RS Aggarwal Solutions: Polynomials (Exercise 2B) | Mathematics (Maths) Class 10 PDF Download

 Page 1


Exercise 2B
1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x
3
2x
2
5x + 6) and verify
the relation between it zeros and coefficients.
Sol:
The given polynomial is p(x) = (x
3
2x
2
5x + 6)
p(3) = (3
3
2 × 3
2
5 × 3 + 6) = (27 18 15 + 6) = 0
p(-2) = [ ( 2
3
) 2 × ( 2)
2
5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0
p(1) = (1
3
2 × 1
2
5 × 1 + 6) = ( 1 2 5 + 6) = 0
3, 2 and 1are the zeroes of p(x),
Let = 3, = 2 and = 1. Then we have:
( ) = (3 2 + 1) = 2 = 
( ) = ( 6 2 + 3) = =
= { 3 × (-2) × 1} = =
2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x
3
10x
2
27x + 10) 
and verify the relation between its zeroes and coefficients.
Sol:
p(x) = (3x
3
10x
2
27x + 10)
p(5) = (3 × 5
3
10 × 5
2
27 × 5 + 10) = (375 250 135 + 10) = 0
p( 2) = [3 × ( 2
3
) 10 × ( 2
2
) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0
p = =
= = = = 0
5, 2 and are the zeroes of p(x).
Let = 5, = 2 and = . Then we have:
( ) = = =
( ) = = =
= = =
Page 2


Exercise 2B
1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x
3
2x
2
5x + 6) and verify
the relation between it zeros and coefficients.
Sol:
The given polynomial is p(x) = (x
3
2x
2
5x + 6)
p(3) = (3
3
2 × 3
2
5 × 3 + 6) = (27 18 15 + 6) = 0
p(-2) = [ ( 2
3
) 2 × ( 2)
2
5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0
p(1) = (1
3
2 × 1
2
5 × 1 + 6) = ( 1 2 5 + 6) = 0
3, 2 and 1are the zeroes of p(x),
Let = 3, = 2 and = 1. Then we have:
( ) = (3 2 + 1) = 2 = 
( ) = ( 6 2 + 3) = =
= { 3 × (-2) × 1} = =
2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x
3
10x
2
27x + 10) 
and verify the relation between its zeroes and coefficients.
Sol:
p(x) = (3x
3
10x
2
27x + 10)
p(5) = (3 × 5
3
10 × 5
2
27 × 5 + 10) = (375 250 135 + 10) = 0
p( 2) = [3 × ( 2
3
) 10 × ( 2
2
) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0
p = =
= = = = 0
5, 2 and are the zeroes of p(x).
Let = 5, = 2 and = . Then we have:
( ) = = =
( ) = = =
= = =
______________________________________________________________________________ 
3. Find a cubic polynomial whose zeroes are 2, -3and 4.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = 2, b = 3 and c = 4
Substituting the values in 1, we get
x
3
(2 3 + 4)x
2
+ ( 6 12 + 8)x ( 24)
x
3
3x
2
10x + 24
4. Find a cubic polynomial whose zeroes are  , 1 and 3.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = , b = 1 and c = 3
Substituting the values in (1), we get
x
3
x
2
+ x
x
3
x
2
4x + 
2x
3
+3x
2
8x + 3
5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken
two at a time and the product of its zeroes as 5, -2 and -24 respectively.
Sol:
We know the sum, sum of the product of the zeroes taken two at a time and the product of
the zeroes of a cubic polynomial then the cubic polynomial can be found as
x
3
(sum of the zeroes)x
2
+ (sum of the product of the zeroes taking two at a time)x 
product of zeroes
Therefore, the required polynomial is
x
3
5x
2
2x + 24
6. If f(x) =
3
3 5 3 x x x is divided by g(x) =
2
2 x
Sol: x 3
x 2 x
3
3x
2
+ 5x 3
x
3
          2x
- +
3x
2
+ 7x 3
3x
2
+ 6
+
7x 9
Quotient q(x) = x 3
Remainder r(x) = 7x 9
Page 3


Exercise 2B
1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x
3
2x
2
5x + 6) and verify
the relation between it zeros and coefficients.
Sol:
The given polynomial is p(x) = (x
3
2x
2
5x + 6)
p(3) = (3
3
2 × 3
2
5 × 3 + 6) = (27 18 15 + 6) = 0
p(-2) = [ ( 2
3
) 2 × ( 2)
2
5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0
p(1) = (1
3
2 × 1
2
5 × 1 + 6) = ( 1 2 5 + 6) = 0
3, 2 and 1are the zeroes of p(x),
Let = 3, = 2 and = 1. Then we have:
( ) = (3 2 + 1) = 2 = 
( ) = ( 6 2 + 3) = =
= { 3 × (-2) × 1} = =
2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x
3
10x
2
27x + 10) 
and verify the relation between its zeroes and coefficients.
Sol:
p(x) = (3x
3
10x
2
27x + 10)
p(5) = (3 × 5
3
10 × 5
2
27 × 5 + 10) = (375 250 135 + 10) = 0
p( 2) = [3 × ( 2
3
) 10 × ( 2
2
) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0
p = =
= = = = 0
5, 2 and are the zeroes of p(x).
Let = 5, = 2 and = . Then we have:
( ) = = =
( ) = = =
= = =
______________________________________________________________________________ 
3. Find a cubic polynomial whose zeroes are 2, -3and 4.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = 2, b = 3 and c = 4
Substituting the values in 1, we get
x
3
(2 3 + 4)x
2
+ ( 6 12 + 8)x ( 24)
x
3
3x
2
10x + 24
4. Find a cubic polynomial whose zeroes are  , 1 and 3.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = , b = 1 and c = 3
Substituting the values in (1), we get
x
3
x
2
+ x
x
3
x
2
4x + 
2x
3
+3x
2
8x + 3
5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken
two at a time and the product of its zeroes as 5, -2 and -24 respectively.
Sol:
We know the sum, sum of the product of the zeroes taken two at a time and the product of
the zeroes of a cubic polynomial then the cubic polynomial can be found as
x
3
(sum of the zeroes)x
2
+ (sum of the product of the zeroes taking two at a time)x 
product of zeroes
Therefore, the required polynomial is
x
3
5x
2
2x + 24
6. If f(x) =
3
3 5 3 x x x is divided by g(x) =
2
2 x
Sol: x 3
x 2 x
3
3x
2
+ 5x 3
x
3
          2x
- +
3x
2
+ 7x 3
3x
2
+ 6
+
7x 9
Quotient q(x) = x 3
Remainder r(x) = 7x 9
 
7. If f(x) = x
4
3x
2
+ 4x + 5 is divided by g(x)= x
2
x + 1
Sol: x
2
+ x 3
x
2
x + 1   x
4
+ 0x
3
3x
2
+ 4x + 5
x
4  
x
3
+ x
2
- +       
x
3
4x
2
+ 4x + 5
x
3
x
2
+   x 
+        
3x
2
+ 3x + 5
3x
2
+ 3x 3
+ +
 8
Quotient q(x) = x
2
+ x 3
Remainder r(x) = 8
8. If f(x) = x
4
5x + 6 is divided by g(x) =  2 x
2
.
Sol:
We can write
f(x) as x
4
+ 0x
3
+ 0x
2
5x + 6 and  g(x) as x
2
+ 2
x
2
2
x
2
+ 2      x
4
+ 0x
3
+ 0x
2
5x + 6
x
4  
2 x
2
- +
2x
2
5x + 6
2x
2
4
+
5x + 10
Quotient q(x) = x
2
2
Remainder r(x) = 5x + 10
Page 4


Exercise 2B
1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x
3
2x
2
5x + 6) and verify
the relation between it zeros and coefficients.
Sol:
The given polynomial is p(x) = (x
3
2x
2
5x + 6)
p(3) = (3
3
2 × 3
2
5 × 3 + 6) = (27 18 15 + 6) = 0
p(-2) = [ ( 2
3
) 2 × ( 2)
2
5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0
p(1) = (1
3
2 × 1
2
5 × 1 + 6) = ( 1 2 5 + 6) = 0
3, 2 and 1are the zeroes of p(x),
Let = 3, = 2 and = 1. Then we have:
( ) = (3 2 + 1) = 2 = 
( ) = ( 6 2 + 3) = =
= { 3 × (-2) × 1} = =
2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x
3
10x
2
27x + 10) 
and verify the relation between its zeroes and coefficients.
Sol:
p(x) = (3x
3
10x
2
27x + 10)
p(5) = (3 × 5
3
10 × 5
2
27 × 5 + 10) = (375 250 135 + 10) = 0
p( 2) = [3 × ( 2
3
) 10 × ( 2
2
) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0
p = =
= = = = 0
5, 2 and are the zeroes of p(x).
Let = 5, = 2 and = . Then we have:
( ) = = =
( ) = = =
= = =
______________________________________________________________________________ 
3. Find a cubic polynomial whose zeroes are 2, -3and 4.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = 2, b = 3 and c = 4
Substituting the values in 1, we get
x
3
(2 3 + 4)x
2
+ ( 6 12 + 8)x ( 24)
x
3
3x
2
10x + 24
4. Find a cubic polynomial whose zeroes are  , 1 and 3.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = , b = 1 and c = 3
Substituting the values in (1), we get
x
3
x
2
+ x
x
3
x
2
4x + 
2x
3
+3x
2
8x + 3
5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken
two at a time and the product of its zeroes as 5, -2 and -24 respectively.
Sol:
We know the sum, sum of the product of the zeroes taken two at a time and the product of
the zeroes of a cubic polynomial then the cubic polynomial can be found as
x
3
(sum of the zeroes)x
2
+ (sum of the product of the zeroes taking two at a time)x 
product of zeroes
Therefore, the required polynomial is
x
3
5x
2
2x + 24
6. If f(x) =
3
3 5 3 x x x is divided by g(x) =
2
2 x
Sol: x 3
x 2 x
3
3x
2
+ 5x 3
x
3
          2x
- +
3x
2
+ 7x 3
3x
2
+ 6
+
7x 9
Quotient q(x) = x 3
Remainder r(x) = 7x 9
 
7. If f(x) = x
4
3x
2
+ 4x + 5 is divided by g(x)= x
2
x + 1
Sol: x
2
+ x 3
x
2
x + 1   x
4
+ 0x
3
3x
2
+ 4x + 5
x
4  
x
3
+ x
2
- +       
x
3
4x
2
+ 4x + 5
x
3
x
2
+   x 
+        
3x
2
+ 3x + 5
3x
2
+ 3x 3
+ +
 8
Quotient q(x) = x
2
+ x 3
Remainder r(x) = 8
8. If f(x) = x
4
5x + 6 is divided by g(x) =  2 x
2
.
Sol:
We can write
f(x) as x
4
+ 0x
3
+ 0x
2
5x + 6 and  g(x) as x
2
+ 2
x
2
2
x
2
+ 2      x
4
+ 0x
3
+ 0x
2
5x + 6
x
4  
2 x
2
- +
2x
2
5x + 6
2x
2
4
+
5x + 10
Quotient q(x) = x
2
2
Remainder r(x) = 5x + 10
 
9. By actual division, show that x
2
3 is a factor of  2x
4
+ 3x
3
2x
2
9x 12.
Sol:
Let f(x) = 2x
4
+ 3x
3
2x
2
9x 12 and g(x) as  x
2
3
2x
2
+ 3x + 4
x
2
3 2x
4
+ 3x
3
2x
2
9x 12
2x
4               
6x
2
- +
3x
3
+ 4x
2
9x 12
3x
3               
9x 
+
4x
2
12
4x
2
12
+
x
Quotient q(x) = 2x
2
+ 3x + 4
Remainder r(x) = 0
Since, the remainder is 0.
Hence, x
2
3 is a factor of 2x
4
+ 3x
3
2x
2
9x 12
10. On dividing 3x
3
+ x
2
+ 2x + 5 is divided by a polynomial g(x), the quotient and remainder are
(3x 5) and (9x + 10) respectively. Find g(x).
Sol:
By using division rule, we have
Dividend = Quotient × Divisor + Remainder
3x
3
+ x
2
+ 2x + 5 = (3x 5)g(x) + 9x + 10
3x
3
+ x
2
+ 2x + 5 9x 10 = (3x 5)g(x)
3x
3
+ x
2
7x 5 =  (3x 5)g(x)
g(x) = 
x
2
+ 2x + 1
3x 5 3x
3
+  x
2
7x 5
3x
3
5x
2
+
6x
2
7x 5
6x
2
10x 
+
3x 5
3x 5
+
X
 g(x) = x
2
+ 2x + 1
Page 5


Exercise 2B
1. Verify that 3, -2, 1 are the zeros of the cubic polynomial p(x) = (x
3
2x
2
5x + 6) and verify
the relation between it zeros and coefficients.
Sol:
The given polynomial is p(x) = (x
3
2x
2
5x + 6)
p(3) = (3
3
2 × 3
2
5 × 3 + 6) = (27 18 15 + 6) = 0
p(-2) = [ ( 2
3
) 2 × ( 2)
2
5 × ( 2) + 6] = ( 8 8 + 10 + 6) = 0
p(1) = (1
3
2 × 1
2
5 × 1 + 6) = ( 1 2 5 + 6) = 0
3, 2 and 1are the zeroes of p(x),
Let = 3, = 2 and = 1. Then we have:
( ) = (3 2 + 1) = 2 = 
( ) = ( 6 2 + 3) = =
= { 3 × (-2) × 1} = =
2. Verify that 5, -2 and are the zeroes of the cubic polynomial p(x) = (3x
3
10x
2
27x + 10) 
and verify the relation between its zeroes and coefficients.
Sol:
p(x) = (3x
3
10x
2
27x + 10)
p(5) = (3 × 5
3
10 × 5
2
27 × 5 + 10) = (375 250 135 + 10) = 0
p( 2) = [3 × ( 2
3
) 10 × ( 2
2
) 27 × ( 2) + 10] = ( 24 40 + 54 + 10) = 0
p = =
= = = = 0
5, 2 and are the zeroes of p(x).
Let = 5, = 2 and = . Then we have:
( ) = = =
( ) = = =
= = =
______________________________________________________________________________ 
3. Find a cubic polynomial whose zeroes are 2, -3and 4.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = 2, b = 3 and c = 4
Substituting the values in 1, we get
x
3
(2 3 + 4)x
2
+ ( 6 12 + 8)x ( 24)
x
3
3x
2
10x + 24
4. Find a cubic polynomial whose zeroes are  , 1 and 3.
Sol:
If the zeroes of the cubic polynomial are a, b and c then the cubic polynomial can be found as
x
3
(a + b + c)x
2
+ (ab + bc + ca)x 
Let a = , b = 1 and c = 3
Substituting the values in (1), we get
x
3
x
2
+ x
x
3
x
2
4x + 
2x
3
+3x
2
8x + 3
5. Find a cubic polynomial with the sum of its zeroes, sum of the products of its zeroes taken
two at a time and the product of its zeroes as 5, -2 and -24 respectively.
Sol:
We know the sum, sum of the product of the zeroes taken two at a time and the product of
the zeroes of a cubic polynomial then the cubic polynomial can be found as
x
3
(sum of the zeroes)x
2
+ (sum of the product of the zeroes taking two at a time)x 
product of zeroes
Therefore, the required polynomial is
x
3
5x
2
2x + 24
6. If f(x) =
3
3 5 3 x x x is divided by g(x) =
2
2 x
Sol: x 3
x 2 x
3
3x
2
+ 5x 3
x
3
          2x
- +
3x
2
+ 7x 3
3x
2
+ 6
+
7x 9
Quotient q(x) = x 3
Remainder r(x) = 7x 9
 
7. If f(x) = x
4
3x
2
+ 4x + 5 is divided by g(x)= x
2
x + 1
Sol: x
2
+ x 3
x
2
x + 1   x
4
+ 0x
3
3x
2
+ 4x + 5
x
4  
x
3
+ x
2
- +       
x
3
4x
2
+ 4x + 5
x
3
x
2
+   x 
+        
3x
2
+ 3x + 5
3x
2
+ 3x 3
+ +
 8
Quotient q(x) = x
2
+ x 3
Remainder r(x) = 8
8. If f(x) = x
4
5x + 6 is divided by g(x) =  2 x
2
.
Sol:
We can write
f(x) as x
4
+ 0x
3
+ 0x
2
5x + 6 and  g(x) as x
2
+ 2
x
2
2
x
2
+ 2      x
4
+ 0x
3
+ 0x
2
5x + 6
x
4  
2 x
2
- +
2x
2
5x + 6
2x
2
4
+
5x + 10
Quotient q(x) = x
2
2
Remainder r(x) = 5x + 10
 
9. By actual division, show that x
2
3 is a factor of  2x
4
+ 3x
3
2x
2
9x 12.
Sol:
Let f(x) = 2x
4
+ 3x
3
2x
2
9x 12 and g(x) as  x
2
3
2x
2
+ 3x + 4
x
2
3 2x
4
+ 3x
3
2x
2
9x 12
2x
4               
6x
2
- +
3x
3
+ 4x
2
9x 12
3x
3               
9x 
+
4x
2
12
4x
2
12
+
x
Quotient q(x) = 2x
2
+ 3x + 4
Remainder r(x) = 0
Since, the remainder is 0.
Hence, x
2
3 is a factor of 2x
4
+ 3x
3
2x
2
9x 12
10. On dividing 3x
3
+ x
2
+ 2x + 5 is divided by a polynomial g(x), the quotient and remainder are
(3x 5) and (9x + 10) respectively. Find g(x).
Sol:
By using division rule, we have
Dividend = Quotient × Divisor + Remainder
3x
3
+ x
2
+ 2x + 5 = (3x 5)g(x) + 9x + 10
3x
3
+ x
2
+ 2x + 5 9x 10 = (3x 5)g(x)
3x
3
+ x
2
7x 5 =  (3x 5)g(x)
g(x) = 
x
2
+ 2x + 1
3x 5 3x
3
+  x
2
7x 5
3x
3
5x
2
+
6x
2
7x 5
6x
2
10x 
+
3x 5
3x 5
+
X
 g(x) = x
2
+ 2x + 1
 
11. Verify division algorithm for the polynomial f(x)= (8 + 20x + x
2
6x
3
) by g(x) =( 2 + 5x 
3x
2
).
Sol:
We can write f(x) as 6x
3
+ x
2
+ 20x + 8 and g(x) as 3x
2
+ 5x + 2
x
2
+ 2x + 1
3x
2
+ 5x + 2 6x
3
+    x
2
+ 20x + 8
6x
3
+10x
2
+ 4x
+
9x
2
+16x + 8
9x
2
+15x + 6
+
x + 2
Quotient = 2x + 3
Remainder = x + 2
By using division rule, we have
Dividend = Quotient × Divisor + Remainder
6x
3
+ x
2
+ 20x + 8 = ( 3x
2
+ 5x + 2) (2x + 3) + x + 2
6x
3
+ x
2
+ 20x + 8 = 6x
3
+ 10x
2
+ 4x 9x
2
+ 15x + 6 + x + 2
6x
3
+ x
2
+ 20x + 8 = 6x
3
+ x
2
+ 20x + 8
12. It is given that 1 is one of the zeroes of the polynomial x
3
+ 2x
2
11x 12. Find all the
zeroes of the given polynomial.
Sol:
Let f(x) = x
3
+ 2x
2
11x 12
Since 1 is a zero of f(x), (x+1) is a factor of f(x).
On dividing f(x) by (x+1), we get
x + 1     x
3
+ 2x
2
11x 12     x
2
+ x + 12
x
3
+   x
2
x
2
11x 12
x
2
+ x 
12x 12
12x 12
+       +
X