CLASS 9 MATHS CHAPTER-2 POLYNOMIALS

Exercise 2.1

Question 1. Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.

(i) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png

(ii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

(iii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

(iv) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.png

(v) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png


Solution:

(i) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png

We can observe that in the polynomial NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, we have x as the only variable and the powers of x in each term are a whole number.

Therefore, we conclude that NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngis a polynomial in one variable.

(ii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

We can observe that in the polynomial NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png, we have y as the only variable and the powers of y in each term are a whole number.

Therefore, we conclude that NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.pngis a polynomial in one variable.

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

We can observe that in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png, we have t as the only variable and the powers of t in each term are not a whole number.

Therefore, we conclude thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.pngis not a polynomial in one variable.

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.png

We can observe that in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.png, we have y as the only variable and the powers of y in each term are not a whole number.

Therefore, we conclude thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.pngis not a polynomial in one variable.

(v)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png

We can observe that in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png, we have x, y and t as the variables and the powers of x, y and t in each term is a whole number.

Therefore, we conclude thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png is a polynomial but not a polynomial in one variable.

Question 2. Write the coefficients of in each of the following :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image008.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image009.png

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image010.png

Solution:

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png

The coefficient ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.pngis 1.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image008.png

The coefficient ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image008.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image011.png.

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image009.png

The coefficient ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image009.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image012.png.

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image010.png

The coefficient ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngin the polynomial NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image010.pngis 0.

Question 3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.
Solution:
The binomial of degree 35 can beNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image013.png.

The binomial of degree 100 can beNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image014.png.

Question 4. Write the degree of each of the following polynomials :
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image015.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image016.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image017.png

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image018.png3


Solution:
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image019.png

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image019.png, the highest power of the variable x is 3.

Therefore, we conclude that the degree of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image019.pngis 3.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png, the highest power of the variable y is 2.

Therefore, we conclude that the degree of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.pngis 2.

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image021.png

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We observe that in the polynomial, the highest power of the variable t is 1.

Therefore, we conclude that the degreeNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image021.png of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image021.pngis 1.

(iv)3

We know that the degree of a polynomial is the highest power of the variable in the polynomial.

We can observe that in the polynomial 3, the highest power of the assumed variable x is 0.

Therefore, we conclude that the degree of the polynomial 3 is 0.

Question 5. 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

Solution :
(i) The degree of x2 + x is 2. So, it is a quadratic polynomial.
(ii) The degree of x – x3 is 3. So, it is a cubic polynomial.
(iii) The degree of y + y2 + 4 is 2. So, it is a quadratic polynomial.
(iv) The degree of 1 + x is 1. So, it is a linear polynomial.
(v) The degree of 3t is 1. So, it is a linear polynomial.
(vi) The degree of r2 is 2. So, it is a quadratic polynomial.
(vii) The degree of 7x3 is 3. So, it is a cubic polynomial.

NCERT Solutions for Class 9 Maths Exercise 2.2

Question 1. Find the value of the polynomial 5x – 4x2 + 3 at
(i) x = 0
(ii) x = – 1
(iii) x = 2

Solution:
Let p(x) = 5x – 4x2 + 3
(i) p(0) = 5(0) – 4(0)2 + 3 = 0 – 0 + 3 = 3
Thus, the value of 5x – 4x2 + 3 at x = 0 is 3.
(ii) p(-1) = 5(-1) – 4(-1)+ 3
= – 5x – 4x+ 3 = -9 + 3 = -6
Thus, the value of 5x – 4x+ 3 at x = -1 is -6.
(iii) p(2) = 5(2) – 4(2)+ 3 = 10 – 4(4) + 3
= 10 – 16 + 3 = -3
Thus, the value of 5x – 4x2 + 3 at x = 2 is – 3.

Question 2. Find p (0), p (1) and p (2) for each of the following polynomials.
(i) p(y) = y2 – y +1
(ii) p (t) = 2 +1 + 2t-t3
(iii) P (x) = x3
(iv) p (x) = (x-1) (x+1)

Solution:
(i) Given that p(y) = y2 – y + 1.
∴ P(0) = (0)2 – 0 + 1 = 0 – 0 + 1 = 1
p(1) = (1)– 1 + 1 = 1 – 1 + 1 = 1
p(2) = (2)2 – 2 + 1 = 4 – 2 + 1 = 3
(ii) Given that p(t) = 2 + t + 2t2 – t3
∴p(0) = 2 + 0 + 2(0)– (0)3
= 2 + 0 + 0 – 0=2
P(1) = 2 + 1 + 2(1)2 – (1)3
= 2 + 1 + 2 – 1 = 4
p( 2) = 2 + 2 + 2(2)2 – (2)3
= 2 + 2 + 8 – 8 = 4
(iii) Given that p(x) = x3
∴ p(0) = (0)3 = 0, p(1) = (1)3 = 1
p(2) = (2)3 = 8
(iv) Given that p(x) = (x – 1)(x + 1)
∴ p(0) = (0 – 1)(0 + 1) = (-1)(1) = -1
p(1) = (1 – 1)(1 +1) = (0)(2) = 0
P(2) = (2 – 1)(2 + 1) = (1)(3) = 3

Question 3. Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1,x = –NCERT Solutions for Class 9 Maths chapter 2-Polynomials
(ii) p (x) = 5x – π, x = NCERT Solutions for Class 9 Maths chapter 2-Polynomials
(iii) p (x) = x2 – 1, x = x – 1
(iv) p (x) = (x + 1) (x – 2), x = – 1,2
(v) p (x) = x2, x = 0
(vi) p (x) = 1x + m, x = – NCERT Solutions for Class 9 Maths chapter 2-Polynomials
(vii) P (x) = 3x2 – 1, x = – \frac { 1 }{ \sqrt { 3 } },NCERT Solutions for Class 9 Maths chapter 2-Polynomials
(viii) p (x) = 2x + 1, x = NCERT Solutions for Class 9 Maths chapter 2-Polynomials

Solution:
(i) We have , p(x) = 3x + 1
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/A3
(ii) We have, p(x) = 5x – π
∴ NCERT Solutions for Class 9 Maths chapter 2-Polynomials
(iii) We have, p(x) = x2 – 1
∴ p(1) = (1)2 – 1 = 1 – 1=0
Since, p(1) = 0, so x = 1 is a zero of x2 -1.
Also, p(-1) = (-1)2 -1 = 1 – 1 = 0
Since p(-1) = 0, so, x = -1, is also a zero of x2 – 1.

(iv) We have, p(x) = (x + 1)(x – 2)
∴ p(-1) = (-1 +1) (-1 – 2) = (0)(- 3) = 0
Since, p(-1) = 0, so, x = -1 is a zero of (x + 1)(x – 2).
Also, p( 2) = (2 + 1)(2 – 2) = (3)(0) = 0
Since, p(2) = 0, so, x = 2 is also a zero of (x + 1)(x – 2).

(v) We have, p(x) = x2
∴ p(o) = (0)2 = 0
Since, p(0) = 0, so, x = 0 is a zero of x2.

(vi) We have, p(x) = lx + m
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/A3a

(vii) We have, p(x) = 3x2 – 1
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/A3b

(viii) We have, p(x) = 2x + 1
∴ NCERT Solutions for Class 9 Maths chapter 2-Polynomials
Since, NCERT Solutions for Class 9 Maths chapter 2-Polynomials≠ 0, so, x = NCERT Solutions for Class 9 Maths chapter 2-Polynomialsis not a zero of 2x + 1.

Question 4. 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 where c and d are real numbers.

Solution:
(i) We have, p(x) = x + 5. Since, p(x) = 0
⇒ x + 5 = 0
⇒ x = -5.
Thus, zero of x + 5 is -5.

(ii) We have, p(x) = x – 5.
Since, p(x) = 0 ⇒ x – 5 = 0 ⇒ x = -5
Thus, zero of x – 5 is 5.

(iii) We have, p(x) = 2x + 5. Since, p(x) = 0
⇒ 2x + 5 =0
⇒ 2x = -5
⇒ x = NCERT Solutions for Class 9 Maths chapter 2-Polynomials
Thus, zero of 2x + 5 is NCERT Solutions for Class 9 Maths chapter 2-Polynomials.

(iv) We have, p(x) = 3x – 2. Since, p(x) = 0
⇒ 3x – 2 = 0
⇒ 3x = 2
⇒ x = NCERT Solutions for Class 9 Maths chapter 2-Polynomials
Thus, zero of 3x – 2 is NCERT Solutions for Class 9 Maths chapter 2-Polynomials

(v) We have, p(x) = 3x. Since, p(x) = 0
⇒ 3x = 0 ⇒ x = 0
Thus, zero of 3x is 0.

(vi) We have, p(x) = ax, a ≠ 0.
Since, p(x) = 0 => ax = 0 => x-0
Thus, zero of ax is 0.

(vii) We have, p(x) = cx + d. Since, p(x) = 0
⇒ cx + d = 0 ⇒ cx = -d ⇒ NCERT Solutions for Class 9 Maths chapter 2-Polynomials
Thus, zero of cx + d is NCERT Solutions for Class 9 Maths chapter 2-Polynomials

NCERT Solutions for Class 9 Maths Exercise 2.3

Question 1. Find the remainder whenis divided by

(i) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

(ii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

(iii) x

(iv) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png

(v) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.png

Solution:
(i) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image008.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image009.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image010.png

=-1+3-3+1

=0

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image011.png, we will get the remainder as 0.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image012.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image013.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image014.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image015.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image016.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image017.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image018.png

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png, we will get the remainder asNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image019.png.

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image021.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image013.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image022.png

=0+0+0+1

=1

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngby x, we will get the remainder as 1.

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image023.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image024.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image025.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image026.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image027.png

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png, we will get the remainder asNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image028.png.

(v)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image029.png

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image030.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.png in the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image013.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image031.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image032.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image033.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image034.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image035.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image036.png

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image029.png, we will get the remainder asNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image037.png.

Question 2. Find the remainder whenis divided by.


Solution:

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image039.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image040.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image039.pngin the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image038.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image041.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image042.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image043.png

= 5a

Therefore, we conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image038.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image039.png, we will get the remainder asNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image044.png.

Question 3. Check whether is a factor of.

Solution:
We know that if the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.pngis a factor ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.png, then on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.png, we must get the remainder as 0.

We need to find the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image047.png

While applying the remainder theorem, we need to put the zero of the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.pnginthe polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image048.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image049.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image050.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image051.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image052.png

We conclude that on dividing the polynomialNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.pngbyNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.png, we will get the remainder asNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image053.png, which is not 0.

Therefore, we conclude thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.pngis not a factor ofNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.png.

NCERT Solutions for Class 9 Maths Exercise 2.4

Question 1. 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 + 3x+ x + 1
(iv) x3 – x2 – (2 +√2 )x + √2

Solution :
The zero of x + 1 is -1.
(i) Let p (x) = x3 + x2 + x + 1
∴ p (-1) = (-1)3 + (-1)2 + (-1) + 1 .
= -1 + 1 – 1 + 1
⇒ p (- 1) = 0
So, (x+ 1) is a factor of x3 + x2 + x + 1.

(ii) Let p (x) = x4 + x3 + x2 + x + 1
∴ P(-1) = (-1)+ (-1)3 + (-1)2 + (-1)+1
= 1 – 1 + 1 – 1 + 1
⇒ P (-1) ≠ 1
So, (x + 1) is not a factor of x4 + x3 + x2 + x+ 1.

(iii) Let p (x) = x4 + 3x3 + 3x2 + x + 1 .
∴ p (-1)= (-1)4 + 3 (-1)3 + 3 (-1)2 + (- 1) + 1
= 1 – 3 + 3 – 1 + 1 = 1
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x4 + 3x3 + 3x2 + x+ 1.

(iv) Let p (x) = x3 – x2 – (2 + √2) x + √2
∴ p (- 1) =(- 1)3– (-1)2 – (2 + √2)(-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
⇒ p (-1) ≠ 0
So, (x + 1) is not a factor of x3 – x2 – (2 + √2) x + √2.

Question 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 + x2 – 2x – 1, g (x) = x + 1
(ii) p(x)= x3 + 3x2 + 3x + 1, g (x) = x + 2
(iii) p (x) = x– 4x+ x + 6, g (x) = x – 3

Solution :

(i) We have, p (x)= 2x3 + x2 – 2x – 1 and g (x) = x + 1
∴ p(-1) = 2(-1)3 + (-1)– 2(-1) – 1
= 2(-1) + 1 + 2 – 1
= -2 + 1 + 2 -1 = 0
⇒ p(-1) = 0, so g(x) is a factor of p(x).

(ii) We have, p(x) x3 + 3x2 + 3x + 1 and g(x) = x + 2
∴ p(-2) = (-2)3 + 3(-2)2+ 3(-2) + 1
= -8 + 12 – 6 + 1
= -14 + 13
= -1
⇒ p(-2) ≠ 0, so g(x) is not a factor of p(x).

(iii) We have, = x3 – 4x2 + x + 6 and g (x) = x – 3
∴ p(3) = (3)3 – 4(3)2 + 3 + 6
= 27 – 4(9) + 3 + 6
= 27 – 36 + 3 + 6 = 0
⇒ p(3) = 0, so g(x) is a factor of p(x).

Question 3. 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 – √2 x + 1
(iv) p (x) = kx– 3x + k

Solution :
For (x – 1) to be a factor of p(x), p(1) should be equal to 0.

(i) Here, p(x) = x2 + x + k
Since, p(1) = (1)+1 + k
⇒ p(1) = k + 2 = 0
⇒ k = -2.

(ii) Here, p (x) = 2x2 + kx + √2
Since, p(1) = 2(1)2 + k(1) + √2
= 2 + k + √2 =0
k = -2 – √2 = -(2 + √2)

(iii) Here, p (x) = kx2 – √2 x + 1
Since, p(1) = k(1)2 – (1) + 1
= k – √2 + 1 = 0
⇒ k = √2 -1

(iv) Here, p(x) = kx2 – 3x + k
p(1) = k(1)2 – 3(1) + k
= k – 3 + k
= 2k – 3 = 0
⇒ k = NCERT Solutions for Class 9 Maths chapter 2-Polynomials

Question 4. Factorise
(i) 12x2 – 7x +1
(ii) 2x2 + 7x + 3
(iii) 6x2 + 5x – 6
(iv) 3x2 – x – 4

Solution :
(i) We have,
12x2 – 7x + 1 = 12x2 – 4x- 3x + 1
= 4x (3x – 1 ) -1 (3x – 1)
= (3x -1) (4x -1)
Thus, 12x2 -7x + 3 = (2x – 1) (x + 3)

(ii) We have, 2x2 + 7x + 3 = 2x2 + x + 6x + 3
= x(2x + 1) + 3(2x + 1)
= (2x + 1)(x + 3)
Thus, 2×2 + 7x + 3 = (2x + 1)(x + 3)

(iii) We have, 6×2 + 5x – 6 = 6×2 + 9x – 4x – 6
= 3x(2x + 3) – 2(2x + 3)
= (2x + 3)(3x – 2)
Thus, 6x2 + 5x – 6 = (2x + 3)(3x – 2)

(iv) We have, 3x– x – 4 = 3x2 – 4x + 3x – 4
= x(3x – 4) + 1(3x – 4) = (3x – 4)(x + 1)
Thus, 3x2 – x – 4 = (3x – 4)(x + 1)

Question 5. Factorise
(i) x3 – 2x2 – x + 2
(ii) x3 – 3x2 – 9x – 5
(iii) x3 + 13x2 + 32x + 20
(iv) 2y3 + y2 – 2y – 1
Solution:

(i) We have, x3 – 2x2 – x + 2
Rearranging the terms, we have x– x – 2x2 + 2
= x(x– 1) – 2(x2 -1) = (x– 1)(x – 2)
= [(x)2 – (1)2](x – 2)
= (x – 1)(x + 1)(x – 2)
[∵ (a– b2) = (a + b)(a-b)]
Thus, x3 – 2x2 – x + 2 = (x – 1)(x + 1)(x – 2)

(ii) We have, x3 – 3x2 – 9x – 5
= x+ x2 – 4x2 – 4x – 5x – 5 ,
= x(x + 1) – 4x(x + 1) – 5(x + 1)
= (x + 1)(x– 4x – 5)
= (x + 1)(x2 – 5x + x – 5)
= (x + 1)[x(x – 5) + 1(x – 5)]
= (x + 1)(x – 5)(x + 1)
Thus, x– 3x2 – 9x – 5 = (x + 1)(x – 5)(x +1)

(iii) We have, x+ 13x2 + 32x + 20
= x3 + x2 + 12x2 + 12x + 20x + 20
= x2(x + 1) + 12x(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)
Thus, x3 + 13x2 + 32x + 20
= (x + 1)(x + 2)(x + 10)

(iv) We have, 2y3 + y2 – 2y – 1
= 2y3 – 2y2 + 3y2 – 3y + y – 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)(y + 1)(2y + 1)
Thus, 2y3 + y2 – 2y – 1
= (y – 1)(y + 1)(2y +1)

NCERT Solutions for Class 9 Maths Exercise 2.5

Question 1. Use suitable identities to find the following products :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png

(ii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

(iii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

(iv) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.png

(v)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image005.png


Solution:

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png.

We need to apply the above identity to find the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image008.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image009.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image010.png

Therefore, we conclude that the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image001.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image011.png.

(ii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png.

We need to apply the above identity to find the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image012.png=NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image013.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image014.png

Therefore, we conclude that the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image002.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image015.png.

(iii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image006.pngNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image007.png.

We need to apply the above identity to find the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image016.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image017.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image018.png

Therefore, we conclude that the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image003.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image019.png.

(iv) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image004.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png.

We need to apply the above identity to find the product

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image022.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image023.png

Therefore, we conclude that the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image021.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image024.png.

(v) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image025.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.png.

We need to apply the above identity to find the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image026.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image027.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image028.png

Therefore, we conclude that the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image026.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image029.png.

Question 2. Evaluate the following products without multiplying directly :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image030.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image031.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image032.png

Solution:
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image030.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image033.png

We can observe that, we can apply the identity

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image034.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image035.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image036.png

=10000+1000+21

=11021

Therefore, we conclude that the value of the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image037.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image038.png.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image039.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image040.png

We can observe that, we can apply the identity

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image034.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image041.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image042.png

=10000-900+20

=9120

Therefore, we conclude that the value of the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image039.pngis NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image043.png.

(iii) NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image044.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image045.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image020.pngwith respect to the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image046.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image047.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image048.png

=10000-16

=9984

Therefore, we conclude that the value of the productNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image044.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image049.png.

Question 3. Factorize the following using appropriate identities :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image050.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image051.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image052.png

Solution:
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image050.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image053.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image054.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image055.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image056.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image051.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image057.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image054.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image055.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image056.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image052.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image060.png

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image055.png
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image061.png

Question 4. Expand each of the following, using suitable identities :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image062.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image063.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image064.png

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image065.png

(v)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image066.png

(vi)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image067.png


Solution:

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image062.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.png.

We need to apply the above identity to expand the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image062.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image069.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image063.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.png.

We need to apply the above identity to expand the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image063.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image070.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image064.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.png.

We need to apply the above identity to expand the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image065.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image072.png

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image065.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.png.

We need to apply the above identity to expand the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image065.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image072.png

(v)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image066.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.png.

We need to apply the above identity to expand the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image066.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image073.png

(vi)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image067.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image074.pngNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image075.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image076.jpg

Question 5. Factorize :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image077.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image078.png

Solution:
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image077.png

The expression NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image079.pngcan also be written as

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image080.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.pngwith respect to the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image081.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image082.png

Therefore, we conclude that after factorizing the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image077.png, we getNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image082.png.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image078.png

We need to factorize the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image078.png.

The expression NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image083.pngcan also be written as

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image084.png

We can observe that, we can apply the identityNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image068.pngwith respect to the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image085.png, to get

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image086.png

Therefore, we conclude that after factorizing the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image078.png, we getNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image086.png.

Question 6. Write the following cubes in expanded form :

(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image087.png

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image088.png

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.png

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image090.png

Solution :
(i)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image087.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image091.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image092.png

Therefore, the expansion of the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image087.pngis NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image093.png.

(ii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image088.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image091.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image095.png

Therefore, the expansion of the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image098.png .

(iii)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image091.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image099.png

Therefore, the expansion of the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image089.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image100.png.

(iv)NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image090.png

We know thatNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image094.png.

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/image099.png

Therefore, the expansion of the expressionNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image090.pngisNCERT Solutions for Class 9 Maths chapter 2-Polynomials/image100.png.

Question 7. Evaluate the following using suitable identities
(i) (99)3
(ii) (102)3
(iii) (998)3

Solution:
(i) We have, 99 = (100 -1)
∴ 993 = (100 – 1)3
= (100)3 – 13 – 3(100)(1)(100 -1)
[Using (a – b)3 = a3 – b3 – 3ab (a – b)]
= 1000000 – 1 – 300(100 – 1)
= 1000000 -1 – 30000 + 300
= 1000300 – 30001 = 970299

(ii) We have, 102 =100 + 2
∴ 1023 = (100 + 2)3
= (100)3 + (2)3 + 3(100)(2)(100 + 2)
[Using (a + b)3 = a3 + b3 + 3ab (a + b)]
= 1000000 + 8 + 600(100 + 2)
= 1000000 + 8 + 60000 + 1200 = 1061208

(iii) We have, 998 = 1000 – 2
∴ (998)3 = (1000-2)3
= (1000)3– (2)3 – 3(1000)(2)(1000 – 2)
[Using (a – b)3 = a3 – b– 3ab (a – b)]
= 1000000000 – 8 – 6000(1000 – 2)
= 1000000000 – 8 – 6000000 +12000
= 994011992

Question 8. Factorise each of the following
(i) 8a3 +b+ 12a2b+6ab2
(ii) 8a3 -b3-12a2b+6ab2
(iii) 27-125a3 -135a+225a2
(iv) 64a3 -27b3 -144a2b + 108ab2
NCERT Solutions for Class 9 Maths chapter 2-Polynomials/ Q8

Solution:
(i) 8a3 +b3 +12a2b+6ab2
= (2a)+ (b)3 + 6ab(2a + b)
= (2a)3 + (b)3 + 3(2a)(b)(2a + b)
= (2 a + b)3
[Using a+ b3 + 3 ab(a + b) = (a + b)3]
= (2a + b)(2a + b)(2a + b)

(ii) 8a3 – b3 – 12a2b + 6ab2
= (2a)3 – (b)3 – 3(2a)(b)(2a – b)
= (2a – b)3
[Using a3 + b+ 3 ab(a + b) = (a + b)3]
= (2a – b) (2a – b) (2a – b)

(iii) 27 – 125a3 – 135a + 225a2
= (3)3 – (5a)3 – 3(3)(5a)(3 – 5a)
= (3 – 5a)3
[Using a+ b+ 3 ab(a + b) = (a + b)3]
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) 64a-27b3 -144a2b + 108ab2
= (4a)3 – (3b)3 – 3(4a)(3b)(4a – 3b)
= (4a – 3b)3
[Using a– b3 – 3 ab(a – b) = (a – b)3]
= (4a – 3b)(4a – 3b)(4a – 3b)

NCERT Solutions for Class 9 Maths chapter 2-Polynomials/ A8

Question 9. Verify
(i) x3 + y3 = (x + y)-(x2 – xy + y2)
(ii) x3 – y3 = (x – y) (x2 + xy + y2)

Solution:
(i) ∵ (x + y)3 = x+ y3 + 3xy(x + y)
⇒ (x + y)3 – 3(x + y)(xy) = x3 + y3
⇒ (x + y)[(x + y)2-3xy] = x3 + y3
⇒ (x + y)(x2 + y2 – xy) = x3 + y3
Hence, verified.

(ii) ∵ (x – y)3 = x– y– 3xy(x – y)
⇒ (x – y)3 + 3xy(x – y) = x3 – y3
⇒ (x – y)[(x – y)+ 3xy)] = x3 – y3
⇒ (x – y)(x2 + y2 + xy) = x3 – y3
Hence, verified.

Question 10. Factorise each of the following
(i) 27y3 + 125z3
(ii) 64m3 – 343n3
[Hint See question 9]

Solution:
(i) We know that
x3 + y3 = (x + y)(x2 – xy + y2)
We have, 27y3 + 125z3 = (3y)3 + (5z)3
= (3y + 5z)[(3y)2 – (3y)(5z) + (5z)2]
= (3y + 5z)(9y2 – 15yz + 25z2)

(ii) We know that
x3 – y3 = (x – y)(x+ xy + y2)
We have, 64m3 – 343n3 = (4m)3 – (7n)3
= (4m – 7n)[(4m)2 + (4m)(7n) + (7n)2]
= (4m – 7n)(16m2 + 28mn + 49n2)

Question 11. Factorise 27x+y3 +z3 -9xyz.

Solution:
We have,
27x3 + y3 + z3 – 9xyz = (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
Using the identity,
x3 + y3 + z– 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
We have, (3x)3 + (y)3 + (z)3 – 3(3x)(y)(z)
= (3x + y + z)[(3x)3 + y3 + z3 – (3x × y) – (y × 2) – (z × 3x)]
= (3x + y + z)(9x2 + y2 + z– 3xy – yz – 3zx)

Question 12. Verify that
x+y3 +z3 – 3xyz = NCERT Solutions for Class 9 Maths chapter 2-Polynomials(x + y+z)[(x-y)2 + (y – z)2 +(z – x)2]

Solution:
R.H.S
NCERT Solutions for Class 9 Maths chapter 2-Polynomials(x + y + z)[(x – y)2+(y – z)2+(z – x)2]
NCERT Solutions for Class 9 Maths chapter 2-Polynomials(x + y + 2)[(x+ y2 – 2xy) + (y2 + z2 – 2yz) + (z2 + x2 – 2zx)]
NCERT Solutions for Class 9 Maths chapter 2-Polynomials(x + y + 2)(x2 + y2 + y2 + z2 + z2 + x2 – 2xy – 2yz – 2zx)
NCERT Solutions for Class 9 Maths chapter 2-Polynomials(x + y + z)[2(x2 + y2 + z2 – xy – yz – zx)]
= 2 x NCERT Solutions for Class 9 Maths chapter 2-Polynomialsx (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= (x + y + z)(x2 + y2 + z2 – xy – yz – zx)
= x+ y3 + z3 – 3xyz = L.H.S.
Hence, verified.

Question 13. If x + y + z = 0, show that x3 + y3 + z3 = 3 xyz.

Solution:
Since, x + y + z = 0
⇒ x + y = -z (x + y)3 = (-z)3
⇒ x3 + y3 + 3xy(x + y) = -z3
⇒ x3 + y3 + 3xy(-z) = -z3 [∵ x + y = -z]
⇒ x3 + y3 – 3xyz = -z3
⇒ x3 + y+ z3 = 3xyz
Hence, if x + y + z = 0, then
x3 + y3 + z= 3xyz

Question 14. Without actually calculating the cubes, find the value of each of the following
(i) (- 12)3 + (7)+ (5)3
(ii) (28)3 + (- 15)3 + (- 13)3

Solution:

(i) We have, (-12)3 + (7)3 + (5)3
Let x = -12, y = 7 and z = 5.
Then, x + y + z = -12 + 7 + 5 = 0
We know that if x + y + z = 0, then, x3 + y+ z3 = 3xyz
∴ (-12)3 + (7)3 + (5)3 = 3[(-12)(7)(5)]
= 3[-420] = -1260

(ii) We have, (28)+ (-15)3 + (-13)3
Let x = 28, y = -15 and z = -13.
Then, x + y + z = 28 – 15 – 13 = 0
We know that if x + y + z = 0, then x3 + y3 + z3 = 3xyz
∴ (28)3 + (-15)+ (-13)3 = 3(28)(-15)(-13)
= 3(5460) = 16380

Question 15. 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 35y+ 13y – 12

Solution:
Area of a rectangle = (Length) x (Breadth)
(i) 25a2 – 35a + 12 = 25a2 – 20a – 15a + 12 = 5a(5a – 4) – 3(5a – 4) = (5a – 4)(5a – 3)
Thus, the possible length and breadth are (5a – 3) and (5a – 4).

(ii) 35y2+ 13y -12 = 35y2 + 28y – 15y -12
= 7y(5y + 4) – 3(5y + 4) = (5 y + 4)(7y – 3)
Thus, the possible length and breadth are (7y – 3) and (5y + 4).

Question 16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?
(i) Volume 3x2 – 12x
(ii) Volume 12ky2 + 8ky – 20k

Solution:
Volume of a cuboid = (Length) x (Breadth) x (Height)
(i) We have, 3x– 12x = 3(x2 – 4x)
= 3 x (x – 4)
∴ The possible dimensions of the cuboid are 3, x and (x – 4).

(ii) We have, 12ky2 + 8ky – 20k
= 4[3ky2 + 2ky – 5k] = 4[k(3y2 + 2y – 5)]
= 4 x k x (3y2 + 2y – 5)
= 4k[3y2 – 3y + 5y – 5]
= 4k[3y(y – 1) + 5(y – 1)]
= 4k[(3y + 5) x (y – 1)]
= 4k x (3y + 5) x (y – 1)
Thus, the possible dimensions of the cuboid are 4k, (3y + 5) and (y -1).

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