NCERT Solutions for Class 9 Math Chapter 2.5 POLYNOMIALS ALL SOLUTION

EXERCISE 2.1 EXERCISE 2.2 EXERCISE 2.3 EXERCISE 2.4

1. Use suitable identities to find the following products:

( i ) $\ \left( x + 4 \right)\left( x + 10 \right)\$

( ii ) $\ \left( x + 8 \right)\left( x - 10 \right)\$

( iii ) $\ \left( 3x + 4 \right)\left( 3x - 5 \right)\$

( iv ) $\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\$

( v ) $\ \left( 3 - 2x \right)\left( 3 + 2x \right)\$

solution :

( i ) $\ \left( x + 4 \right)\left( x + 10 \right)\$

Here we can use identity IV :

$\ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \$

$\ ( x + 4 ) ( x + 10) \$

$\ = x^2 + ( 4 + 10 )x + 4×10 \$

$\ = x^2 + 14x + 40 \$

( ii ) $\ \left( x + 8 \right)\left( x - 10 \right)\$

Here we can use identity IV :

$\ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \$

$\ ( x + 8 ) ( x - 10) \$

$\ = x^2 + ( 8 + (-10) )x + 8×(-10) \$

$\ = x^2 - 2x - 80 \$

( iii ) $\ \left( 3x + 4 \right)\left( 3x - 5 \right)\$

Here we can use identity IV :

$\ ( x + a ) ( x + b ) = x^2 + ( a + b )x + ab \$

$\ ( 3x + 4 ) ( 3x - 5) \$

$\ = (3x)^2 + ( 4 + (-5) )3x + 4×(-5) \$

$\ = 9x^2 - 3x - 20 \$

( iv ) $\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\$

Here we can use identity III :

$\ ( x + a ) ( x - a ) = x^2 - a^2 \$

$\ ( y^2 + \frac{3} {2} ) ( y^2 - \frac{3} {2}) \$

$\ = (y^2)^2 - (\frac{3} {2})^2 \$

$\ = y^4 - \frac{9} {4} \$

( v ) $\ \left( 3 - 2x \right)\left( 3 + 2x \right)\$

Here we can use identity III :

$\ ( x + a ) ( x - a ) = x^2 - a^2 \$

$\ ( 3 - 2x ) ( 3 + 2x) \$

$\ = 3^2 - (2x)^2 \$

$\ = 9 - 4x^2 \$

2. Evaluate the following products without multiplying directly:

( i ) $\ 103 \times 107\$

( ii ) $\ 95 \times 96\$

( iii ) $\ 104 \times 96\$

solution :

( i ) $\ 103 \times 107\$

$\ 103 × 107 = (100 + 3) (100 + 7) \$

$\ = (100)^2 + (3 + 7)(100) + (3×7) \$ using identity IV

$\ = 10000 + 1000 + 21 \$

$\ = 11021 \$

( ii ) $\ 95 \times 96\$

$\ 95 × 96 = (100 + (-5)) (100 + (-4)) \$

$\ = (100)^2 + ((-5) + (-4)) \$ $\ (100) + ((-5)×(-4)) \$

using identity IV

$\ = 10000 + (-900) + 20 \$

$\ = 10000 - 900 + 20\$

$\ = 9120 \$

( iii ) $\ 104 \times 96\$

$\ 104 × 96 = (100 + 4) (100 - 4) \$

$\ = (100)^2 - 4^2 \$ using identity III

$\ = 10000 - 16 \$

$\ = 9984 \$

3. Factorise the following using appropriate identities:

( i ) $\ 9 x^2 + 6xy + y^2 \$

( ii ) $\ 4 y^2 - 4y + 1 \$

( iii ) $\ x^2 - \frac{ y^2 }{ 100 }\$

solution :

( i ) $\ 9 x^2 + 6xy + y^2 \$

Here you can see that

$\ 9 x^2 + 6xy + y^2 \$

$\ 9x^2 = (3x)^2, \$ $\ y^2 = (y)^2, \$ $\ 6xy = 2 (3x) (y) \$

Comparing the given expression with $\ x^2 + 2xy + y^2 ,\$ we observe that $\ x = 3x \$ and $\ y = y \$ .

Using Identity I, we get

$\ 9 x^2 + 6xy + y^2 \$ $\ = ( 3x + y )^2 \$

$\ = ( 3x + y ) ( 3x + y ) \$

( ii ) $\ 4 y^2 - 4y + 1 \$

Here you can see that

$\ 4 y^2 - 4y + 1 \$

$\ 4y^2 = (2y)^2, \$ $\ 1 = (1)^2, \$ $\ 4y = 2 (2y) (1) \$

Comparing the given expression with $\ x^2 - 2xy + y^2 ,\$ we observe that $\ x = 2y \$ and $\ y = 1 \$ .

Using Identity II, we get

$\ 4 y^2 - 4y + 1 \$ $\ = ( 2y - 1 )^2 \$

$\ = ( 2y - 1 ) ( 2y - 1 ) \$

( iii ) $\ x^2 - \frac{ y^2 }{ 100 }\$

Here you can see that

$\ x^2 - \frac{ y^2 }{ 100 }\$

$\ x^2 = (x)^2, \$ $\ \frac {y^2} {100} = (\frac {y} {10})^2, \$

Comparing the given expression with $\ x^2 - y^2 ,\$ we observe that $\ x = x \$ and $\ y = \frac {y} {10} \$ .

Using Identity III, we get

$\ x^2 - \frac{ y^2 }{ 100 }\$ $\ = ( x + \frac {y} {10} )(x - \frac {y} {10}) \$

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

( i ) $\ \left( x + 2y + 4z \right)^2 \$

( ii ) $\ \left( 2x - y + z \right)^2 \$

( iii ) $\ \left( -2x + 3y + 2z \right)^2 \$

( iv ) $\ \left( 3a - 7b - c \right)^2 \$

( v ) $\ \left( -2x + 5y - 3z \right)^2 \$

( vi ) $\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\$

solution :

( i ) $\ \left( x + 2y + 4z \right)^2 \$

Comparing the given expression with $\ ( x + 2y + 4z )^2 \$ , we find that

$\ x = x,\$ $\ y = 2y \$ and $\ z = 4z \$ .

Therefore, using Identity V, we have

$\ ( x + 2y + 4z )^2 \$

$\ = (x)^2 + (2y)^2 + (4z)^2 \$ $\ + 2 (x) (2y) + 2 (2y) (4z) + 2 (4z) (x) \$

$\ = x^2 + 4y^2 + 16z^2 \$ $\ + 4xy + 16yz + 8xz \$

( ii ) $\ \left( 2x - y + z \right)^2 \$

Comparing the given expression with $\ ( 2x - y + z )^2 \$ , we find that

$\ x = 2x,\$ $\ y = -y \$ and $\ z = z \$ .

Therefore, using Identity V, we have

$\ ( 2x - y + z )^2 \$

$\ = (2x)^2 + (-y)^2 + (z)^2 \$ $\ + 2 (2x) (-y) + 2 (-y) (z) + 2 (z) (2x) \$

$\ = 4x^2 + y^2 + z^2 \$ $\ - 4xy - 2yz + 4xz \$

( iii ) $\ \left( -2x + 3y + 2z \right)^2 \$

Comparing the given expression with $\ ( -2x + 3y + 2z )^2 \$ , we find that

$\ x = -2x,\$ $\ y = 3y \$ and $\ z = 2z \$ .

Therefore, using Identity V, we have

$\ ( -2x + 3y + 2z )^2 \$

$\ = (-2x)^2 + (3y)^2 + (2z)^2 \$ $\ + 2 (-2x) (3y) + 2 (3y) (2z) + 2 (2z) (-2x) \$

$\ = 4x^2 + 9y^2 + 4z^2 \$ $\ - 12xy + 12yz - 8xz \$

( iv ) $\ \left( 3a - 7b - c \right)^2 \$

Comparing the given expression with $\ ( 3a - 7b - c )^2 \$ , we find that

$\ x = 3a,\$ $\ y = -7b \$ and $\ z = -c \$ .

Therefore, using Identity V, we have

$\ ( 3a - 7b - c )^2 \$

$\ = (3a)^2 + (-7b)^2 + (-c)^2 \$ $\ + 2 (3a) (-7b) + 2 (-7b) (-c) + 2 (-c) (3a) \$

$\ = 9a^2 + 49b^2 + c^2 \$ $\ - 42ab + 14bc - 6ac \$

( v ) $\ \left( -2x + 5y - 3z \right)^2 \$

Comparing the given expression with $\ ( -2x + 5y - 3z )^2 \$ , we find that

$\ x = -2x,\$ $\ y = 5y \$ and $\ z = -3z \$ .

Therefore, using Identity V, we have

$\ ( -2x + 5y - 3z )^2 \$

$\ = (-2x)^2 + (5y)^2 + (-3z)^2 \$ $\ + 2 (-2x) (5y) + 2 (5y) (-3z) + 2 (-3z) (-2x) \$

$\ = 4x^2 + 25y^2 + 9z^2 \$ $\ - 20xy - 30yz + 12xz \$

( vi ) $\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\$

Comparing the given expression with $\ [ \frac {1} {4}a - \frac {1} {2}b + 1 ]^2 \$ , we find that

$\ x = \frac {1} {4}a,\$ $\ y = - \frac {1} {2}b \$ and $\ z = 1 \$ .

Therefore, using Identity V, we have

$\ [ \frac {1} {4}a - \frac {1} {2}b + 1 ]^2 \$

$\ = (\frac {1} {4}a)^2 + (-\frac {1} {2}b)^2 + (1)^2 \$ $\ + 2 (\frac {1} {4}a) (- \frac {1} {2}b) + 2 (- \frac {1} {2}b) (1) + 2 (1) (\frac {1} {4a}) \$

$\ = \frac {1} {16}a^2 + \frac {1} {4}b^2 + 1 \$ $\ - \frac {1} {4}ab - b + \frac {1} {2}a \$

5. Factorise:

( i ) $\ 4 x^2 + 9 y^2 + 16 z^2 \$ $\ + 12xy - 24yz - 16xz\$

( ii ) $\ 2 x^2 + y^2 + 8 z^2 \$ $\ - 2\sqrt 2 xy - 4\sqrt 2 yz - 8xz\$

solution :

( i ) $\ 4 x^2 + 9 y^2 + 16 z^2 \$ $\ + 12xy - 24yz - 16xz\$

We have $\ 4 x^2 + 9 y^2 + 16 z^2 \$ $\ + 12xy - 24yz - 16xz\$

$\ = (2x)^2 + (3y)^2 + (-4z)^2 \$ $\ + 2 (2x) (3y) + 2 (3y) (-4z) + 2 (-4z) (2x) \$

⇒ $\ [ 2x + 3y + (-4z) ]^2 \$ ( Using Identity V )

⇒ $\ ( 2x + 3y - 4z )^2 \$ $\ = ( 2x + 3y - 4z ) \$ $\ ( 2x + 3y - 4z ) \$

( ii ) $\ 2 x^2 + y^2 + 8 z^2 \$ $\ - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz\$

We have $\ 2 x^2 + y^2 + 8 z^2 \$ $\ - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz\$

$\ = (- \sqrt 2 x)^2 + (y)^2 + (2 \sqrt 2 z)^2 \$ $\ + 2 (- \sqrt 2 x) (y) + 2 (y) (2 \sqrt 2 z) + 2 (2 \sqrt 2z) (- \sqrt 2 x) \$

$\ [ (- \sqrt 2 x) + y + 2 \sqrt 2 z ]^2 \$ ( Using Identity V )

$\ (- \sqrt 2 x + y = 2 \sqrt 2 z )^2 \$ $\ = (- \sqrt 2 x + y = 2 \sqrt 2 z ) \$ $\ (- \sqrt 2 x + y = 2 \sqrt 2 z ) \$

6. Write the following cubes in expanded form:

( i ) $\ \left( 2x + 1 \right)^3 \$

( ii ) $\ \left( 2a - 3b \right)^3 \$

( iii ) $\ \left[ \frac{3}{2}x + 1 \right]^3 \$

( iv ) $\ \left[ x + \frac{2}{3}y \right]^3 \$

solution :

( i ) $\ \left( 2x + 1 \right)^3 \$

Comparing the given expression with $\ (x + y)^3 \$ , we find that

$\ x = 2x and y = 1. \$

So, using Identity VI, we have:

$\ ( 2x + 1 )^3 \$ $\ = (2x)^3 + (1)^3 + 3 (2x) (1) ( 2x + 1 ) \$

$\ 8x^3 + 1 + 6x ( 2x + 1 ) \$

$\ 8x^3 + 1 + 12x^2 + 6x \$

$\ 8x^3 + 12x^2 + 6x + 1 \$

( ii ) $\ \left( 2a - 3b \right)^3 \$

Comparing the given expression with $\ (x + y)^3 \$ , we find that

$\ x = 2a and y = 3b. \$

So, using Identity VII, we have:

$\ \left( 2a - 3b \right)^3 \$ $\ = (2a)^3 - (3b)^3 - 3 (2a) (3b) ( 2a - 3b ) \$

$\ 8a^3 - 27b^3 - 18ab ( 2a - 3b ) \$

$\ 8a^3 - 27b^3 - 36a^2b + 54ab^2 \$

( iii ) $\ \left[ \frac{3}{2}x + 1 \right]^3 \$

Comparing the given expression with $\ (x + y)^3 \$ , we find that

$\ x = \frac {3} {2}x and y = 1. \$

So, using Identity VI, we have:

$\ \left[ \frac{3}{2}x + 1 \right]^3 \$ $\ = (\frac {3} {2}x)^3 + (1)^3 + 3 (\frac {3} {2}x) (1) ( \frac {3} {2}x + 1 ) \$

$\ \frac {27} {8}x^3 + 1 + \frac {9} {2}x ( \frac {3} {2}x + 1 ) \$

$\ \frac {27} {8}x^3 + 1 + \frac {27} {4}x^2 + \frac {9} {2}x \$

$\ \frac {27} {8}x^3 + \frac {27} {4}x^2 + \frac {9} {2}x + 1 \$

( iv ) $\ \left[ x - \frac{2}{3}y \right]^3 \$

Comparing the given expression with $\ (x + y)^3 \$ , we find that

$\ x = x and y = \frac {2} {3}y . \$

So, using Identity VII, we have:

$\ \left[ x - \frac{2}{3}y \right]^3 \$ $\ = (x)^3 - (\frac {2} {3}y)^3 - 3 (x) (\frac {2} {3}y) ( x - \frac {2} {3}y ) \$

$\ x^3 - \frac {8} {27} y^3 - 2xy ( x - \frac {2} {3}y ) \$

$\ x^3 - \frac {8} {27} y^3 - 2x^2y + \frac {4} {3}xy^2 \$

7. Evaluate the following using suitable identities:

( i ) $\ \left( 99 \right)^3 \$

( ii ) $\ \left( 102 \right)^3 \$

( iii ) $\ \left( 998 \right)^3 \$

solution :

( i ) $\ \left( 99 \right)^3 \$

We have

$\ (99)^3 = ( 100 - 1 )^3\$

$\ = (100)^3 - (1)^3 - 3 (100) (1) (100 - 1) \$ ( Using identity VII )

$\= 1000000 - 1 - 300 (100 - 1)\$

$\= 999999 - 30000 + 300\$

$\= 969999 + 300\$

$\= 970299\$

( ii ) $\ \left( 102 \right)^3 \$

We have

$\ (102)^3 = ( 100 + 2 )^3\$

$\ = (100)^3 + (2)^3 + 3 (100) (2) (100 + 2) \$ ( Using identity VI )

$\= 1000000 + 8 + 600 (100 + 2)\$

$\= 1000008 + 60000 + 1200\$

$\= 1060008 + 1200\$

$\= 1061208\$

( iii ) $\ \left( 998 \right)^3 \$

We have

$\ (998)^3 = ( 1000 - 2 )^3\$

$\ = (1000)^3 - (2)^3 - 3 (1000) (2) (1000 - 2) \$ ( Using identity VII )

$\= 1000000000 - 8 - 6000 (1000 - 2)\$

$\= 999999992 - 6000000 + 12000\$

$\= 993999992 + 12000\$

$\= 994011992\$

8. Factorise each of the following:

( i ) $\ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \$

( ii ) $\ 8a^3 - b^3 - 12a^2b + 6ab^2 \$

( iii ) $\ 27 - 125a^3 - 135a + 225a^2 \$

( iv ) $\ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \$

( v ) $\ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \$

solution :

( i ) $\ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \$

The given expression can be written as

$\(2a)^3 + (b)^3 + 3 (4a^2) (b) + 3 ( 2a ) ( b^2 )\$

$\(2a)^3 + (b)^3 + 3 (2a)^2 (b) + 3 (2a) (b)^2\$

$\(2a + b )^3\$ ( Using identity VI )

$\ (2a+b) \$ $\ (2a+b) \$ $\ (2a+b) \$

( ii ) $\ 8a^3 - b^3 - 12a^2b + 6ab^2 \$

The given expression can be written as

$\(2a)^3 - (b)^3 - 3 (4a^2) (b) + 3 ( 2a ) ( b^2 )\$

$\(2a)^3 - (b)^3 - 3 (2a) (b) (2a - b)\$

$\(2a - b )^3\$ ( Using identity VII )

$\ (2a-b) \$ $\ (2a-b) \$ $\ (2a-b) \$

( iii ) $\ 27 - 125a^3 - 135a + 225a^2 \$

The given expression can be written as

$\(3)^3 - (5a)^3 - 3 (3^2) (5a) + 3 ( 3 ) ( (5a)^2 )\$

$\(3)^3 - (5a)^3 - 3 (3) (5a) (3 - 5a) \$

$\(3 5a )^3\$ ( Using identity VII )

$\ (3 - 5a) \$ $\ (3 - 5a) \$ $\ (3-5a) \$

( iv ) $\ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \$

The given expression can be written as

$\(4a)^3 - (3b)^3 - 3 ((4a)^2) (3b) + 3 ( 4a ) ( (3b)^2 )\$

$\(4a)^3 - (3b)^3 - 3 (4a) (3b) (2a - 3b) \$

$\(4a - 3b )^3\$ ( Using identity VII )

$\ (4a-3b) \$ $\ (4a-3b) \$ $\ (4a-3b) \$

( v ) $\ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \$

The given expression can be written as

$\(3p)^3 - (\frac {1} {6})^3 - 3 ((3p)^2) (\frac {1} {6}) + 3 ( 3p ) ( (\frac {1} {6})^2 )\$

$\(3p)^3 - (\frac {1} {6})^3 - 3 (3p) (\frac {1} {6}) (3p - \frac {1} {6}) \$

$\(3p - \frac {1} {6} )^3\$ ( Using identity VII )

$\ (3p - \frac{1} {6}) \$ $\ (3p - \frac{1} {6}) \$ $\ (3p - \frac{1} {6}) \$

9. Verify :

( i ) $\ x^3 + y^3 = ( x + y ) \$ $\ ( x^2 - xy + y^2 ) \$

( ii ) $\ x^3 - y^3 = ( x - y ) \$ $\( x^2 + xy + y^2 ) \$

solution :

( i ) $\ x^3 + y^3 = ( x + y ) \$ $\ ( x^2 - xy + y^2 ) \$

We know that,

$\(x+y)^3 = x^3 +y^3 +3xy(x+y) \$

⇒ $\x^3 + y^3 = (x+y)^3 -3xy(x+y)\$

⇒ $\x^3 + y^3 = (x+y) [(x+y)^2 - 3xy]\$ ( Taking $\(x+y)\$ common )

⇒ $\x^3+y^3=(x+y)[x^2+y^2+2xy-3xy] \$

⇒ $\x^3+y^3=(x+y)(x^2-xy+y^2) \$

Proved.

( ii ) $\ x^3 - y^3 = ( x - y ) \$ $\( x^2 + xy + y^2 ) \$

We know that,

$\(x-y)^3 = x^3 -y^3 -3xy(x-y) \$

⇒ $\x^3 - y^3 = (x-y)^3 +3xy(x-y)\$

⇒ $\x^3 - y^3 = (x-y) [(x-y)^2 + 3xy]\$ ( Taking $\(x-y)\$ common )

⇒ $\x^3+-y^3=(x-y)[x^2+y^2-2xy+3xy] \$

⇒ $\x^3-y^3=(x-y)(x^2+xy+y^2) \$

Proved.

10. Factorise each of the following:

( i ) $\ 27y^3 + 125z^3 \$

( ii ) $\ 64m^3 - 343n^3 \$

solution :

( 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-xy+y^2) \$

$\27y^3+125z^3\$ = $\ (3y)^3+(5z)^3 \$

= $\(3y+5z)((3x)^2-(3x)(5z)+(5z)^2) \$

= $\(3y+5z)(9y^2 -15yz+25z^2)\$

( ii ) $\ 64m^3 - 343n^3 \$

The expression, $\64m^3-343n^3\$ can be written as $\ (4m)^3-(7n)^3 \$

$\64m^3-343n^3\$ = $\ (4m)^3+(7n)^3 \$

We know that,

⇒ $\x^3-y^3=(x-y)(x^2+xy+y^2) \$

$\64m^3-343n^3\$ = $\ (4m)^3-(7n)^3 \$

= $\(4m-7n)((4m)^2+(4m)(7n)+(7n)^2) \$

= $\(4m-7n)(16m^2 +28mn+49n^2)\$

11. Factorise : $\ 27x^3 + y^3 + z^3 - 9xyz \$

solution :

Factorise : $\ 27x^3 + y^3 + z^3 - 9xyz \$

Here, we have

$\ 27x^3 + y^3 + z^3 - 9xyz \$

$\(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)\$ $\(9x^2+y^2+z^2-3xy-yz-3xz)\$

12. Verify that $\ x^3 + y^3 + z^3 - 3xyz \$ $\ = \frac{1} {2} ( x + y + z ) \$ $\ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \$

solution :

Verify that $\ x^3 + y^3 + z^3 - 3xyz \$ $\ = \frac{1} {2} ( x + y + z ) \$ $\ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \$

We know that,

$\x^3+y^3+z^3-3xyz = \$ $\(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\$

⇒ $\ x^3 + y^3 + z^3 - 3xyz \$ $\ = \frac{1} {2} ( x + y + z ) \$ $\ [2 (x^2 + y^2 + z^2-xy-yz-zx) ] \$

= $\ \frac{1} {2} ( x + y + z ) \$ $\ (2x^2 +2y^2 +2z^2 -2xy-2yz-2zx) \$

= $\ \frac{1} {2} ( x + y + z ) \$ $\ [(x^2+y^2-2xy)(y^2+z^2-2yz)(z^2+x^2-2zx)] \$

= $\ x^3 + y^3 + z^3 - 3xyz \$ $\ = \frac{1} {2} ( x + y + z ) \$ $\ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \$

13. If $\ x + y + z = 0 \$ , show that $\ x^3 + y^3 + z^3 = 3xyz \$ .

solution :

If $\ x + y + z = 0 \$ , show that $\ x^3 + y^3 + z^3 = 3xyz \$ .

We know that,

$\x^3+y^3+z^3-3xyz = \$ $\(x+y+z)(x^2+y^2+z^2-xy-yz-zx)\$

Now, according to the question, let $\ (x+y+z) = 0, \$

then, $\x^3+y^3+z^3 -3xyz = \$ $\ (0)(x^2+y^2+z^2–xy–yz–xz) \$

⇒ $\x^3+y^3+z^3–3xyz = 0 \$

⇒ $\x^3+y^3+z^3=3xyz \$

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 ) $\ ( -12 )^3 + ( 7 )^3 + ( 5 )^3 \$

Let a $\ = −12 \$

$\b = 7\$

$\c = 5\$

We know that if $\x+y+z = 0,\$ then $\x^3+y^3+z^3=3xyz \$ .

Here, $\−12+7+5=0\$

$\(−12)^3+(7)^3+(5)^3 = 3xyz\$

$\= 3×(-12)×7×5\$

$\= -1260\$

( ii ) $\ ( 28 )^3 + ( -15 )^3 + ( -13 )^3 \$

Let a $\ = 28 \$

$\b = -15\$

$\c = -13\$

We know that if $\x+y+z = 0,\$ then $\x^3+y^3+z^3=3xyz \$ .

Here, $\28+(-15)+(-13)=0\$

$\(28)^3+(-15)^3+(-13)^3 = 3xyz\$

$\= 3×(28)×(-15)×(-13)\$

$\= 16380\$

15. Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:

( i ) $\ Area : 25a^2 - 35a + 12 \$

( ii ) $\ Area : 35y^2 - 13y - 12 \$

solution :

( i ) $\ Area : 25a^2 - 35a + 12 \$

By splitting method:

If we can find two numbers p and q such that,

$\p+q=-35\$ and p q = 25 × 12 = 300

then we can get the factor.

So, let us look for the pair of factors of 300.

Some are − 15 and − 20.

of these pairs, − 15 and − 20 will giue us p + q = − 35.

So, $\ 25a^2-35a+12\$ $\ = 25a^2 +[(-15)+(-20)]a+12\$

= $\25a^2-15a-20a+12\$

= $\5a(5a-3)-4(5a-3)\$

= $\(5a-4) (5a-3)\$

= Possible expression for length $\= (5a–4)\$

= Possible expression for breadth $\= (5a –3)\$

( ii ) $\ Area : 35y^2 - 13y - 12 \$

By splitting method:

If we can find two numbers p and q such that,

$\p+q=-13\$ and p q = 35 × − 12 = −420

then we can get the factor.

So, let us look for the pair of factors of − 420.

Some are − 15 and 28.

of these pairs, − 15 and 28 will giue us p + q = − 13.

So, $\ 35y^2-13y-12\$ $\ = 35y^2 +[(-15)+28]y-12\$

= $\35y^2-15y+28y-12\$

= $\5y(7y-3)-4(7y-3)\$

= $\(5y-4) (7y-3)\$

= Possible expression for length $\= (5y–4)\$

= Possible expression for breadth $\= (7y –3)\$

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

( i ) $\ Volume : 3x^2 - 12x \$

( ii ) $\ Volume : 12ky^2 + 8ky - 20k \$

solution :

( i ) $\ Volume : 3x^2 - 12x \$

$\3x^2–12x\$ can be written as $\3x(x–4)\$

Taking 3x common both the terms.

Possible expression for length = 3

Possible expression for breadth = x

Possible expression for height = (x – 4)

( ii ) $\ Volume : 12ky^2 + 8ky - 20k \$

$\12ky^2+8ky–20k\$ can be written as $\4k(3y^2+2y–5)\$

Taking 4k common both the terms.

Here, $\4k(3y^2+2y-5)\$

By splitting method:

$\4k (3y^2+2y-5)\$ is written as $\4k (3y^2+5y-3y-5)\$

$\4k [ y(3y+5)-1(3y+5)]\$

$\4k [(3y+5) (y-1)]\$

$\4k (3y+5) (y-1)\$

Possible expression for length = 4k

Possible expression for breadth = ( 3y + 5 )

Possible expression for height = (y – 1)

EXERCISE 2.1 EXERCISE 2.2 EXERCISE 2.3 EXERCISE 2.4

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NCERT Solutions for Class 9 Math Chapter 2.5 POLYNOMIALS ALL SOLUTION

1. Use suitable identities to find the following products:

( i ) (x+4)(x+10)\ \left( x + 4 \right)\left( x + 10 \right)\

( ii ) (x+8)(x-10)\ \left( x + 8 \right)\left( x - 10 \right)\

( iii ) (3x+4)(3x-5)\ \left( 3x + 4 \right)\left( 3x - 5 \right)\

( iv ) (y2+32)(y2-32)\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\

( v ) (3-2x)(3+2x)\ \left( 3 - 2x \right)\left( 3 + 2x \right)\

solution :

2. Evaluate the following products without multiplying directly:

( i ) 103×107\ 103 \times 107\

( ii ) 95×96\ 95 \times 96\

( iii ) 104×96\ 104 \times 96\

solution :

3. Factorise the following using appropriate identities:

( i ) 9x2+6xy+y2\ 9 x^2 + 6xy + y^2 \

( ii ) 4y2-4y+1\ 4 y^2 - 4y + 1 \

( iii ) x2-y2100\ x^2 - \frac{ y^2 }{ 100 }\

solution :

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

( i ) (x+2y+4z)2\ \left( x + 2y + 4z \right)^2 \

( ii ) (2x-y+z)2\ \left( 2x - y + z \right)^2 \

( iii ) (-2x+3y+2z)2\ \left( -2x + 3y + 2z \right)^2 \

( iv ) (3a-7b-c)2\ \left( 3a - 7b - c \right)^2 \

( v ) (-2x+5y-3z)2\ \left( -2x + 5y - 3z \right)^2 \

( vi ) [14x-12b+1]\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\

solution :

5. Factorise:

( i ) 4x2+9y2+16z2\ 4 x^2 + 9 y^2 + 16 z^2 \ +12xy-24yz-16xz \ + 12xy - 24yz - 16xz\

( ii ) 2x2+y2+8z2\ 2 x^2 + y^2 + 8 z^2 \ -2√2xy-4√2yz-8xz \ - 2\sqrt 2 xy - 4\sqrt 2 yz - 8xz\

solution :

6. Write the following cubes in expanded form:

( i ) (2x+1)3\ \left( 2x + 1 \right)^3 \

( ii ) (2a-3b)3\ \left( 2a - 3b \right)^3 \

( iii ) [32x+1]3\ \left[ \frac{3}{2}x + 1 \right]^3 \

( iv ) [x+23y]3\ \left[ x + \frac{2}{3}y \right]^3 \

solution :

7. Evaluate the following using suitable identities:

( i ) (99)3\ \left( 99 \right)^3 \

( ii ) (102)3\ \left( 102 \right)^3 \

( iii ) (998)3\ \left( 998 \right)^3 \

solution :

8. Factorise each of the following:

( i ) 8a3+b3+12a2b+6ab2 \ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \

( ii ) 8a3-b3-12a2b+6ab2 \ 8a^3 - b^3 - 12a^2b + 6ab^2 \

( iii ) 27-125a3-135a+225a2 \ 27 - 125a^3 - 135a + 225a^2 \

( iv ) 64a3-27b3-144a2b+108ab2 \ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \

( v ) 27p3-1216-92p2+14p \ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \

solution :

9. Verify :

( i ) x3+y3=(x+y) \ x^3 + y^3 = ( x + y ) \ (x2-xy+y2)\ ( x^2 - xy + y^2 ) \

( ii ) x3-y3=(x-y) \ x^3 - y^3 = ( x - y ) \ (x2+xy+y2) \( x^2 + xy + y^2 ) \

solution :

10. Factorise each of the following:

( i ) 27y3+125z3 \ 27y^3 + 125z^3 \

( ii ) 64m3-343n3 \ 64m^3 - 343n^3 \

solution :

11. Factorise : 27x3+y3+z3-9xyz \ 27x^3 + y^3 + z^3 - 9xyz \

solution :

12. Verify that x3+y3+z3-3xyz\ x^3 + y^3 + z^3 - 3xyz \ =12(x+y+z) \ = \frac{1} {2} ( x + y + z ) \ [(x-y)2+(y-z)2+(z-x)2] \ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \

solution :

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

solution :

14. Without actually calculating the cubes, find the value of each of the following:

( i ) (-12)3+(7)3+(5)3 \ ( -12 )^3 + ( 7 )^3 + ( 5 )^3 \

( ii ) (28)3+(-15)3+(-13)3 \ ( 28 )^3 + ( -15 )^3 + ( -13 )^3 \

solution :

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 \ Area : 25a^2 - 35a + 12 \

( ii ) Area:35y2-13y-12 \ Area : 35y^2 - 13y - 12 \

solution :

16. What are the possible expressions for the dimensions of the cuboids whose volumes are given below?

( i ) Volume:3x2-12x \ Volume : 3x^2 - 12x \

( ii ) Volume:12ky2+8ky-20k \ Volume : 12ky^2 + 8ky - 20k \

solution :

vishwakarma k.

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NCERT Solutions for Class 9 Math Chapter 1.1 NUMBER SYSTEM ALL SOLUTION

( i ) $\ \left( x + 4 \right)\left( x + 10 \right)\$

( ii ) $\ \left( x + 8 \right)\left( x - 10 \right)\$

( iii ) $\ \left( 3x + 4 \right)\left( 3x - 5 \right)\$

( iv ) $\ \left( y^2 + \frac{3}{2} \right)\left( y^2 - \frac{3}{2} \right)\$

( v ) $\ \left( 3 - 2x \right)\left( 3 + 2x \right)\$

( i ) $\ 103 \times 107\$

( ii ) $\ 95 \times 96\$

( iii ) $\ 104 \times 96\$

( i ) $\ 9 x^2 + 6xy + y^2 \$

( ii ) $\ 4 y^2 - 4y + 1 \$

( iii ) $\ x^2 - \frac{ y^2 }{ 100 }\$

( i ) $\ \left( x + 2y + 4z \right)^2 \$

( ii ) $\ \left( 2x - y + z \right)^2 \$

( iii ) $\ \left( -2x + 3y + 2z \right)^2 \$

( iv ) $\ \left( 3a - 7b - c \right)^2 \$

( v ) $\ \left( -2x + 5y - 3z \right)^2 \$

( vi ) $\ \left[ \frac{1}{4}x - \frac{1}{2}b + 1 \right]\$

( i ) $\ 4 x^2 + 9 y^2 + 16 z^2 \$ $\ + 12xy - 24yz - 16xz\$

( ii ) $\ 2 x^2 + y^2 + 8 z^2 \$ $\ - 2\sqrt 2 xy - 4\sqrt 2 yz - 8xz\$

( i ) $\ \left( 2x + 1 \right)^3 \$

( ii ) $\ \left( 2a - 3b \right)^3 \$

( iii ) $\ \left[ \frac{3}{2}x + 1 \right]^3 \$

( iv ) $\ \left[ x + \frac{2}{3}y \right]^3 \$

( i ) $\ \left( 99 \right)^3 \$

( ii ) $\ \left( 102 \right)^3 \$

( iii ) $\ \left( 998 \right)^3 \$

( i ) $\ 8 a^3 + b^3 + 12 a^2b + 6 ab^2 \$

( ii ) $\ 8a^3 - b^3 - 12a^2b + 6ab^2 \$

( iii ) $\ 27 - 125a^3 - 135a + 225a^2 \$

( iv ) $\ 64a^3 - 27b^3 - 144a^2b + 108ab^2 \$

( v ) $\ 27p^3 - \frac{1} {216} - \frac{9} {2}p^2 + \frac{1} {4}p \$

( i ) $\ x^3 + y^3 = ( x + y ) \$ $\ ( x^2 - xy + y^2 ) \$

( ii ) $\ x^3 - y^3 = ( x - y ) \$ $\( x^2 + xy + y^2 ) \$

( i ) $\ 27y^3 + 125z^3 \$

( ii ) $\ 64m^3 - 343n^3 \$

11. Factorise : $\ 27x^3 + y^3 + z^3 - 9xyz \$

12. Verify that $\ x^3 + y^3 + z^3 - 3xyz \$ $\ = \frac{1} {2} ( x + y + z ) \$ $\ [ ( x - y )^2 + ( y - z )^2 + ( z - x )^2 ] \$

13. If $\ x + y + z = 0 \$ , show that $\ x^3 + y^3 + z^3 = 3xyz \$ .

( i ) $\ ( -12 )^3 + ( 7 )^3 + ( 5 )^3 \$

( ii ) $\ ( 28 )^3 + ( -15 )^3 + ( -13 )^3 \$

( i ) $\ Area : 25a^2 - 35a + 12 \$

( ii ) $\ Area : 35y^2 - 13y - 12 \$

( i ) $\ Volume : 3x^2 - 12x \$

( ii ) $\ Volume : 12ky^2 + 8ky - 20k \$