RS Aggarwal Solutions: Polynomials (Exercise 2B)

# RS Aggarwal Solutions: Polynomials (Exercise 2B) - Mathematics (Maths) Class 10

``` 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
```

## Mathematics (Maths) Class 10

115 videos|478 docs|129 tests

## FAQs on RS Aggarwal Solutions: Polynomials (Exercise 2B) - Mathematics (Maths) Class 10

 1. How do I find the degree of a polynomial? Ans. To find the degree of a polynomial, you need to look at the term with the highest exponent. The degree is the value of that exponent. For example, if the highest exponent in a polynomial is 3, then the degree of the polynomial is 3.
 2. How do I add or subtract polynomials? Ans. To add or subtract polynomials, you need to combine like terms. Like terms have the same variables raised to the same powers. Simply add or subtract the coefficients of the like terms while keeping the variables and their exponents the same.
 3. How do I multiply polynomials? Ans. To multiply polynomials, you can use the distributive property or the FOIL method. If you have two polynomials, multiply each term of the first polynomial by each term of the second polynomial, and then combine like terms if possible.
 4. What is the remainder theorem? Ans. The remainder theorem states that if a polynomial f(x) is divided by x-a, then the remainder is equal to f(a). In other words, if you divide a polynomial by a linear factor (x-a), the remainder you get is the value of the polynomial when you substitute x=a.
 5. How do I solve polynomial equations? Ans. To solve polynomial equations, set the equation equal to zero and try to factor it if possible. Then, set each factor equal to zero and solve for the values of x. These values are the solutions to the polynomial equation. If factoring is not possible, you may need to use other methods such as the quadratic formula or synthetic division.

## Mathematics (Maths) Class 10

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