CBSE Class 9

These NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.3

Question 1.
In which quadrant or on which axis do each of the points (-2, 4), (3, -1), (-1, 0), (1, 2) and (-3, -5) lie? Verify your answer by locating them on the Cartesian plane.
Solution:
The point (-2, 4) lies on the IInd quadrant in the cartesian plane because in point (-2, 4), x is negative and y is positive.

Again, the point (3, -1) lies on the IVth quadrant in the cartesian plane because, in point (3, -1), x is positive y is negative.

Again, the point (-1, 0) lies on the x-axis because in point (-1, 0) the value of y is zero.

Again the point (1, 2) lie on the Ist quadrant because in points (1, 2) both x and y are positive.

The point (-3, -5) lie on IIIrd quadrant in the cartesian plane because in point (-3, -5) both x and y are negative.

Question 2.
Plot the point (x, y) given in the following table on the plane, choosing suitable units of distance on the axes.

Solution:
The pairs of numbers given in the table can be represented by the points (-2, 8), (-1, 7), (0, -1.25), (1, 3), and (3, -1) we use the scale 1 cm = 1 unit on the axes.

These NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.2

Question 1.
Write the answer to each of the following questions:
(i) What is the name of horizontal and vertical lines drawn to determine the position of any point in the cartesian plane?
(ii) What is the name of each part of the plane formed by these two lines?
(iii) Write the name of the point where these two lines intersect.
Solution:
(i) The name of the horizontal line drawn to determine the position of any point in the cartesian plane is the x-axis. The name of the vertical line is the y-axis.

(ii) The name of each part of the plane formed by the x-axis and y-axis is called quadrants (one-fourth part) numbered I, II, III, and IV anticlockwise from OX.

(iii) Name of the point where these two lines intersect is the origin.

Question 2.
See fig. 3.14 and write the following:
(i) The coordinates of B.
(ii) The coordinates of C.
(iii) The point identified by the coordinates (-3, -5)
(iv) The point identified by the coordinates (2, -4)
(v) The abscissa of point D.
(vi) The ordinate of the point H.
(vii) The coordinates of the point L.
(viii) The coordinates of the point M.

Solution:
According to Fig. 3.14.
(i) The coordinate of B is (-5, 2).

(ii) The coordinate of C is (5, -5).

(iii) The point identified by the coordinates (-3, -5) is E.

(iv) The point identified by the coordinate (2, -4) is G.

(v) The abscissa of the pint D is 6.

(vi) The ordinate of the point H is -3.

(vii) The coordinate of the point L is (0, 5).

(viii) The coordinate of the point M is (-3, 0).

These NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Ex 3.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry Exercise 3.1

Question 1.
How will you describe the position of a table lamp on your study table to another person?
Solution:

Consider the lamp is placed on the table. Take the space taken by the lamp as a point and draw perpendiculars on length AB and breadth BC of the table. Now measure the distance.
Let the length of perpendiculars be 20 cm and 30 cm from length AB and breadth BC respectively. So, the position of the lamp from (AB, BC) edges will be (20, 30).

Question 2.
(Street Plan): A city has two main roads meeting at the center of the city. These two roads are along the North-South direction and East-West direction. All other streets of the city-run parallel to main roads and are 200 m apart There are about 5 streets in each direction. Using 1 cm = 200 m, draw the model of the city in your notebook. Represent roads/streets by single lines.
There are many cross streets in your modal. A particular cross-street is made by two streets, one running in the North-South direction and another in the East-West direction. Each cross street is referred to in the following manner. If the 2nd street running in the North-South direction and 5lh in the East-West direction meet at some crossing then we will call this cross-street (2, 5). Using this convention, find
(i) How many cross-streets can be referred to as (4, 3)
(ii) How many cross-streets can be referred to as (3, 4).
Solution:
According to the question street plan drawn as

Both the cross streets are marked in the above figure. They are only one found because of the two reference lines we have used for locating them.

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.5

Question 1.
Use suitable identities to find the following products:
(i) (x + 4) (x + 10)
(ii) (x + 8) (x – 10)
(iii) (3x + 4) (3x – 5)
(iv) (y² + \(\frac{3}{2}\)) (y² – \(\frac{3}{2}\))
(v) (3 – 2x) (3 + 2x)
Solution:
(i) We have given, (x + 4)(x + 10)
We know that
(x + a) (x + b) = x² + (a + b)x + ab
Therefore,
(x + 4) (x + 10)
= x² + (4 + 10)x + 4 × 10
= x² + 14x + 40

(ii) We have given (x + 8) (x – 10)
or, (x + 8) (x + (-10))
We know that,
(x + a) (x + b) = x² + (a + b)x + ab
(x + 8) (x + (-10))
= x² + (8 + (-10))x + 8 × (-10)
= x² – 2x – 80

(iii) We have given, (3x + 4)(3x – 5)
or, (3x + 4) (3x + (- 5))
We know that
(x + a) (x + b) = x² + (a + b)x + ab
(3x + 4)(3x – 5)
= (3x)² + (4 + (-5))3x + 4 × (-5)
= 9x² – 3x – 20

(iv) We have given (y² + \(\frac{3}{2}\)) (y² – \(\frac{3}{2}\))
We know that (x + y)(x – y) = x² – y²
\(\left(y^{2}+\frac{3}{2}\right)\left(y^{2}-\frac{3}{2}\right)=\left(y^{2}\right)^{2}-\left(\frac{3}{2}\right)^{2}\)
= \(y^{4}-\frac{9}{4}\)

(v) We have given (3 – 2x)(3 + 2x)
We know that (x – y)(x + y) = x² – y²
(3 – 2x)(3 + 2x)
= (3)² – (2x)²
= 9 – 4x²

Question 2.
Evaluate the following products without multiplying directly.
(i) 103 × 107
(ii) 95 × 96
(iii) 104 × 96
Solution:
(i) We have given, 103 × 107
We can also write
(100 + 3) × (100 + 7)
We know that
(x + a) (x + b) = x² + (a + b)x + ab
(100 + 3) (100 + 7)
= (100)² + (3 + 7) × 100 + 3 × 7
= 10000 + 1000 + 21
= 11021

(ii) We have given, 95 × 96
We can also write (90 + 5) × (90 + 6)
We know that
(x + a) (x + b) = x² + (a + b)x + ab
(90 + 5) × (90 + 6)
= (90)² + (5 + 6) × 90 + 5 × 6
= 8100 + 990 + 30
= 9120

(iii) We have given, 104 × 96
We can also write (100 + 4) × (100 – 4)
We know that (x + y)(x – y) = x² – y²
(100 + 4)(100 – 4)
= (100)² – (4)²
= 10000 – 16
= 9984

Question 3.
Factorise the following using appropriate identities:
(i) 9x² + 6xy + y²
(ii) 4y² – 4y + 1
(iii) \(x^{2}-\frac{y^{2}}{100}\)
Solution:
(i) We have given, 9x² + 6xy + y²
We can also write (3x)² + 2 . 3x . y + (y)²
We know that
a² + 2ab + b² = (a + b)²
(3x)² + 2(3x)(y) + (y)²
= (3x + y)²
= (3x + y)(3x + y)

(ii) We have, 4y² – 4y + 1
We can also write (2y)² – 2 . 2y . 1 + (1)²
We know that
x² – 2xy + y² = (x – y)²
(2y)² – 2 . 2y . 1 + (1)²
= (2y – 1)²
= (2y – 1)(2y – 1)

(iii) We have given, \(x^{2}-\frac{y^{2}}{100}\)
We can also write, \((x)^{2}-\left(\frac{y}{100}\right)^{2}\)
We know that
a² – b² = (a + b)(a – b)
\((x)^{2}-\left(\frac{y}{100}\right)^{2}=\left(x+\frac{y}{10}\right)\left(x-\frac{y}{10}\right)\)

Question 4.
Expand each of the following using suitable identities:
(i) (x + 2y + 4z)²
(ii) (2x – y + z)²
(iii) (-2x + 3y + 2z)²
(iv) (3a – 7b – c)²
(iv) (-2x + 5y – 3z)²
(v) \(\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}\)
Solution:
(i) We have given that, (x + 2y + 4z)²
We know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,
(x + 2y + 4z)²
= (x)² + (2y)² + (4z)² + 2 . x . 2y + 2 . 2y . 4z + 2 . 4z . x
= x² + 4y² + 16z² + 4xy + 16yz + 8zx

(ii) We have given that (2x – y + z)²
We can also write, (2x + (-y) + z)²
Now, we know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,
(2x + (-y) + z)²
= (2x)² + (-y)² + (z)² + 2 . 2x . (-y) + 2 . (-y) . z + 2 . z . 2x
= 4x² + y² + z² – 4xy – 2yz + 4zx

(iii) We have given that, (-2x + 3y + 2z)²
We can also write, [(-2x) + 3y + 2z]²
Now, we know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,
((-2x) + 3y + 2z)² = (-2x)² + (3y)² + (2z)² + 2 . (-2x) . 3y + 2 . 3y . 2z + 2 . 2z . (-2x)
= 4x² + 9y² + 4z² – 12xy + 12yz – 8zx

(iv) We have given that, (3a – 7b – c)²
We can also write (3a + (-7b) + (-c)²
Now, we know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,
(3a + (-7b) + (-c))²
= (3a)² + (-7b)² + (-c)² + 2 . 3a . (-7b) + 2 . (-7b) . (-c) + 2 . (-c) . 3a
= 9a² + 49b² + c² – 42ab + 14bc – 6ca

(v) We have given that, (-2x + 5y – 3z)²
We can also write [(-2x) + 5y + (-3z)]²
Now, we know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,
[(-2x) + 5y + (-3z)]²
= (-2x)² + (5y)² + (-3z)² + 2 . (-2x). 5y + 2 . 5y . (-3z) + 2 . (-3z) . (-2x)
= 4x² + 25y² + 9z² – 20xy – 30yz + 12zx

(vi) We have given that, \(\left[\frac{1}{4} a-\frac{1}{2} b+1\right]^{2}\)
We can also write \(\left[\frac{1}{4} a\left(-\frac{1}{2} b\right)+1\right]^{2}\)
Now, we know that
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
Therefore,

Question 5.
Factorise:
(i) 4x² + 9y² + 16z² + 12xy – 24yz – 16xz
(ii) 2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
Solution:
(i) We have given
4x² + 9y² + 16z² + 12xy – 24yz – 16xz
We can also write
(2x)² + (3y)² + (-4z)² + 2 . 2x . 3y + 2 . 3y . (-4z) + 2 . (-4z) . 2x
Now we know that
a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
Therefore,
(2x)² + (3y)² + (-4z)² + 2 . 2x . 3y + 2 . 3y (-4z) + 2 . (-4z). 2x
= (2x + 3y + (-4z))²
= (2x + 3y – 4z)²
= (2x + 3y – 4z) (2x + 3y – 4z)

(ii) We have given that
2x² + y² + 8z² – 2√2xy + 4√2yz – 8xz
We can also write
(-√2x)² + (y)² + (2√2z)² + 2 . (-√2x) . y + 2 . y . (2√2z) + 2 . (2√2 z) . (-√2x)
Now, we know that
a² + b² + c² + 2ab + 2bc + 2ca = (a + b + c)²
Therefore,
(-√2x)² + (y)² + (2√2z)² + 2 . (-√2x) . y + 2 . y . (2√2z) + 2 . (2√2z) . (-√2x)
= ((-√2x) + y + 2√2z)²
= (-√2x + y + 2√2z) (-√2x + y + 2√2z)

Question 6.
Write the following cubes in expanded front:
(i) (2x + 1)³
(ii) (2a – 3b)³
(iii) \(\left[\frac{3}{2} x+1\right]^{3}\)
(iv) \(\left[x-\frac{2}{3} y\right]\)
Solution:
(i) We have given that, (2x + 1)³
We know that
(a + b)³ = a³ + b³ + 3ab(a + b)
Therefore,
(2x + 1)³ = (2x)³ + (1)³ + 3 . (2x)(1)(2x + 1)
= 8x³ + 1 + 6x(2x + 1)
= 8x³ + 1 + 12x³ + 6x
= 8x³ + 12x³ + 6x + 1

(ii) We have given, (2a – 3b)³
We know that (a – b)³ = a³ – b³ – 3ab(a – b)
(2a – 3b)³
= 8a³ – 27b³ – 18ab(2a – 3b)
= 8a³ – 27b³ – 36a²b + 54ab²

(iii) We have given that, \(\left[\frac{3}{2} x+1\right]^{3}\)
We know that
(a + b)³ = a³ + b³ + 3ab(a + b)
Therefore,

(iv) We have given that, \(\left[x-\frac{2}{3} y\right]\)
We know that
(a – b)³ = a³ – b³ – 3ab(a – b)
Therefore,

Question 7.
Evaluate the following using suitable identities:
(i) (99)³
(ii) (102)³
(iii) (998)³
Solution:
(i) We have given, (99)³
We can also write, (100 – 1)³
We know that (a – b)³ = a³ – b³ – 3ab(a – b)
(100 – 1)³ = (100)³ – (1)³ – 3 . 100 . 1 (100 – 1)
= 10,00,000 – 1 – 30000 + 300
= 9,70,299

(ii) We have given that, (102)³
We can also write (100 + 2)³
Now, we know that
(a + b)³ = a³ + b³ + 3ab(a + b)
(100 + 2)³ = (100)³ + (2)³+ 3 . 100 . 2 (100 + 2)
= 10,00,000 + 8 + 600(100 + 2)
= 10,00,000 + 8 + 60000 + 1200
= 10,61,208

(iii) We have given that (998)³
We can also write (1000 – 2)³
Now, we know that (a – b)³ = a³ – b³ – 3ab(a – b)
(1000 – 2)³ = (1000)³ – (2)³ – 3 . 1000 . 2(1000 – 2)
= 1,00,00,00,000 – 8 – 6000 (100 – 2)
= 99,40,11,992

Question 8.
Factarise each of the following:
(i) 8a³ + b³ + 12a²b + 6ab²
(ii) 8a³ – b³ – 12a²b + 6ab²
(iii) 27 – 125a³ – 135a + 225a³
(iv) 64a³ – 27b³ -144a²b + 108ab²
(v) \(27 \mathrm{P}^{3}-\frac{1}{216}-\frac{9}{2} \mathrm{P}^{2}+\frac{1}{4} \mathrm{P}\)
Solution:
(i) We have given that
8a³ + b³ + 12a²b + 6ab²
We can also write
(2a)³ + (b)³ + 6ab(2a + b)
or, (2a)³ + b³ + 3 . 2a . b(2a + b)
We know that
a³ + b³ + 3ab(a + b) = (a + b)³
(2a)³ + (b)³ + 3 . 2a . b (2a + b) = (2a + b)³

(ii) We have given that
8a³ – b³– 12a²b + 6ab²
We can also write
(2a)³ – (b)³ – 6ab(2a – b)
or, (2a)³ – (b)³ – 3 . 2a . b(2a – b)
We know that
a³ – b³ – 3ab(a – b) = (a – b)³
(2a)³ – (b)³ – 3 . 2a . b(2a – b) = (2a – b)³

(iii) We have given that
27 – 125a³ – 135a + 225a²
We can also write (3)³ – (5a)³ – 45a(3 – 5a)
or, (3)³ – (5a)³ – 3 . 3 . 5a(3 – 5a)
Now, we know that,
a³ – b³ – 3ab(a – b) = (a – b)³
(3)³ – (5a)³ – 3 . 3 . 5a(3 – 5a)
= (3 – 5a)³
= (3 – 5a) (3 – 5a) (3 – 5a)

(iv) We have given that
64a³ – 27b³ – 144a²b + 108ab²
We can also write
(4a)³ – (3b)³ – 36ab(4a – 3b)
or, (4a)³ – (3b)³ – 3 . 4a . 3b(4a – 3b)
Now, we know that
a³ – b³ – 3ab(a – b) = (a – b)³
(4a)³ – (3b)³ – 3 . 4a . 3b(4a – 3b)
= (4a – 3b)³

(v) We have given that
\(27 \mathrm{P}^{3}-\frac{1}{216}-\frac{9}{2} \mathrm{P}^{2}+\frac{1}{4} \mathbf{P}\)
We can also write,
\((3 P)^{3}-\left(\frac{1}{6}\right)^{3}-3.3 P \cdot \frac{1}{6}\left(3 P-\frac{1}{6}\right)\)
We know that
a³ – b³ – 3ab(a – b) = (a – b)³
\((3 P)^{3}-\left(\frac{1}{6}\right)^{3}-3.3 P \cdot \frac{1}{6}\left(3 P-\frac{1}{6}\right)=\left(3 P-\frac{1}{6}\right)^{3}\)

Question 9.
Verify
(i) x³ + y³ = (x + y) (x² – xy + y²)
(ii) x³ – y³ = (x – y) (x² + xy + y²)
Solution:
(i) R.H.S.
(x + y) (x² – xy + y²)
By actual multiplication we have
x³ – x²y + xy² + x²y – xy² + y³
= x³ + y³
= L.H.S

(ii) R.H.S.
(x – y)(x² + xy + y²)
By actual multiplication we have
x³ + x²y + xy² – x²y – xy² – y³
= x³ – y³
= L.H.S.

Question 10.
Factorise each of the following:
(i) 27y³ + 125z³
(ii) 64m³ – 343n³
Solution:
(i) We have given, 27y³ + 125z³
We can also write, (3y)³ + (5z)³
We know that
a³ + b³ = (a + b)(a² – ab + b²)
Therefore,
(3y)³ + (5z)³ = (3y + 5z) ((3y)² – (3y).(5z) + (5z)²)
= (3y + 5z) (9y² – 15yz + 25z²)

(ii) We have given, 64m³ – 343n³
We can also write (4m)³ – (7n)³
We know that,
a³ – b³ = (a – b)(a² + ab + b²)
Therefore,
(4m)³ – (7n)³ = (4m – 7n)((4m)² + (4m)(7n) + (7n)²)
= (4m – 7n)(16m² + 28mn + 49n²)

Question 11.
Factorise: 27x³ + y³ + z³ – 9xyz
Solution:
We have given,
27x³ + y³ + z³ – 9xyz
We can also write,
(3x)³ + (y)³ + (z)³ – 3 . (3x) . (y) . (z)
We know that
a³ + b³ + c³ – 3abc = (a + b + c) (a² + b² + c² – ab – bc – ca)
Therefore,
(3x)³ + (y)³ + (z)³ – 3 . (3x) (y) (z)
= (3x + y + z) [(3x)² + (y)² + (z)² – (3x)(y) -(y)(z) – (z)(3x)]
= (3x + y + z) (9x² + y² + z² – 3xy – yz – 3zx)

Question 12.
Verify that x³ + y³ + z³ – 3xyz = 1/2 (x + y + z) [(x – y)² + (y – z)² + (z – x)²]
Solution:
We have given L.H.S. is x³ + y³ + z³ – 3xyz
We know that
x³ + y³ + z³ – 3xyz = (x + y + z)(x² + y² + z² – xy – yz – zx)
Multiply and divide 2 we get
= (x + y + z)² × \(\frac{1}{2}\) (x² + y² + z² – xy – yz – zx)
= \(\frac{1}{2}\) (x + y + z)(2x² + 2y² + 2z² – 2xy – 2yz – 2zx)
= \(\frac{1}{2}\) (x + y + z) (x² + y² – 2xy + y² + z² – 2yz + z² + x² – 2zx)
= \(\frac{1}{2}\) (x + y + z)((x – y)² + (y – z)² + (z – x)²)
Hence verified.

Question 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 + z² – xy – yz – zx)
Here, we have given x + y + z = 0
x³ + y³ + z³ – 3xyz = 0 × (x² + y² + z² – xy – yz – zx)
or, x³ + y³ + z³ – 3xyz = 0
x³ + y³ + z³ = 3xyz

Question 14.
Without actually calculating the cubes, find the value of each of the following:
(i) (-12)³ + (7)³ + (5)³
(ii) (28)³ + (-15)³ + (-13)³
Solution:
(i) We have given that (-12)³ + (7)³ + (5)³
First we check the value of x + y + z
We have -12 + 7 + 5 = 0
We know that if x + y + z = 0 then x³ + y³ + z³ = 3xyz
(-12)³ + (7)³ + (5)³ = 3 × (-12) × 7 × 5 = -1260

(ii) We have given that
(28)³ + (-15)³ + (-13)³
First we check the value of x + y + z
We have 28 + (-15) + (-13) = 0
We know that if x + y + z = 0 then,
x³ + y³ + z³ = 3xyz
(28)³ + (-15)³ + (-13)³
= 3 × 28 × (-15) × (-13)
= 16,380

Question 15.
Give possible expressions of the length and breadth of each of the following rectangles in which their areas are given.
(i) Area: 25a² – 35a + 12
(ii) Area: 35y² + 13y – 12
Solution:
(i) We know that
Area of rectangle = length × breadth …. (i)
But, we have given that
Area of rectangle = 25a² – 35a + 12 …. (ii)
from equation (i) and (ii)
l × b = 25a² – 35a + 12
= 25a² – 20a – 15a + 12
= 5a(5a – 4) – 3(5a – 4)
l × b = (5a – 4)(5a – 3)
length (l) = (5a – 4) and breadth (b) = (5a – 3)

(ii) We know that
Area of rectangle = length × breadth ……. (i)
But we have given that
Area of rectangle = 35y² + 13y – 12 ……. (ii)
From equation (i) and (ii)
length × breadth = 35y² + 13y – 12
= 35y² + 28y – 15y – 12
= 7y(5y + 4) – 3(5y + 4)
or, length × breadth = (5y + 4) (7y – 3)
Therefore,
length = (5y + 4)
breadth = (7y – 3)

Question 16.
What are the possible expression for the dimensions of the cuboids whose volumes are given below:
(i) volume = 3x² – 12x
(ii) volume = 12ky² + 8ky – 20k
Solution:
(i) We know that
Volume of cuboid = length × breadth × height ……. (i)
But we have given that
Volume of cuboid = 3x² – 12x …… (ii)
From equation (i) and (ii)
length × breadth × height = 3x² – 12x = 3x(x – 4)
or, length × breadth × height = 3 × x × (x – 4)
length of cuboid = 3
breadth of cuboid = x
height of cuboid = (x – 4)

(ii) We know that,
Volume of cuboid = length × breadth × height …….. (i)
But, we have given that Volume of cuboid = 12ky² + 8ky – 20k ……. (ii)
From equation (i) and (ii)
length × breadth × height
= 12ky² + 8ky – 20k
= 4k (3y² + 2y – 5)
= 4k (3y² + 5y – 3y – 5)
= 4k[y(3y + 5) – 1(3y + 5)]
or, length × breadth × height = 4k(3y + 5)(y – 1)
length of cuboid = 4k
breadth of cuboid = (3y + 5)
and height of cuboid = (y – 1)

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.4

Question 1.
Which of the following polynomials has (x + 1) a factor:
(i) x³ + x² + x + 1
(ii) x⁴ + x³ + x² + x + 1
(iii) x⁴ + 3x³ + 3x² + x +1
(iv) x³ – x² – (2 + √2)x + √2
Solution:
(i) Here, p(x) = x³ + x² + x + 1
and the zero of x + 1 is -1.
So, p(-1) = (-1)³ + (-1)² + (-1) + 1
= -1 + 1 – 1 + 1
= 0
Here, remainder is zero, therefore x + 1 is the factor of x³ + x² + x + 1.

(ii) Here, p(x) = x⁴ + x³ + x² + x + 1
and the zero of x + 1 is -1.
So p(-1) = (-1)⁴ + (-1)³ + (-1)² + (-1) + 1
= 1 – 1 + 1 – 1 + 1
= 1
Here, remainder is 1, therefore x + 1 is not the factor of x⁴ + x³ + x² + x + 1.

(iii) Here, p(x) = x⁴ + 3x³ + 3x² + x + 1
and the zero of x + 1 is -1
So, p(-1) = (-1)⁴ + 3(-1)³ + 3(-1)² + (-1) + 1
= 1 – 3 + 3 – 1 + 1
= 1
Here, remainder is 1, therefore x + 1 is not the factor of x⁴ + 3x³ + 3x² + x + 1.

(iv) Here, p(x) = x³ – x² – (2 + √2)x + √2 and the zero of x + 1 is -1
So, p(-1) = (-1)³ – (-1)² – (2 + √2) (-1) + √2
= -1 – 1 + 2 + √2 + √2
= 2√2
Here, remainder is 2√2 therefore x + 1 is not the 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) pix) = x³ – 4x² + x + 6, g(x) = x – 3
Solution:
(i) Here, p(x) = 2x³ + x² – 2x – 1 and the zero of g(x) = x + 1 is -1
So p(-1) = 2(-1)³ + (-1)² – 2(-1) – 1
= -2 + 1 + 2 – 1
= 0
Here, remainder is zero, therefore g(x) = x + 1 is the factor of p(x) = 2x³ + x² – 2x – 1

(ii) Here, p(x) = x³ + 3x² + 3x + 1
and the zero of g(x) = x + 2 is -2.
So, p(-2) = (-2)³ + 3(-2)² + 3(-2) + 1
= -8 + 12 – 6 + 1
= -1
Here, remainder is -1, therefore g(x) = x + 2 is not the factor of p(x) = x³ + 3x² + 3x + 1

(iii) Here, p(x) = x³ – 4x² + x + 6
and the zero of g(x) = x – 3 is 3.
So, p(3) = (3)³ – 4(3)² + 3 + 6
= 27 – 36 + 3 + 6
= 0
Here, remainder is 0, therefore g(x) = x – 3 is the factor of p(x) = x³ – 4x² + x + 6.

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 – √2x + 1
(iv) p(x) = kx² – 3x + k
Solution:
(i) We have give that x – 1 is the factor of p(x) = x² + x + k
So, (1)² + 1 + k = 0
or, 1 + 1 + k = 0
2 + k = 0
k = -2
Therefore, the value of k in p(x) = x² + x + k is -2.

(ii) We have given that x – 1 is the factor of P(x) = 2x² + kx + √2
P(1) = 0
Now, P(1) = 2(1)² + k(1) + √2
So, 0 = 2 + k + √2
or, k = -2 – √2
or, k = -(2 + √2)
Therefore the value of k in p(x) = 2x² + kx + 2 is -(2 + √2)

(iii) We have given that x – 1 is a factor of
p(x) = kx² – √2x + 1
p(1) = 0
Now, p(1) = k(1)² – √2(1) + 1
So, k – √2 + 1 = 0
k = √2 – 1
Therefore, the value of it in p(x) = kx² – √2x + 1 is √2 – 1.

(iv) We have given that x – 1 is the factor of p(x) = kx² – 3x + k
P(1) = 0
Now, p(1) = k(1)² – 3(1) + k
So, k – 3 + k = 0
or, 2k – 3 = 0
k = \(\frac{3}{2}\)
Therefore the value of k in p(x) = kx² – 3x + k is \(\frac{3}{2}\)

Question 4.
Factorise
(i) 12x² – 7x + 1
(ii) 2x² + 7x + 3
(iii) 6x² + 5x – 6
(iv) 3x² – x – 4
Solution:
(i) we have given 12x² – 7x + 1
We can also write 12x² – 4x – 3x – 1
or, 4x(3x – 1) – 1(3x – 1)
or, (3x – 1)(4x – 1)

(ii) We have given 2x² + 7x + 3
We can also write 2x² + 6x + x + 3
or, 2x(x + 3) + 1(x + 3)
or, (x + 3) (2x + 1)

(iii) We have given 6x² + 5x – 6
We can also write 6x² + 9x – 4x – 6
or, 3x(2x + 3) – 2(2x + 3)
or, (2x + 3) (3x – 2)

(iv) We have given 3x² – x – 4
We can also write 3x² – 4x + 3x – 4
or, x(3x – 4) + 1(3x – 4)
or, (3x – 4)(x + 1)

Question 5.
Factorise:
(i) x³ – 2x² – x + 2
(ii) x³ – 3x² – 9x – 5
(iii) x³ – 13x² + 20
(iv) 2y³ + y² – 2y – 1
Solution:
(i) Let p(x) = x³ – 2x² – x + 2
The factors of 2 are ±1, ±2
By trial, we find that p(2) = 0
So, (x – 2) must be the factor of p(x)
Therefore, x³ – 2x² – x + 2
= (x – 2)(x² – 1)
= (x – 2)(x + 1)(x – 1)
[∵ a² – b² = (a + b)(a – b)]
∴ x³ – 2x² – x + 2 = (x – 2)(x – 1)(x + 1)

(ii) Let P(x) = x³ – 3x² – 9x – 5
The factors of -5 are ±1, ±5
By trial, we find that P(-1) = 0
So, (x + 1) is the fact or of p(x)
Therefore,
x³ – 3x² – 9x – 5
= (x + 1) (x² – 4x – 5)
= (x + 1) [x² – 5x + x – 5]
= (x + 1) [x(x – 5) + 1(x – 5)]
∴ x³ – 3x² – 9x – 5 = (x + 1)(x + 1)(x – 5)

(iii) Let P(x) = x³ + 13x² + 32x + 20
The factor of 20 are +1, ±2, ±4, ±5, ±10, ±20.
By trial, we find that P(-1) = 0
So, (x +1) must be the factor of P(x) Therefore,
x³ + 12x² + 32x + 20
= (x + 1) (x² + 12x + 20)
= (x + 1) [x² + 10x + 2x + 20]
= (x + 1) [x(x +10) + 2(x + 10)]
∴ x³ + 13x² + 32x + 20 = (x + 1)(x + 10) (x + 2)

(iv) Let P(y) = 2y³ + y² – 2y – 1
The factor of -1 are ±1
By trial, we find that P(1) = 0
So, (y – 1) must be the factor of 2y³ + y² – 2y – 1
Therefore,
2y³ + y² – 2y – 1
= (y – 1) (2y² + 3y + 1)
= (y – 1) [2y² + 2y + y + 1]
= (y – 1) [2y(y + 1) + 1(y + 1)]
∴ 2y³ + y² – 2y – 1 = (y – 1)(y + 1)(2y + 1)

These NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 9 Maths Chapter 2 Polynomials Exercise 2.3

Question 1.
Find the remainder when x³ + 3x² + 3x + 1 is divided by
(i) x + 1
(ii) x – \(\frac{1}{2}\)
(iii) x
(iv) x + p
(v) 5 + 2x
Solution:
(i) By long division, we have:

Here, the remainder is 0.

(ii) By long division, we have:

Here, the remainder is \(\frac{27}{8}\)

(iii) By long division, we have

Here, the remainder is 1

(iv) we have given
x³ + 3x² + 3x + 1 ÷ x + π
or, x³ + 3x² + 3x + 1 ÷ x + \(\frac{22}{7}\) (∵ π = \(\frac{22}{7}\))
By long division,

Here, the remainder is \(-\frac{3669}{49}\)

(v) We have given
x³ + 3x² + 3x + 1 ÷ 5 + 2x
or, x³ + 3x² + 3x + 1 ÷ 2x + 5
By long division, we have

Here, remainder is \(-\frac{27}{8}\)

Question 2.
Find the remainder when x³ – ax² + 6x – a is divided by x – a.
Solution:
Here, p(x) = x³ – ax² + 6x – a, and the zero of x – a is a
So, p(a) = a³ – a(a)² + 6(a) – a
= a³ – a³ + 6a – a
= 5a
So, 5a is the remainder when x³ – ax + 6x – a is divided by x – a.

Question 3.
Check whether 7 + 3x is a factor of 3x³ + 7x.
Solution:
We know that 7 + 3x is a factor of 3x³ + 7x only if 7 + 3x divides 3x³ + 7x leaving no remainder.
Here, p(x) = 3x³ + 7x and the zero of 7 + 3x is \(-\frac{7}{3}\). so,

Here, the remainder is \(\frac{-490}{9}\).
Therefore 7 + 3x is not the factor of 3x³ + 7x.