x+1
(iv) x^{3 }â€“ x^{2}â€“ (2+âˆš2)x +âˆš2
Solution:
Let p(x) =Â x^{3}â€“x^{2}â€“(2+âˆš2)x +âˆš2
The zero of x+1 is -1.
p(âˆ’1) = (-1)^{3}â€“(-1)^{2}â€“(2+âˆš2)(-1) + âˆš2 = âˆ’1âˆ’1+2+âˆš2+âˆš2
= 2âˆš2 â‰ 0
âˆ´By factor theorem, x+1 is not a factor ofÂ x^{3}â€“x^{2}â€“(2+âˆš2)x +âˆš2
2. Use the Factor Theorem to determine whetherÂ g(x) is a factor ofÂ p(x) in each of the following cases:
(i) p(x) = 2x^{3}+x^{2}â€“2xâ€“1,Â g(x) =Â x+1
Solution:
p(x) = 2x^{3}+x^{2}â€“2xâ€“1,Â g(x) =Â x+1
g(x) = 0
â‡’ x+1 = 0
â‡’ x = âˆ’1
âˆ´Zero of g(x) is -1.
Now,
p(âˆ’1) = 2(âˆ’1)^{3}+(âˆ’1)^{2}â€“2(âˆ’1)â€“1
= âˆ’2+1+2âˆ’1
= 0
âˆ´By factor theorem, g(x) is a factor of p(x).
(ii) p(x)=x^{3}+3x^{2}+3x+1,Â g(x) =Â x+2
Solution:
p(x) = x^{3}+3x^{2}+3x+1,Â g(x) =Â x+2
g(x) = 0
â‡’ x+2 = 0
â‡’ x = âˆ’2
âˆ´ Zero of g(x) is -2.
Now,
p(âˆ’2) = (âˆ’2)^{3}+3(âˆ’2)^{2}+3(âˆ’2)+1
= âˆ’8+12âˆ’6+1
= âˆ’1 â‰ 0
âˆ´By factor theorem, g(x) is not a factor of p(x).
(iii) p(x)=x^{3}â€“4x^{2}+x+6,Â g(x) =Â xâ€“3
Solution:
p(x) = x^{3}â€“4x^{2}+x+6,Â g(x) =Â xÂ -3
g(x) = 0
â‡’ xâˆ’3 = 0
â‡’ x = 3
âˆ´ Zero of g(x) is 3.
Now,
p(3) = (3)^{3}âˆ’4(3)^{2}+(3)+6
= 27âˆ’36+3+6
= 0
âˆ´By factor theorem, g(x) is a factor of p(x).
3. Find the value ofÂ k, ifÂ xâ€“1 is a factor ofÂ p(x) in each of the following cases:
(i) p(x) = x^{2}+x+k
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
â‡’ (1)^{2}+(1)+k = 0
â‡’ 1+1+k = 0
â‡’ 2+k = 0
â‡’ k = âˆ’2
(ii) p(x) = 2x^{2}+kx+âˆš2
Solution:
If x-1 is a factor of p(x), then p(1)=0
â‡’ 2(1)^{2}+k(1)+âˆš2 = 0
â‡’ 2+k+âˆš2 = 0
â‡’ k = âˆ’(2+âˆš2)
(iii) p(x) = kx^{2}â€“âˆš2x+1
Solution:
If x-1 is a factor of p(x), then p(1)=0
By Factor Theorem
â‡’ k(1)^{2}-âˆš2(1)+1=0
â‡’ k = âˆš2-1
(iv) p(x)=kx^{2}â€“3x+k
Solution:
If x-1 is a factor of p(x), then p(1) = 0
By Factor Theorem
â‡’ k(1)^{2}â€“3(1)+k = 0
â‡’ kâˆ’3+k = 0
â‡’ 2kâˆ’3 = 0
â‡’ k= 3/2
4. Factorize:
(i) 12x^{2}â€“7x+1
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -7 and product =1Ã—12 = 12
We get -3 and -4 as the numbers [-3+-4=-7 and -3Ã—-4 = 12]
12x^{2}â€“7x+1= 12x^{2}-4x-3x+1
= 4x(3x-1)-1(3x-1)
= (4x-1)(3x-1)
(ii) 2x^{2}+7x+3
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 7 and product = 2Ã—3 = 6
We get 6 and 1 as the numbers [6+1 = 7 and 6Ã—1 = 6]
2x^{2}+7x+3 = 2x^{2}+6x+1x+3
= 2x (x+3)+1(x+3)
= (2x+1)(x+3)
(iii) 6x^{2}+5x-6Â
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = 5 and product = 6Ã—-6 = -36
We get -4 and 9 as the numbers [-4+9 = 5 and -4Ã—9 = -36]
6x^{2}+5x-6 = 6x^{2}+9xâ€“4xâ€“6
= 3x(2x+3)â€“2(2x+3)
= (2x+3)(3xâ€“2)
(iv) 3x^{2}â€“xâ€“4Â
Solution:
Using the splitting the middle term method,
We have to find a number whose sum = -1 and product = 3Ã—-4 = -12
We get -4 and 3 as the numbers [-4+3 = -1 and -4Ã—3 = -12]
3x^{2}â€“xâ€“4 = 3x^{2}â€“xâ€“4
= 3x^{2}â€“4x+3xâ€“4
= x(3xâ€“4)+1(3xâ€“4)
= (3xâ€“4)(x+1)
5. Factorize:
(i) x^{3}â€“2x^{2}â€“x+2
Solution:
LetÂ p(x) = x^{3}â€“2x^{2}â€“x+2
Factors of 2 are Â±1 and Â± 2
Now,
p(x) = x^{3}â€“2x^{2}â€“x+2
p(âˆ’1) = (âˆ’1)^{3}â€“2(âˆ’1)^{2}â€“(âˆ’1)+2
= âˆ’1âˆ’2+1+2
= 0
Therefore, (x+1) is the factor ofÂ p(x)
Now, Dividend = Divisor Ã— Quotient + Remainder
(x+1)(x^{2}â€“3x+2) = (x+1)(x^{2}â€“xâ€“2x+2)
= (x+1)(x(xâˆ’1)âˆ’2(xâˆ’1))
= (x+1)(xâˆ’1)(x-2)
(ii) x^{3}â€“3x^{2}â€“9xâ€“5
Solution:
LetÂ p(x) =Â x^{3}â€“3x^{2}â€“9xâ€“5
Factors of 5 are Â±1 and Â±5
By trial method, we find that
p(5)Â = 0
So,Â (x-5)Â is factor ofÂ p(x)
Now,
p(x) =Â x^{3}â€“3x^{2}â€“9xâ€“5
p(5) =Â (5)^{3}â€“3(5)^{2}â€“9(5)â€“5
= 125âˆ’75âˆ’45âˆ’5
= 0
Therefore, (x-5) is the factor ofÂ Â p(x)
Now, Dividend = Divisor Ã— Quotient + Remainder
(xâˆ’5)(x^{2}+2x+1) = (xâˆ’5)(x^{2}+x+x+1)
= (xâˆ’5)(x(x+1)+1(x+1))
= (xâˆ’5)(x+1)(x+1)
(iii) x^{3}+13x^{2}+32x+20
Solution:
LetÂ p(x) =Â x^{3}+13x^{2}+32x+20
Factors of 20 are Â±1, Â±2, Â±4, Â±5, Â±10 and Â±20
By trial method, we find that
p(-1)Â = 0
So,Â (x+1)Â is factor ofÂ p(x)
Now,
p(x)=Â x^{3}+13x^{2}+32x+20
p(-1) =Â (âˆ’1)^{3}+13(âˆ’1)^{2}+32(âˆ’1)+20
= âˆ’1+13âˆ’32+20
= 0
Therefore, (x+1) is the factor ofÂ p(x)
Now, Dividend = Divisor Ã— QuotientÂ +Remainder
(x+1)(x^{2}+12x+20) = (x+1)(x^{2}+2x+10x+20)
= (xâˆ’5)x(x+2)+10(x+2)
= (xâˆ’5)(x+2)(x+10)
(iv) 2y^{3}+y^{2}â€“2yâ€“1
Solution:
LetÂ p(y) =Â 2y^{3}+y^{2}â€“2yâ€“1
Factors =Â 2Ã—(âˆ’1)= -2 are Â±1 and Â±2
By trial method, we find that
p(1)Â = 0
So,Â (y-1)Â is factor ofÂ p(y)
Now,
p(y) =Â 2y^{3}+y^{2}â€“2yâ€“1
p(1) =Â 2(1)^{3}+(1)^{2}â€“2(1)â€“1
= 2+1âˆ’2
= 0
Therefore, (y-1) is the factor ofÂ p(y)
Now, Dividend = Divisor Ã— QuotientÂ + Remainder
(yâˆ’1)(2y^{2}+3y+1) = (yâˆ’1)(2y^{2}+2y+y+1)
= (yâˆ’1)(2y(y+1)+1(y+1))
= (yâˆ’1)(2y+1)(y+1)
Exercise 2.5 Page: 48
1. Use suitable identities to find the following products:
(i) (x+4)(xÂ +10)Â
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) = x^{2}+(4+10)x+(4Ã—10)
= x^{2}+14x+40
(ii) (x+8)(xÂ â€“10)Â Â Â Â Â
Solution:
Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab
[Here, a = 8 and b = âˆ’10]We get,
(x+8)(xâˆ’10) = x^{2}+(8+(âˆ’10))x+(8Ã—(âˆ’10))
= x^{2}+(8âˆ’10)xâ€“80
= x^{2}âˆ’2xâˆ’80
(iii) (3x+4)(3xâ€“5)
Solution:
Using the identity, (x+a)(x+b) = x ^{2}+(a+b)x+ab
[Here, x = 3x, a = 4 and b = âˆ’5]We get,
(3x+4)(3xâˆ’5) = (3x)^{2}+[4+(âˆ’5)]3x+4Ã—(âˆ’5)
= 9x^{2}+3x(4â€“5)â€“20
= 9x^{2}â€“3xâ€“20
(iv) (y^{2}+3/2)(y^{2}-3/2)
Solution:
Using the identity, (x+y)(xâ€“y) = x^{2}â€“y^{ 2}
[Here, x = y^{2}and y = 3/2]We get,
(y^{2}+3/2)(y^{2}â€“3/2) = (y^{2})^{2}â€“(3/2)^{2}
= y^{4}â€“9/4
2. Evaluate the following products without multiplying directly:
(i) 103Ã—107
Solution:
103Ã—107= (100+3)Ã—(100+7)
Using identity, [(x+a)(x+b) = x^{2}+(a+b)x+ab
Here, x = 100
a = 3
b = 7
We get, 103Ã—107 = (100+3)Ã—(100+7)
= (100)^{2}+(3+7)100+(3Ã—7))
= 10000+1000+21
= 11021
(ii) 95Ã—96 Â
Solution:
95Ã—96 = (100-5)Ã—(100-4)
Using identity, [(x-a)(x-b) = x^{2}-(a+b)x+ab
Here, x = 100
a = -5
b = -4
We get, 95Ã—96 = (100-5)Ã—(100-4)
= (100)^{2}+100(-5+(-4))+(-5Ã—-4)
= 10000-900+20
= 9120
(iii) 104Ã—96
Solution:
104Ã—96 = (100+4)Ã—(100â€“4)
Using identity, [(a+b)(a-b)= a^{2}-b^{2}]
Here, a = 100
b = 4
We get, 104Ã—96 = (100+4)Ã—(100â€“4)
= (100)^{2}â€“(4)^{2}
= 10000â€“16
= 9984
3. Factorize the following using appropriate identities:
(i) 9x^{2}+6xy+y^{2}
Solution:
9x^{2}+6xy+y^{2 }= (3x)^{2}+(2Ã—3xÃ—y)+y^{2}
Using identity, x^{2}+2xy+y^{2 }= (x+y)^{2}
Here, x = 3x
y = y
9x^{2}+6xy+y^{2 }= (3x)^{2}+(2Ã—3xÃ—y)+y^{2}
= (3x+y)^{2}
= (3x+y)(3x+y)
(ii) 4y^{2}âˆ’4y+1
Solution:
4y^{2}âˆ’4y+1 = (2y)^{2}â€“(2Ã—2yÃ—1)+1
Using identity, x^{2} â€“ 2xy + y^{2 }= (x â€“ y)^{2}
Here, x = 2y
y = 1
4y^{2}âˆ’4y+1 = (2y)^{2}â€“(2Ã—2yÃ—1)+1^{2}
= (2yâ€“1)^{2}
= (2yâ€“1)(2yâ€“1)
(iii) Â x^{2}â€“y^{2}/100
Solution:
x^{2}â€“y^{2}/100 = x^{2}â€“(y/10)^{2}
Using identity, x^{2}-y^{2 }= (x-y)(x+y)
Here, x = x
y = y/10
x^{2}â€“y^{2}/100 = x^{2}â€“(y/10)^{2}
= (xâ€“y/10)(x+y/10)
4. Expand each of the following, using suitable identities:
(i) (x+2y+4z)^{2}
(ii) (2xâˆ’y+z)^{2}
(iii) (âˆ’2x+3y+2z)^{2}
(iv) (3a â€“7bâ€“c)^{2}
(v) (â€“2x+5yâ€“3z)^{2}
((1/4)a-(1/2)b +1)^{2}
Solution:
(i) (x+2y+4z)^{2}
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = x
y = 2y
z = 4z
(x+2y+4z)^{2 }= x^{2}+(2y)^{2}+(4z)^{2}+(2Ã—xÃ—2y)+(2Ã—2yÃ—4z)+(2Ã—4zÃ—x)
= x^{2}+4y^{2}+16z^{2}+4xy+16yz+8xz
(ii) (2xâˆ’y+z)^{2}Â
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = 2x
y = âˆ’y
z = z
(2xâˆ’y+z)^{2 }= (2x)^{2}+(âˆ’y)^{2}+z^{2}+(2Ã—2xÃ—âˆ’y)+(2Ã—âˆ’yÃ—z)+(2Ã—zÃ—2x)
= 4x^{2}+y^{2}+z^{2}â€“4xyâ€“2yz+4xz
(iii) (âˆ’2x+3y+2z)^{2}
Solution:
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = âˆ’2x
y = 3y
z = 2z
(âˆ’2x+3y+2z)^{2 }= (âˆ’2x)^{2}+(3y)^{2}+(2z)^{2}+(2Ã—âˆ’2xÃ—3y)+(2Ã—3yÃ—2z)+(2Ã—2zÃ—âˆ’2x)
= 4x^{2}+9y^{2}+4z^{2}â€“12xy+12yzâ€“8xz
(iv) (3a â€“7bâ€“c)^{2}
Solution:
Using identity (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = 3a
y = â€“ 7b
z = â€“ c
(3a â€“7bâ€“ c)^{2 }= (3a)^{2}+(â€“ 7b)^{2}+(â€“ c)^{2}+(2Ã—3a Ã—â€“ 7b)+(2Ã—â€“ 7b Ã—â€“ c)+(2Ã—â€“ c Ã—3a)
= 9a^{2} + 49b^{2 }+ c^{2}â€“ 42ab+14bcâ€“6ca
(v) (â€“2x+5yâ€“3z)^{2}
Solution:
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = â€“2x
y = 5y
z = â€“ 3z
(â€“2x+5yâ€“3z)^{2 }= (â€“2x)^{2}+(5y)^{2}+(â€“3z)^{2}+(2Ã—â€“2x Ã— 5y)+(2Ã— 5yÃ—â€“ 3z)+(2Ã—â€“3z Ã—â€“2x)
= 4x^{2}+25y^{2 }+9z^{2}â€“ 20xyâ€“30yz+12zx
(vi) ((1/4)a-(1/2)b+1)^{2}
Solution:
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
Here, x = (1/4)a
y = (-1/2)b
z = 1
5. Factorize:
(i) 4x^{2}+9y^{2}+16z^{2}+12xyâ€“24yzâ€“16xz
(ii ) 2x^{2}+y^{2}+8z^{2}â€“2âˆš2xy+4âˆš2yzâ€“8xz
Solution:
(i) 4x^{2}+9y^{2}+16z^{2}+12xyâ€“24yzâ€“16xz
Using identity, (x+y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}
4x^{2}+9y^{2}+16z^{2}+12xyâ€“24yzâ€“16xzÂ = (2x)^{2}+(3y)^{2}+(âˆ’4z)^{2}+(2Ã—2xÃ—3y)+(2Ã—3yÃ—âˆ’4z)+(2Ã—âˆ’4zÃ—2x)
= (2x+3yâ€“4z)^{2}
= (2x+3yâ€“4z)(2x+3yâ€“4z)
(ii) 2x^{2}+y^{2}+8z^{2}â€“2âˆš2xy+4âˆš2yzâ€“8xz
Using identity, (x +y+z)^{2} = x^{2}+y^{2}+z^{2}+2xy+2yz+2zx
We can say that, x^{2}+y^{2}+z^{2}+2xy+2yz+2zx = (x+y+z)^{2}
2x^{2}+y^{2}+8z^{2}â€“2âˆš2xy+4âˆš2yzâ€“8xz
= (-âˆš2x)^{2}+(y)^{2}+(2âˆš2z)^{2}+(2Ã—-âˆš2xÃ—y)+(2Ã—yÃ—2âˆš2z)+(2Ã—2âˆš2Ã—âˆ’âˆš2x)
= (âˆ’âˆš2x+y+2âˆš2z)^{2}
= (âˆ’âˆš2x+y+2âˆš2z)(âˆ’âˆš2x+y+2âˆš2z)
6. Write the following cubes in expanded form:
(i) (2x+1)^{3}
(ii) (2aâˆ’3b)^{3}
(iii) ((3/2)x+1)^{3}
(iv) (xâˆ’(2/3)y)^{3}
Solution:
(i) (2x+1)^{3}
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
(2x+1)^{3}= (2x)^{3}+1^{3}+(3Ã—2xÃ—1)(2x+1)
= 8x^{3}+1+6x(2x+1)
= 8x^{3}+12x^{2}+6x+1
(ii) (2aâˆ’3b)^{3}
Using identity,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)
(2aâˆ’3b)^{3 }= (2a)^{3}âˆ’(3b)^{3}â€“(3Ã—2aÃ—3b)(2aâ€“3b)
= 8a^{3}â€“27b^{3}â€“18ab(2aâ€“3b)
= 8a^{3}â€“27b^{3}â€“36a^{2}b+54ab^{2}
(iii) ((3/2)x+1)^{3}
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
((3/2)x+1)^{3}=((3/2)x)^{3}+1^{3}+(3Ã—(3/2)xÃ—1)((3/2)x +1)
(iv) Â (xâˆ’(2/3)y)^{3}
Using identity, (x â€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)
7. Evaluate the following using suitable identities:Â
(i) (99)^{3}
(ii) (102)^{3}
(iii) (998)^{3}
Solutions:
(i) (99)^{3}
Solution:
We can write 99 as 100â€“1
Using identity, (x â€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)
(99)^{3 }= (100â€“1)^{3}
= (100)^{3}â€“1^{3}â€“(3Ã—100Ã—1)(100â€“1)
= 1000000 â€“1â€“300(100 â€“ 1)
= 1000000â€“1â€“30000+300
= 970299
(ii) (102)^{3}
Solution:
We can write 102 as 100+2
Using identity,(x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
(100+2)^{3 }=(100)^{3}+2^{3}+(3Ã—100Ã—2)(100+2)
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200
= 1061208
(iii) (998)^{3}
Solution:
We can write 99 as 1000â€“2
Using identity,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)
(998)^{3 }=(1000â€“2)^{3}
=(1000)^{3}â€“2^{3}â€“(3Ã—1000Ã—2)(1000â€“2)
= 1000000000â€“8â€“6000(1000â€“ 2)
= 1000000000â€“8- 6000000+12000
= 994011992
8. Factorise each of the following:
(i) 8a^{3}+b^{3}+12a^{2}b+6ab^{2}
(ii) 8a^{3}â€“b^{3}â€“12a^{2}b+6ab^{2}
(iii) 27â€“125a^{3}â€“135a +225a^{2}Â Â
(iv) 64a^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}
(v) 27p^{3}â€“(1/216)âˆ’(9/2) p^{2}+(1/4)p
Solutions:
(i) 8a^{3}+b^{3}+12a^{2}b+6ab^{2}
Solution:
The expression, 8a^{3}+b^{3}+12a^{2}b+6ab^{2} can be written as (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}
8a^{3}+b^{3}+12a^{2}b+6ab^{2 }= (2a)^{3}+b^{3}+3(2a)^{2}b+3(2a)(b)^{2}
= (2a+b)^{3}
= (2a+b)(2a+b)(2a+b)
Here, the identity, (x +y)^{3} = x^{3}+y^{3}+3xy(x+y) is used.
Â
(ii) 8a^{3}â€“b^{3}â€“12a^{2}b+6ab^{2}
Solution:
The expression, 8a^{3}â€“b^{3}âˆ’12a^{2}b+6ab^{2} can be written as (2a)^{3}â€“b^{3}â€“3(2a)^{2}b+3(2a)(b)^{2}
8a^{3}â€“b^{3}âˆ’12a^{2}b+6ab^{2 }= (2a)^{3}â€“b^{3}â€“3(2a)^{2}b+3(2a)(b)^{2}
= (2aâ€“b)^{3}
= (2aâ€“b)(2aâ€“b)(2aâ€“b)
Here, the identity,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y) is used.
Â
(iii) 27â€“125a^{3}â€“135a+225a^{2}Â
Solution:
The expression, 27â€“125a^{3}â€“135a +225a^{2} can be written as 3^{3}â€“(5a)^{3}â€“3(3)^{2}(5a)+3(3)(5a)^{2}
27â€“125a^{3}â€“135a+225a^{2 }=
3^{3}â€“(5a)^{3}â€“3(3)^{2}(5a)+3(3)(5a)^{2}
= (3â€“5a)^{3}
= (3â€“5a)(3â€“5a)(3â€“5a)
Here, the identity, (xâ€“y)^{3} = x^{3}â€“y^{3}-3xy(xâ€“y) is used.
(iv) 64a3â€“27b3â€“144a^{2}b+108ab^{2}
Solution:
The expression, 64a^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}can be written as (4a)^{3}â€“(3b)^{3}â€“3(4a)^{2}(3b)+3(4a)(3b)^{2}
64a^{3}â€“27b^{3}â€“144a^{2}b+108ab^{2}=
(4a)^{3}â€“(3b)^{3}â€“3(4a)^{2}(3b)+3(4a)(3b)^{2}
=(4aâ€“3b)^{3}
=(4aâ€“3b)(4aâ€“3b)(4aâ€“3b)
Here, the identity, (x â€“ y)^{3} = x^{3} â€“ y^{3} â€“ 3xy(x â€“ y) is used.
(v) 7p^{3}â€“ (1/216)âˆ’(9/2) p^{2}+(1/4)p
Solution:
The expression, 27p^{3}â€“(1/216)âˆ’(9/2) p^{2}+(1/4)p
can be written as (3p)^{3}â€“(1/6)^{3}â€“3(3p)^{2}(1/6)+3(3p)(1/6)^{2}
27p^{3}â€“(1/216)âˆ’(9/2) p^{2}+(1/4)p =
(3p)^{3}â€“(1/6)^{3}â€“3(3p)^{2}(1/6)+3(3p)(1/6)^{2}
= (3pâ€“16)^{3}
= (3pâ€“16)(3pâ€“16)(3pâ€“16)
9. Verify:
(i) x^{3}+y^{3 }= (x+y)(x^{2}â€“xy+y^{2})
(ii) x^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})
Solutions:
(i) x^{3}+y^{3 }= (x+y)(x^{2}â€“xy+y^{2})
We know that, (x+y)^{3} = x^{3}+y^{3}+3xy(x+y)
â‡’ x^{3}+y^{3 }= (x+y)^{3}â€“3xy(x+y)
â‡’ x^{3}+y^{3 }= (x+y)[(x+y)^{2}â€“3xy]
Taking (x+y) common â‡’ x^{3}+y^{3 }= (x+y)[(x^{2}+y^{2}+2xy)â€“3xy]
â‡’ x^{3}+y^{3 }= (x+y)(x^{2}+y^{2}â€“xy)
(ii) x^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})Â
We know that,(xâ€“y)^{3} = x^{3}â€“y^{3}â€“3xy(xâ€“y)
â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)^{3}+3xy(xâ€“y)
â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)[(xâ€“y)^{2}+3xy]
Taking (x+y) common â‡’ x^{3}âˆ’y^{3 }= (xâ€“y)[(x^{2}+y^{2}â€“2xy)+3xy]
â‡’ x^{3}+y^{3 }= (xâ€“y)(x^{2}+y^{2}+xy)
10. Factorize each of the following:
(i) 27y^{3}+125z^{3}
(ii) 64m^{3}â€“343n^{3}
Solutions:
(i) 27y^{3}+125z^{3}
The expression, 27y^{3}+125z^{3 }can be written as (3y)^{3}+(5z)^{3}
27y^{3}+125z^{3 }= (3y)^{3}+(5z)^{3}
We know that, x^{3}+y^{3 }= (x+y)(x^{2}â€“xy+y^{2})
27y^{3}+125z^{3 }= (3y)^{3}+(5z)^{3}
= (3y+5z)[(3y)^{2}â€“(3y)(5z)+(5z)^{2}]
= (3y+5z)(9y^{2}â€“15yz+25z^{2})
(ii) 64m^{3}â€“343n^{3}
The expression, 64m^{3}â€“343n^{3}can be written as (4m)^{3}â€“(7n)^{3}
64m^{3}â€“343n^{3 }=
(4m)^{3}â€“(7n)^{3}
We know that, x^{3}â€“y^{3 }= (xâ€“y)(x^{2}+xy+y^{2})
64m^{3}â€“343n^{3 }= (4m)^{3}â€“(7n)^{3}
= (4m-7n)[(4m)^{2}+(4m)(7n)+(7n)^{2}]
= (4m-7n)(16m^{2}+28mn+49n^{2})
11. Factorise:Â 27x^{3}+y^{3}+z^{3}â€“9xyzÂ
Solution:
The expression27x^{3}+y^{3}+z^{3}â€“9xyzÂ can be written as (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)
27x^{3}+y^{3}+z^{3}â€“9xyzÂ = (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)
We know that, x^{3}+y^{3}+z^{3}â€“3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}â€“xy â€“yzâ€“zx)
27x^{3}+y^{3}+z^{3}â€“9xyzÂ = (3x)^{3}+y^{3}+z^{3}â€“3(3x)(y)(z)
= (3x+y+z)[(3x)^{2}+y^{2}+z^{2}â€“3xyâ€“yzâ€“3xz]
= (3x+y+z)(9x^{2}+y^{2}+z^{2}â€“3xyâ€“yzâ€“3xz)
12. Verify that:
x^{3}+y^{3}+z^{3}â€“3xyz = (1/2) (x+y+z)[(xâ€“y)^{2}+(yâ€“z)^{2}+(zâ€“x)^{2}]
Solution:
We know that,
x^{3}+y^{3}+z^{3}âˆ’3xyz = (x+y+z)(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)
â‡’ x^{3}+y^{3}+z^{3}â€“3xyz = (1/2)(x+y+z)[2(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)]
= (1/2)(x+y+z)(2x^{2}+2y^{2}+2z^{2}â€“2xyâ€“2yzâ€“2xz)
= (1/2)(x+y+z)[(x^{2}+y^{2}âˆ’2xy)+(y^{2}+z^{2}â€“2yz)+(x^{2}+z^{2}â€“2xz)]
= (1/2)(x+y+z)[(xâ€“y)^{2}+(yâ€“z)^{2}+(zâ€“x)^{2}]
13. IfÂ x+y+z =Â 0, show thatÂ x^{3}+y^{3}+z^{3 }= 3xyz.
Solution:
We know that,
x^{3}+y^{3}+z^{3}-3xyz = (x +y+z)(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)
Now, according to the question, let (x+y+z) =Â 0,
then, x^{3}+y^{3}+z^{3 }-3xyz = (0)(x^{2}+y^{2}+z^{2}â€“xyâ€“yzâ€“xz)
â‡’ x^{3}+y^{3}+z^{3}â€“3xyz = 0
â‡’ x^{3}+y^{3}+z^{3 }= 3xyz
Hence Proved
14. Without actually calculating the cubes, find the value of each of the following:
(i) (âˆ’12)^{3}+(7)^{3}+(5)^{3}
(ii) (28)^{3}+(âˆ’15)^{3}+(âˆ’13)^{3}
Solution:
(i) (âˆ’12)^{3}+(7)^{3}+(5)^{3}
Let a = âˆ’12
b = 7
c = 5
We know that ifÂ x+y+z =Â 0, then x^{3}+y^{3}+z^{3}=3xyz.
Here, âˆ’12+7+5=0
(âˆ’12)^{3}+(7)^{3}+(5)^{3 }= 3xyz
= 3Ã—-12Ã—7Ã—5
= -1260
(ii) (28)^{3}+(âˆ’15)^{3}+(âˆ’13)^{3}
Solution:
(28)^{3}+(âˆ’15)^{3}+(âˆ’13)^{3}
Let a = 28
b = âˆ’15
c = âˆ’13
We know that ifÂ x+y+z =Â 0, then x^{3}+y^{3}+z^{3 }= 3xyz.
Here, x+y+zÂ = 28â€“15â€“13 = 0
(28)^{3}+(âˆ’15)^{3}+(âˆ’13)^{3 }= 3xyz
= 0+3(28)(âˆ’15)(âˆ’13)
= 16380
15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:Â
(i) Area :Â 25a^{2}â€“35a+12
(ii) Area :Â 35y^{2}+13yâ€“12
Solution:
(i) Area :Â 25a^{2}â€“35a+12
Using the splitting the middle term method,
We have to find a number whose sum = -35 and product =25Ã—12=300
We get -15 and -20 as the numbers [-15+-20=-35 and -15Ã—-20=300]
25a^{2}â€“35a+12 = 25a^{2}â€“15aâˆ’20a+12
= 5a(5aâ€“3)â€“4(5aâ€“3)
= (5aâ€“4)(5aâ€“3)
Possible expression for lengthÂ = 5aâ€“4
Possible expression for breadthÂ = 5a â€“3
(ii) Area :Â 35y^{2}+13yâ€“12
Using the splitting the middle term method,
We have to find a number whose sum = 13 and product = 35Ã—-12 = 420
We get -15 and 28 as the numbers [-15+28 = 13 and -15Ã—28=420]
35y^{2}+13yâ€“12 = 35y^{2}â€“15y+28yâ€“12
= 5y(7yâ€“3)+4(7yâ€“3)
= (5y+4)(7yâ€“3)
Possible expression for lengthÂ =Â (5y+4)
Possible expression for breadthÂ =Â (7yâ€“3)
16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?Â
(i) Volume :Â 3x^{2}â€“12x
(ii) Volume :Â 12ky^{2}+8kyâ€“20k
Solution:
(i) Volume :Â 3x^{2}â€“12x
3x^{2}â€“12xÂ can be written as 3x(xâ€“4) by taking 3x out of both the terms.
Possible expression for lengthÂ =Â 3
Possible expression for breadthÂ =Â x
Possible expression for heightÂ =Â (xâ€“4)
(ii) Volume:
12ky^{2}+8kyâ€“20k
12ky^{2}+8kyâ€“20kÂ can be written as 4k(3y^{2}+2yâ€“5) by taking 4k out of both the terms.
12ky^{2}+8kyâ€“20k = 4k(3y^{2}+2yâ€“5)
[Here, 3y^{2}+2yâ€“5 can be written as 3y^{2}+5yâ€“3yâ€“5 using splitting the middle term method.]= 4k(3y^{2}+5yâ€“3yâ€“5)
= 4k[y(3y+5)â€“1(3y+5)]
= 4k(3y+5)(yâ€“1)
Possible expression for lengthÂ = 4k
Possible expression for breadthÂ =Â (3y +5)
Possible expression for heightÂ =Â (y -1)
NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Summary
As this is one of the important Chapters in Class 9 Maths, it comes under the unit â€“ Algebra and has a weightage of 12 marks in Class 9 Maths CBSE Term II examination. 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 â€“ 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 â€“ Discussion of Linear, Quadratic and Cubic Polynomial.
- Zeroes of a Polynomial â€“ A zero of a polynomial need not be zero and can have more than one zero.
- Real Numbers and their Decimal Expansions â€“ 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 â€“ In this the solutions for 2 problems in Exercise 2.4.
- Operations on Real Numbers â€“ Here you explore some of the operations like addition, subtraction, multiplication, and division on irrational numbers.
- Laws of Exponents for Real Numbers â€“ Use these laws of exponents to solve the questions.
Key Advantages of NCERT Solutions for Class 9 Maths Chapter 2 â€“ Polynomials
- These NCERT Solutions for Class 9 Maths helps you solve and revise the updated term wise CBSE syllabus of Class 9 for 2021-22.
- 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 Class 10 CBSE Term II Maths exams.
In order to improve the problem solving skills among the Class 9 students, we at BYJUâ€™S have provided solutions for other textbooks of the CBSE Board. To learn more clearly about Polynomials, students can access the solutions link provided below.
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