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Page No 143:

Question 1:

From a rectangular board of length 8 centimetres and breadth 3 centimetres, a small rectangular piece is cut off.


 


 

One side of the piece cut off is of length 3 centimetres itself. Denoting the length of the other side as x, write down an algebraic expression which gives the area of the remaining piece.


 

Answer:

Length of the rectangular board = 8 cm

Breadth of the rectangular board = 3 cm

A small rectangular piece of breadth as 3 cm and length as x cm is cut off from the rectangular board.

Length of the remaining piece = (8 − x) cm

Breadth of the remaining piece = 3 cm

Area of the remaining piece = Length of the remaining piece × Breadth of the remaining piece

= (8 − x) × 3 cm2

= (24 − 3x) cm2

Thus, the algebraic expression for the area of the remaining piece is 24 − 3x.


 

Page No 143:

Question 2:

Find algebraic expressions which give the perimeters and areas of all rectangles with length 1 centimetre more than the breadth.


 

Answer:

Let the breadth of the rectangle be x cm.

Therefore, the length of the rectangle is (x + 1) cm.

Area of the rectangle = Length × Breadth 

= (x + 1) × x cm2

= (x2 + x) cm2

Perimeter of the rectangle = 2 (Length + Breadth)

= 2(x + 1 + x) cm

= 2(2x + 1) cm

= (4x + 2) cm

Thus, the algebraic expressions for the area and perimeter of all the rectangles with length 1 cm more than the breadth are x2 + x and 4x + 2 respectively.


 

Page No 143:

Question 3:

The length and breadth of a rectangle are 10 centimetres and 4 centimetres. It is made larger, increasing the length and breadth by the same amount.


 

Find algebraic expressions to compute the perimeters and areas of such enlarged rectangles in terms of the increase in the lengths of sides.


 

Answer:

Length of the rectangle = 10 cm

Breadth of the rectangle = 4 cm

Let the length and the breadth increase by x cm.

New length = (10 + x) cm

New breadth = (4 + x) cm

Perimeter of the enlarged rectangle = 2 (Length + Breadth)

= 2(10 + x + 4 + x) cm

= 2(2x + 14) cm

= (4x + 28) cm

Area of the enlarged rectangle = Length × Breadth 

= (10 + x) × (4 + x) cm2

= 10(4 + x) + x(4 + x) cm2

= (40 + 10x + 4x + x2) cm2

= {40 + (10 + 4)x + x2} cm2

= (40 + 14x + x2) cm2

Thus, the algebraic expressions for the area and perimeter of the enlarged rectangle are 40 + 14x + x2 and 4x + 28 respectively.


 

Now, let the length and breadth decrease by x cm.

New length = (10 x) cm

New breadth = (4 x) cm

Perimeter of the new rectangle = 2 (Length + Breadth)

= 2(10 x + 4 x) cm

= 2(14 2x) cm

= (28 4x) cm

Area of the new rectangle = Length × Breadth 

= (10 x) × (4 x) cm2

= 10(4 x) x(4 x) cm2

= (40 10x 4x + x2) cm2

= {40 (10 + 4)x + x2} cm2

= (40 14x + x2) cm2

Thus, the algebraic expressions for the area and perimeter of the new rectangle are 40 14x + x2 and 28 4x respectively which are not equal to the algebraic expressions for the area and perimeter of the enlarged rectangle.

Hence, we cannot use the same algebraic expressions in both the cases.


 



Page No 144:

Question 1:

A stone thrown upwards with a speed of 20 metres per second loses speed at the rate of 9.8 metres every second. Find an algebraic expression which gives the speed of the stone at various times.


 

Answer:

Initial speed of the stone = 20 metres per second

Speed of the stone is decreased at the rate of 9.8 metres per second.

After 1 second, speed of the stone = (20 − 1 × 9.8) metres per second

After 2 seconds, speed of the stone = (20 − 2 × 9.8) metres per second

After 3 seconds, speed of the stone = (20 − 3 × 9.8) metres per second

After t seconds, speed of the stone = (20 − t × 9.8) metres per second

= (20 − 9.8t) metres per second

Algebraic expression for the speed of the stone at various times is 20 − 9.8t.


 



Page No 147:

Question 1:

The surface area and volume of a rectangular block in which the length is 1 centimetre more than the width and the height is 1 centimetre more than the length.


 

Answer:

Let the width of the rectangular block be x cm.

Length of the rectangular block = (x + 1) cm

Height of the rectangular block = (x + 1 + 1) cm = (x + 2) cm

Volume of the rectangular block = Length × Width × Height

= (x + 1) × x × (x + 2) cm3

= (x2 + x) × (x + 2) cm3 

= {x2(x + 2) + x(x + 2)} cm3

= (x3 + 2x2 + x2 + 2x) cm3

= {x3 + (2 + 1) x2 + 2x} cm3

= (x3 + 3x2 + 2x) cm3

Surface area of the rectangular block = 2(lb + bh + hl)

= 2{(x + 1)x + x(x + 2) + (x + 2)(x + 1)} cm2

= 2{x2 + x + x2 + 2x + x2 + x + 2x + 2} cm2

= 2{(1 + 1 + 1)x2 + (1 + 2 + 1 + 2)x + 2} cm2

= 2(3x2 + 6x + 2) cm2

= (6x2 + 12x + 4) cm2

Yes, the algebraic expressions for the volume and surface area of the rectangular block are polynomials because the exponents of the variables in the expressions are natural numbers.


 

Page No 147:

Question 2:

The percentage of acid in a solution of 7 litres of water and 3 litres of acid, when more acid is added.


 

Answer:

Quantity of acid in the solution = 3 litres

Quantity of water in the solution = 7 litres

Let x litres of the acid be added in the solution.

Total quantity of acid in the solution = (3 + x) litres

Total quantity of the solution = (7 + 3 + x) litres = (10 + x) litres

Percentage of acid in the solution =

The algebraic expression for percentage of acid in the solution is not a polynomial because the variable, x is in the denominator.


 

Page No 147:

Question 3:

The sum of a number and its square root.


 

Answer:

Let the number be x.

Square root of x =

Sum of the number and its square root = x +

=

The algebraic expression for the sum of a number and its square root is not a polynomial because the exponent of x, i.e,is not a natural number.


 

Page No 147:

Question 4:

The product of the number got by adding the square root of a number to itself, and the number got by subtracting its square root from itself (See the section Sum and difference, of the lesson Algebra, in the Class 8 text book)


 

Answer:

Let the number be x.

Square root of x =

Sum of the number and its square root = x +

Difference of the number and its square root = x

Required product =

The algebraic expression for the required product is a polynomial because the exponents of the variable in the expression are natural numbers.


 



Page No 149:

Question 1:

For p(x) = 2x23x + 1, find p(0), p(1), p(−1).


 

Answer:


 

Page No 149:

Question 2:

If p(x) = 3x2ax + 1 and we want p(1) = 2, what number should we take in the place of a?


 

Answer:


 

Page No 149:

Question 3:

If p(x) = 2x3 + ax27x + b and we want p(1) = 3 and p(2) = 19 what should be a and b?


 


 

Answer:

Putting the value of b from equation (1):

4a + (8 − a) = 17

3a + 8 = 17

3a = 17 8

3a = 9

Putting the value of a in equation (1):

b = 8 − 3 = 5

Thus, the values of a and b are 3 and 5 respectively.


 



Page No 151:

Question 1:

Now for each pair of polynomials p(x) and q(x) given below, find p(x) + q(x) and p(x) − q(x). Also compute p(1) + q(1) and p(1) − q(1) for each.

(i)

(ii)

(iii)

(iv)

(v)


 

Answer:

(i)

Given: p(x) = 2x2 − 5x + 1, q(x) = x2 + 3x + 2

p(x) + q(x) = (2x2 − 5x + 1) + (x2 + 3x + 2)

= 2x2 − 5x + 1 + x2 + 3x + 2

= (2 + 1)x2 + (−5 + 3) x + (1 + 2)

= 3x2 − 2x + 3

p(x) q(x) = (2x2 − 5x + 1) (x2 + 3x + 2)

= 2x2 − 5x + 1 x2 3x 2

= (2 1) x2 (5 + 3) x + (1 2)

= x2 − 8x 1

Now, p(1) + q(1) = 3(1)2 − 2(1) + 3 

= 3 − 2 + 3

= 6 2

= 4

p(1) q(1) = (1)2 − 8(1) − 1 

= 1 − 8 − 1

= 8


 

(ii)

Given: p(x) = 2x2 + x + 1, q(x) = x2 + x + 2

p(x) + q(x) = (2x2 + x + 1) + (x2 + x + 2)

= 2x2 + x + 1 + x2 + x + 2

= (2 + 1)x2 + (1 + 1)x + (1 + 2)

= 3x2 + 2x + 3

p(x) q(x) = (2x2 + x + 1) (x2 + x + 2)

= 2x2 + x + 1 x2 x 2

= (2 1)x2 + (1 1)x + (1 2)

= x2 1

Now, p(1) + q(1) = 3(1)2 + 2(1) + 3 

= 3 + 2 + 3

= 8

p(1) q(1) = (1)2 − 1 

= 1 − 1

= 0


 

(iii)

Given: p(x) = 5x + 1, q(x) = x2 + x + 2

p(x) + q(x) = (5x + 1) + (x2 + x + 2)

= 5x + 1 + x2 + x + 2

= x2 + (5 + 1)x + (1 + 2)

= x2 + 6x + 3

p(x) q(x) = (5x + 1) (x2 + x + 2)

= 5x + 1 x2 x 2

= x2 + (5 1) x + (1 2)

= x2 + 4x 1

Now, p(1) + q(1) = (1)2 + 6(1) + 3 

= 1 + 6 + 3

= 10

p(1) q(1) = (1)2 + 4(1) − 1 

= 1 + 4 − 1 

= 4 − 2

= 2


 

(iv)

Given: p(x) = x2 3x + 2, q(x) = x2 + 3x 2

p(x) + q(x) = (x2 3x + 2) + (x2 + 3x 2)

= x2 3x + 2 + x2 + 3x 2

= (1 + 1)x2 + (3 + 3)x + (2 2)

= 2x2

p(x) q(x) = (x2 3x + 2) (x2 + 3x 2)

= x2 3x + 2 x2 3 x + 2

= (1 1)x2 (3 + 3)x + (2 + 2)

= −6x + 4

Now, p(1) + q(1) = 2(1)2 = 2

p(1) q(1) = −6(1) + 4 

= −6 + 4

=


 

(v)

Given: p(x) = 4x2 2x + 3, q(x) = 4x2 + 2x + 2

p(x) + q(x) = (4x2 2x + 3) + (4x2 + 2x + 2)

= 4x2 2x + 3 4x2 + 2x + 2

= (4 4)x2 + (2 + 2)x + (3 + 2)

= 5

p(x) q(x) = (4x2 2x + 3) (4x2 + 2x + 2)

= 4x2 2x + 3 + 4x2 2x 2

= (4 + 4)x2 (2 + 2)x + (3 2)

= 8x2 4x + 1

Now, p(1) + q(1) = 5 

p(1) q(1) = 8(1)2 4(1) + 1

= 8 − 4 + 1

= 9

= 5


 



Page No 153:

Question 1:

Simplify each of the following as a single polynomial.

(i) (3x + 3) (3x + 2)

(ii) (3x − 4) (4x − 3)

(iii)

(iv)

(v)

(vi)

(vii)


 

Answer:

(i)

(3x + 3)(3x + 2)

= (3x × 3x) + (3x × 2) + (3 × 3x) + (3 × 2)

= 9x2 + 6x + 9x + 6

= 9x2 + (6 + 9)x + 6

= 9x2 + 15x + 6


 

(ii)

(3x 4) (4x 3)

= (3x × 4x) (3x × 3) (4 × 4x) + (4 × 3)

= 12x2 9x 16x + 12

= 12x2 (9 + 16)x + 12

= 12x2 25x + 12


 

(iii)

(2x − 3) (4x2 + 6x + 9)

= (2x × 4x2) + (2x × 6x) + (2x × 9) (3 × 4x2) (3 × 6x) (3 × 9)

= 8x3 + 12x2 + 18x12x218x − 27 

= 8x3 + (12 − 12)x2 + (18 − 18)x − 27 

= 8x3 − 27


 

(iv)

(2x + 3) (4x26x + 9)

= (2x × 4x2) − (2x × 6x) + (2x × 9) + (3 × 4x2) (3 × 6x) + (3 × 9)

= 8x312x2 + 18x + 12x218x + 27 

= 8x3 + (−12 + 12)x2 + (18 − 18)x + 27 

= 8x3 + 27


 

(v)

(x − 1)(x2 + x + 1)

= (x × x2) + (x × x) + (x × 1) − (1 × x2) − (1 × x) − (1 × 1)

= x3 + x2 + xx2x − 1 

= x3 + (1 − 1)x2 + (1 − 1)x − 1 

= x3 − 1 


 

(vi)

(x + 1)(x2x + 1)

= (x × x2) − (x × x) + (x × 1) + (1 × x2) − (1 × x) + (1 × 1)

= x3x2 + x + x2x + 1 

= x3 + (−1 + 1)x2 + (1 − 1)x + 1 

= x3 + 1


 

(vii)

(2x + 2)(5x + 4) − (2x + 3)(4x + 5)

= (2x × 5x) + (2x × 4) + (2 × 5x) + (2 × 4) − (2x × 4x) − (2x × 5) − (3 × 4x) − (3 × 5)

= 10x2 + 8x + 10x + 8 − 8x2 − 10x − 12x − 15 

= (10 − 8)x2 + (8 + 10 − 10 − 12)x + (8 − 15)

= 2x2 − 4x − 7


 

Page No 153:

Question 2:

In each of the first six problems, take p(x) as the first polynomial, q(x) as the second polynomial and compute the following.

p(1) q(1); p(0) q(0); p(1) q(−1); p(−1) q(1)


 

Answer:

(1)

p(x) = (3x + 3)

q(x) = (3x + 2)

p(x) q(x) = 9x2 + 15x + 6

(i) p(1) q(1) = 9(1)2 + 15(1) + 6 

= 9 + 15 + 6

= 30

(ii) p(0) q(0) = 9(0)2 + 15(0) + 6 = 6

(iii) p(1) = 3 × 1 + 3 

= 3 + 3

= 6

q(1) = 3 × (1) + 2

= 3 + 2 

= 1

p(1) q(1) = 6 × (1) = 6

(iv) p(1) = 3 × (1) + 3

= 3 + 3 

= 0

q(1) = 3 × 1 + 2 

= 3 + 2

= 5

p(1) q(1) = 0 × 5 = 0


 

(2)

p(x) = (3x 4)

q(x) = (4x 3)

p(x) q(x) = 12x2 25x + 12

(i) p(1) q(1) = 12(1)2 25(1) + 12 

= 12 − 25 + 12

= 24 25

= 1 

(ii) p(0) q(0) = 12(0)2 25(0) + 12 = 12

(iii) p(1) = 3 × 1 − 4 

= 3 − 4

= 1

q(1) = 4 × (1) − 3 

= 4 − 3 

= 7

p(1) q(1) = 1 × (7) = 7

(iv) p(1) = 3 × (1) − 4 

= 3 − 4 

= 7

q(1) = 4 × 1 − 3 

= 4 − 3

= 1

p(1) q(1) = 7 × 1 = 7


 

(3)

p(x) = (2x − 3) 

q(x) = (4x2 + 6x + 9)

p(x) q(x) = 8x3 − 27

(i) p(1) q(1) = 8(1)3 − 27 

= 8 − 27

= 19

(ii) p(0) q(0) = 8(0)3 − 27 = 27

(iii) p(1) = 2 × 1 − 3 

= 2 − 3

= 1

q(1) = 4(1)2 + 6(1) + 9 

= 4 − 6 + 9

= 13 − 6

= 7

p(1) q(1) = 1 × 7 = 7

(iv) p(1) = 2 × (1)

= 2 3

= 5

q(1) = 4(1)2 + 6(1) + 9 

= 4 + 6 + 9

= 19

p(1) q(1) = 5 × 19 = 95


 

(4)

p(x) = (2x + 3) 

q(x) = (4x26x + 9)

p(x) q(x) = 8x3 + 27

(i) p(1) q(1) = 8(1)3 + 27 = 8 + 27 = 35

(ii) p(0) q(0) = 8(0)3 + 27 = 27

(iii) p(1) = 2 × 1 + 3 

= 2 + 3

= 5

q(1) = 4(1)2 − 6(1) + 9 

= 4 + 6 + 9

= 19

p(1) q(1) = 5 × 19 = 95

(iv) p(1) = 2 × (1) + 3 

= 2 + 3 

= 1

q(1) = 4(1)2 − 6(1) + 9 

= 4 − 6 + 9

= 13 − 6

= 7

p(1) q(1) = 1 × 7 = 7


 

(5)

p(x) = (x − 1)

q(x) = (x2 + x + 1)

p(x) q(x) = x3 − 1 

(i) p(1) q(1) = (1)3 − 1 

= 1 − 1

= 0

(ii) p(0) q(0) = (0)3 − 1 

= −1

(iii) p(1) = 1 − 1 = 0

q(1) = (1)2 + (1) + 1 

= 1 1 + 1 

= 2 − 1

= 1

p(1) q(1) = 0 × 1 = 0

(iv) p(1) = 1 1=

q(1) = (1)2 + (1) + 1 

= 1 + 1 + 1

= 3

p(1) q(1) = 2 × 3 = 6


 

(6)

p(x) = (x + 1)

q(x) = (x2x + 1)

p(x) q(x) = x3 + 1

(i) p(1) q(1) = (1)3 + 1 

= 1 + 1

= 2

(ii) p(0) q(0) = (0)3 + 1 

= 0 + 1

= 1

(iii) p(1) = 1 + 1 = 2

q(1) = (1)2 − (1) + 1 

= 1 + 1 + 1

= 3

p(1) q(1) = 2 × 3 = 6

(iv) p(1) = 1 + 1= 0 

q(1) = (1)2 − 1 + 1 

= 2 − 1

= 1

p(1) q(1) = 0 × 1 = 0


 



Page No 154:

Question 1:

Simplify each of the following and write as a single polynomial.

(i)

(ii)

(iii)

(iv)

(v)

(v)


 

Answer:

(i)

(2x + 1)(3x + 4) + (4x + 3)(3x + 4)

= {(2x + 1) + (4x + 3)}(3x + 4) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= (2x + 4x + 1 + 3)(3x + 4)

= (6x + 4)(3x + 4)

= 18x2 + 24x + 12x + 16

= 18x2 + 36x + 16


 

(ii)

(2x + 1)(3x + 4) + (2x 1)(3x + 4)

= {(2x + 1) + (2x 1)}(3x + 4) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= (2x + 2x + 1 1)(3x + 4)

= 4x(3x + 4)

= 12x2 + 16x 


 

(iii)

(2x + 1)(3x + 4) + (1 2x)(3x + 4)

= {(2x + 1) + (1 2x)}(3x + 4) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= (2x 2x + 1 + 1)(3x + 4)

= 2(3x + 4)

= 6x + 8


 

(iv)

(3x + 4)2 − (2x − 1) (3x + 4)

= (3x + 4)(3x + 4) − (2x − 1)(3x + 4) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= {(3x + 4) (2x 1)}(3x + 4)

= (3x 2x + 4 + 1)(3x + 4)

= (x + 5) (3x + 4)

= 3x2 + 4x + 15x + 20

= 3x2 + 19x + 20


 

(v)

(2x + 1)(3x + 4) + (3x + 4)

= (2x + 1)(3x + 4) + 1(3x + 4)

= {(2x + 1) + 1}(3x + 4) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= (2x + 1 + 1)(3x + 4)

= (2x + 2)(3x + 4)

= 6x2 + 8x + 6x + 8

= 6x2 + 14x + 8


 

(vi)

(x2 + 3x + 1) (2x + 1) + (x23x − 1) (2x + 1)

= {(x2 + 3x + 1) + (x23x − 1)}(2x + 1) [Since p(x)r(x) + q(x)r(x) = (p(x) + q(x))r(x)]

= (x2 + 3x + 1 + x23x − 1)(2x + 1)

= 2x2(2x + 1)

= 4x3 + 2x2


 



Page No 155:

Question 1:

What is the degree of ?


 

Answer:

Let p(x) = x2 + 3x + 1 and q(x) = x3 + 2x − 4 

Degree of (x2 + 3x + 1) (x3 + 2x − 4) = Degree of p(x) q(x)

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x).

Also, the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in p(x) = 2

Degree of p(x) = 2

The largest exponent of x in q(x) = 3

Degree of q(x) = 3

Degree of p(x) q(x) = 2 + 3 = 5

Degree of (x2 + 3x + 1) (x3 + 2x − 4) = 5


 

Page No 155:

Question 2:

If the degree of is to be 5, what should be the degree of p(x)?


 

Answer:

Let q(x) = 3x2 4x + 2 

Degree of p(x) (3x2 4x + 2) = Degree of p(x) q(x) = 5

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x)

Also, the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in q(x) = 2

Degree of q(x) = 2

Degree of p(x) = Degree of p(x) q(x) Degree of q(x)

= 5 − 2

= 3


 

Page No 155:

Question 3:

If the degree of p(x) is 4 and the degree of q(x) is 2, the n what is the degree of p(x) + q(x)?


 

Answer:

Degree of p(x) = 4 

Degree of q(x) = 2

Between p(x) and q(x), p(x) has a higher degree. Therefore, the degree of {p(x) + q(x)} is equal to the degree of p(x).

Degree of {p(x) + q(x)} = 4


 


 

Page No 155:

Question 4:

For each pair of polynomials p(x) and q(x) given below find the degree of p(x) + q(x) and p(x)q(x).


 

(i)

(ii)

(iii)

(iv)


 

Answer:

(i)

Given: p(x) = x3 2x2 + 4x + 1 and q(x) = 5x2 + 3x + 2 

We know that the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in p(x) = 3

Degree of p(x) = 3

The largest exponent of x in q(x) = 2

Degree of q(x) = 2

Between p(x) and q(x), p(x) has a higher degree. Therefore, the degree of {p(x) + q(x)} is equal to the degree of p(x).

Degree of {p(x) + q(x)} = 3

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x)

Degree of p(x) q(x) = 3 + 2 = 5


 

(ii)

Given: p(x) = x3 2x2 + 4x + 1 and q(x) = 2x2 4x + 2 

We know that the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in p(x) = 3

Degree of p(x) = 3

The largest exponent of x in q(x) = 2

Degree of q(x) = 2

Between p(x) and q(x), p(x) has a higher degree. Therefore, the degree of {p(x) + q(x)} is equal to the degree of p(x).

Degree of {p(x) + q(x)} = 3

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x)

Degree of p(x) q(x) = 3 + 2 = 5


 

(iii)

Given: p(x) = x3 2x2 + 4x + 1 and q(x) = x3 + 1 

We know that the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in p(x) = 3

Degree of p(x) = 3

The largest exponent of x in q(x) = 3

Degree of q(x) = 3

It can be observed that both p(x) and q(x) have the same degree.

However, the coefficients of the variable having the largest exponent in both p(x) and q(x) are same in magnitude but opposite in sign. So, they will cancel out while the addition of the polynomials p(x) and q(x) is carried out.

Therefore, the degree of {p(x) + q(x)} is the next largest exponent of the variable in p(x) or q(x).

The next largest exponent is of x2.

Degree of {p(x) + q(x)} = 2

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x)

Degree of p(x) q(x) = 3 + 3 = 6


 

(iv)

Given: p(x) = x3 2x2 + 4x + 1 and q(x) = x3 + 2x2 + 1 

We know that the degree of a polynomial is the largest exponent of the variable in the polynomial.

The largest exponent of x in p(x) = 3

Degree of p(x) = 3

The largest exponent of x in q(x) = 3

Degree of q(x) = 3

It can be observed that both p(x) and q(x) have the same degree.

However, the coefficients of the variable having the largest exponent in both p(x) and q(x) are same in magnitude but opposite in sign. So, they will cancel out while the addition of the polynomials p(x) and q(x) is carried out.

Similarly, the coefficients of x2 will cancel out.

Therefore, the degree of {p(x) + q(x)} is the next largest exponent of the variable in p(x) or q(x).

The next largest exponent is of x.

Degree of {p(x) + q(x)} = 1

We know that degree of p(x) q(x) = degree of p(x) + degree of q(x)

Degree of p(x) q(x) = 3 + 3 = 6


 



Page No 159:

Question 1:

Find the quotient and remainder when each of the polynomials below is divided by 2x + 1.

(i)

(ii)

(iii)

(iv)

(v)

(vi)


 

Answer:

(i) 

6x2 + 7x + 3

= (6x2 + 7x + 2) + 1

= (6x2 + 4x + 3x + 2) + 1

= 2x(3x + 2) + 1(3x + 2) + 1

= (3x + 2)(2x + 1) + 1

Now,

Quotient is 3x + 2 and remainder is 1.


 

(ii)

6x2 + 7x 1

= (6x2 + 7x + 2) 3

= (6x2 + 4x + 3x + 2) 3

= 2x(3x + 2) + 1(3x + 2) 3

= (3x + 2)(2x + 1) 3

Now,

Quotient is 3x + 2 and remainder is 3.


 

(iii)

6x2 + 9x + 3

= 6x2 + 6x + 3x + 3

= 6x(x + 1) + 3(x + 1)

= (6x + 3)(x + 1)

= 3(2x + 1)(x + 1) 

Now,

Quotient is 3x + 3 and remainder is 0.


 

(iv)

6x2 + 9x + 4

= 6x2 + 9x + 3 + 1

= (6x2 + 6x + 3x + 3) + 1 

= 6x(x + 1) + 3(x + 1) + 1

= (6x + 3)(x + 1) + 1

= 3(2x + 1)(x + 1) + 1

Now,

Quotient is 3x + 3 and remainder is 1.


 

(v)

6x2 + 9x + 2

= 6x2 + 9x + 3 1

= (6x2 + 6x + 3x + 3)

= 6x(x + 1) + 3(x + 1) 1

= (6x + 3)(x + 1) 1

= 3(2x + 1)(x + 1) 1

Now,

Quotient is 3x + 3 and remainder is 1.


 

(vi)

Now,

Quotient is and remainder is 0.


 



Page No 162:

Question 1:

Now in each of the following, find the quotient and remainder on dividing p(x) by q(x) and write then in the form p(x) = u(x) q(x) + v(x).

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)


 

Answer:

(i)

Given: p(x) = 6x2 + 7x + 1, q(x) = 3x − 1 

When 6x2 + 7x + 1 is divided by 3x − 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and the remainder be c.

(ax + b)(3x − 1) + c = 6x2 + 7x + 1

3ax2ax + 3bxb + c = 6x2 + 7x + 1

3ax2 + (a + 3b) x + (b + c) = 6x2 + 7x + 1

On comparing the coefficients, we get:

3a = 6, a + 3b = 7 and b + c = 1

3a = 6

a = 2

a + 3b = 7

⇒ −2 + 3b = 7 

3b = 7 + 2 

3b = 9

b = 3

b + c = 1

⇒ −3 + c = 1

c = 1 + 3 = 4

Hence, the quotient is 2x + 3 and the remainder is 4.

Thus, u(x) = 2x + 3 and v(x) = 4.

p(x) = u(x) q(x) + v(x)

6x2 + 7x + 1 = (2x + 3)(3x − 1) + 4


 

(ii)

Given: p(x) = 8x2 18x 5, q(x) = 2x − 5 

When 8x2 18x 5 is divided by 2x − 5, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and remainder be c.

(ax + b)(2x − 5) + c = 8x2 18x 5

2ax25ax + 2bx5b + c = 8x2 18x 5

2ax2 + (−5a + 2b)x + (−5b + c) = 8x2 18x 5

On comparing the coefficients, we get:

2a = 8, − 5a + 2b = 18 and −5b + c = 5

2a = 8

a = 4

5a + 2b = 18

⇒ −5(4) + 2b = 18 

⇒ −20 + 2b = 18

2b = 18 + 20

2b = 2

b = 1

5b + c = 5

⇒ −5(1) + c = 5

⇒ −5 + c = 5

c = 5 + 5 = 0

Hence, the quotient is 4x + 1 and the remainder is 0.

Thus, u(x) = 4x + 1 and v(x) = 0.

p(x) = u(x) q(x) + v(x)

8x2 18x 5 = (4x + 1) (2x − 5) + 0


 

(iii)

Given: p(x) = x2 2, q(x) = 2x − 4 

When x2 2 is divided by 2x − 4, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and the remainder be c.

(ax + b) (2x − 4) + c = x2 2 … (1)

2ax24ax + 2bx4b + c = x2 2

2ax2 + (−4a + 2b) x + (−4b + c) = x2 2

On comparing the coefficients, we get:

2a = 1, −4a + 2b = 0 and −4b + c = 2

2a = 1

a =

4a + 2b = 0

4b + c = 2

⇒ −4(1) + c = 2

⇒ −4 + c = 2

c = 2 + 4 = 2

Hence, the quotient is and the remainder is 2.

Thus, u(x) = and v(x) = 2

p(x) = u(x) q(x) + v(x)

x2 2 = (2x − 4) + 2


 

(iv)

Given: p(x) = 4x2 + 3x + 1, q(x) = 4x + 1 

When 4x2 + 3x + 1 is divided by 4x + 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant.

Let the quotient be ax + b and remainder be c.

(ax + b)(4x + 1) + c = 4x2 + 3x + 1

4ax2 + ax + 4bx + b + c = 4x2 + 3x + 1

4ax2 + (a + 4b)x + (b + c) = 4x2 + 3x + 1

On comparing the coefficients, we get:

4a = 4, a + 4b = 3 and b + c = 1

4a = 4

a = 1

a + 4b = 3

1 + 4b = 3 

4b = 3

4b = 2

b =

b + c = 1

+ c = 1

c = 1 =

Hence, the quotient is x +and the remainder is .

Thus, u(x) = x + and v(x) =

p(x) = u(x) q(x) + v(x)

4x2 + 3x + 1 = (4x + 1) +


 

(v)

Given: p(x) = x3 + x2 3x + 2, q(x) = x

When x3 + x2 3x + 2 is divided by x 1, the quotient must be a polynomial of degree 2 and the remainder must be a constant.

Let the quotient be ax2 + bx + c and the remainder be d.

(ax2 + bx + c)(x 1) + d = x3 + x2 3x + 2

ax3ax2 + bx2bx + cxc + d = x3 + x2 3x + 2

ax3 + (a + b)x2 + (b + c)x + (c + d) = x3 + x2 3x + 2

On comparing the coefficients, we get:

a = 1, a + b = 1, b + c = 3 and c + d = 2

a + b = 1

⇒ −1 + b = 1 

b = 1 + 1 = 2 

b + c = 3

⇒ −2 + c = 3

c = 3 + 2 = 1

c + d = 2

⇒ −(1) + d = 2

1 + d = 2

d = 2 − 1 = 1

Hence, the quotient is x2 + 2x 1 and the remainder is 1.

Thus, u(x) = x2 + 2x 1 and v(x) = 1.

p(x) = u(x) q(x) + v(x)

x3 + x2 3x + 2 = (x2 + 2x 1)(x 1) + 1


 

(vi)

Given: p(x) = x3 1, q(x) = x

When x3 1 is divided by x 1, the quotient must be a polynomial of degree 2 and the remainder must be a constant.

Let the quotient be ax2 + bx + c and the remainder be d.

(ax2 + bx + c)(x 1) + d = x3 1

ax3ax2 + bx2bx + cxc + d = x3 1

ax3 + (a + b)x2 + (b + c)x + (c + d) = x3 1

On comparing the coefficients, we get:

a = 1, a + b = 0, b + c = 0 and c + d = 1

a + b = 0

⇒ −1 + b = 0 

b = 0 + 1 = 1 

b + c = 0

⇒ −1 + c = 0

c = 0 + 1 = 1

c + d = 1

⇒ −1 + d = 1

d = 1 + 1 = 0

Hence, the quotient is x2 + x + 1 and the remainder is 0.

Thus, u(x) = x2 + x + 1 and v(x) = 0

p(x) = u(x) q(x) + v(x)

x3 1 = (x2 + x + 1)(x 1) + 0


 

(vii)

Given: p(x) = x3 + 2x2 + 4x + 2, q(x) = x2 + x + 1

When x3 + 2x2 + 4x + 2 is divided by x2 + x + 1, the quotient must be a polynomial of degree 1 and the remainder must be a constant or a polynomial of degree 1.

Let the quotient be ax + b and the remainder be cx + d.

(ax + b)(x2 + x + 1) + (cx + d) = x3 + 2x2 + 4x + 2

(ax + b)(x2 + x + 1) + (cx + d) = x3 + 2x2 + 4x + 2

ax3 + ax2 + ax + bx2 + bx + b + cx + d = x3 + 2x2 + 4x + 2

ax3 + (a + b)x2 + (a + b + c)x + (b + d) = x3 + 2x2 + 4x + 2

On comparing the coefficients, we get:

a = 1, a + b = 2, a + b + c = 4 and b + d = 2

a + b = 2

1 + b = 2 

b = 2 1 = 1 

a + b + c = 4

1 + 1 + c = 4

c = 4 2 = 2

b + d = 2

1 + d = 2

d = 2 − 1 = 1

Hence, the quotient is x + 1 and the remainder is 2x + 1.

Thus, u(x) = x + 1 and v(x) = 2x + 1.

p(x) = u(x) q(x) + v(x)

x3 + 2x2 + 4x + 2 = (x + 1)(x2 + x + 1) + (2x + 1)


 



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