Kerala Syllabus Class 9 Maths Chapter 10 Solutions Polynomials

Students often refer to Kerala Syllabus 9th Standard Maths Textbook Solutions Chapter 10 Polynomials Extra Questions and Answers Notes to clear their doubts.

Kerala SCERT Class 9 Maths Chapter 10 Solutions Polynomials

Polynomials Class 9 Kerala Syllabus Questions and Answers

Kerala State Syllabus 9th Standard Maths Chapter 10 Polynomials Solutions Questions and Answers

Class 9 Maths Chapter 1 Kerala Syllabus – Algebra of Measurements

Intext Questions And Answers

Question 1.
Let’s start with a question

i) What is the perimeter of the rectangle shown in the figure?
ii) What is the new perimeter if the sides of the rectangle are extended by 1 cm? Also, what is the new perimeter if the sides of the rectangle are extended by 2 cm?
iii) What is the new perimeter if the sides of the rectangle are extended by 3 cm?
iv) What is the new perimeter if the sides of the rectangle are extended by x cm?
Answer:
i) Perimeter = 2( 3 + 2) = 10 cm

ii) If the sides are extended by 1cm ,then
New length = 3 + 1 = 4 cm
New breadth = 2+ 1 = 3 cm
New perimeter = 2(4 + 3) = 14 cm
We can think about this in another way, since all four sides increase by 1 centimeter each, the total increase is 4 centimeters.
Therefore, the new perimeter = 10 + 4 = 14 cm
Similarly, if all four sides increase by 2 centimeters each, the total increase is 8 centimeters.
Therefore, the new perimeter =10 + 8 = 18 centimeters.

iii) If all four sides increase by 3 centimeters each, the total increase is 12 centimeters.
Therefore, the new perimeter =10 + 4 × 3 = 22 cm

iv) If all four sides increase by x centimeters each, the total increase is 4x centimeters.
Therefore, the new perimeter = 10 + 4x centimeters.

Textual Questions And Answers

Question 1.
In all rectangles with one side 1 centimetre less than the other, denote the length of the shorter side as x centimetres.
i) Denote their perimeters as p(x) centimetres and write the relation between x and p(x) as an equation.
ii) Denote their areas as a(x) square centimetres and write the relation between x and a(x) as an equation
iii) Compute p (1), p (2), p (3), p (4), p (5) Do you see any pattern?
iv) Compute a (1), a (2), a (3), a (4), a (5) Do you see any pattern?
Answer:
i) Let x be the shorter side, then the other side will be (x + 1)
Perimeter = 2 (x + x + 1) = 4x + 2 cm
p(x) = 4x + 2

ii) Area = x(x + 1) = x² + x cm²

iii) p(x) = 4x + 2
p(1) = 4 × 1 + 2 = 6
p (2) = 4 × 2 + 2= 10
p (3) = 4 × 3 + 2 = 14
p (4) = 4 × 4 + 2= 18
p (5) = 4 × 5 + 2 = 22
Perimeter is a sequence increasing by 4

iv) a(x) = x² + x a(1) = 1² + 1= 2
a (2) = 2² + 2 = 6
a (3) = 3² + 3 = 12
a (4) = 4² + 4 = 20
a(5) = 5² + 5 = 30
Area is a sequence of consecutive even numbers.

Question 2.
From the four corners of a rectangle, small squares of the same size are cut off and the tabs are raised up to make a box as in the picture below:

i) Denote the length of the sides of the squares as x centimetres and write the lengths of the three edges of the box in terms of x
ii) Denote the volume of the box as v(x) cubic centimetres and write the relation between x and v(x) as an equation
iii) Compute v (\(\frac{1}{2}\)), v (1) and v (1\(\frac{1}{2}\))
Answer:
i) If x cm is the length of the small squares that are cut off, then the length of the maked box by folding it up = 7 – x – x = (7 – 2x) cm
Width = 5 – x – x = (5 – 2x) cm
Height = x cm

ii) V(x)= (7 – 2x)(5 – 2x)(x)
= (35 – 24x + 4x²)(x)
= 4x³ – 24x² + 35x

iii) v(\(\frac{1}{2}\)) = (7 – 2 × \(\frac{1}{2}\))(5 – 2 × \(\frac{1}{2}\))(\(\frac{1}{2}\))
= 6 × 4 × \(\frac{1}{2}\)
=12 cm³

v(1) = (7 – 2)(5 – 2)(1)
= 5 × 3 × 1 = 15 cm³

v(1\(\frac{1}{2}\)) = (7 – 2 × (1\(\frac{1}{2}\))) (5 – 2 × (1\(\frac{1}{2}\)))(1\(\frac{1}{2}\))
= 4 × 2 × \(\frac{3}{2}\)
= 12 cm³

Question 3.
Consider all rectangles that can be made with a rope of length 1 metre. Denote the length of one side as x centimetres and the area enclosed by the rope as a(x) square centimetres.
i) Write the relation between x and a(x) as an equation
ii) Why are a (10) and a (40) the same number?
iii) To get the same number as a(x) when x is taken as two different numbers, what should be the relation between the numbers?
Answer:
i) If one side is x cm ,then the other side is 50 — x cm
Area, a(x) = x (50 – x) = 50x – x² cm²

ii) a(10) = 10(50 – 10) = 10 × 40 = 400
a(40) = 40(50 – 40) = 10 × 40 = 400
10 and 40 are the sides of rectangle, so a (10) and a (40) are same.

iii) Numbers must be sides of rectangle. If we add the two numbers together we get the sum as half the length of the wire (50 cm) used to make it.

Class 9 Maths Kerala Syllabus Chapter 10 Solutions – Special Expressions

Textual Questions And Answers

Question 1.
In each of the problems below, check whether the relation between the specified measurements is a polynomial. Give reasons for your assertions.
i) The relation between the length of the sides of a square park and the area of a 1 metre wide path around it.
ii) The relation between the amount of acid added to a mixture of 3 litres of acid and 7 litres of water, and the change in the percent of acid in the mixture
iii) Two poles of heights 3 metres and 4 metres stand 5 metres apart. A rope is to be stretched from the top of one post to some point on the ground and then stretched to the top of the other pole:

The distance from the foot of one pole to the point on the ground where the rope is fixed, and the total length of the rope.
Answer:
i)

Area of path = Area of large square – Area of small square
= (x + 2)² – x²
= 4x + 4
= 4(x + 1)
This is a polynomial because exponent here is a positive integer.

ii) The solution is \(\frac{3}{10}\) acidic. When x liters of acid are added, the solution’s volume also increases by x litres. The resulting percentage of acid and solution is \(\frac{3+x}{10+x}\) × 100. This expression is not a polynomial because the exponent is not a positive integer.

iii)

Total length of the rope = \(\sqrt{x^2+9}+\sqrt{(5-x)^2+16}\)
This is not a polynomial because it involves square root.

Question 2.
Write each of the following operations as an algebraic expression. Check which of them are polynomials, giving reasons
i) Sum of a number and its reciprocal
ii) Sum of a number and its square root
iii) The product of the sum of a number and its square root, and the difference of the square root and the number
Answer:
Let the number be x
i) x + \(\frac{1}{x}\) It is not a polynomial
ii) x + √x,It is not a polynomial
iii) (x – √x)(x + √x) = x² – x , It is a polynomial

Question 4.
For each of the polynomial p(x) given below, compute p (1) and p (10)
i) p(x) = 2x + 5
ii) p(x) = 3x² + 6x + 1
iii) p(x) = 4x³ + 2x² + 3x + 7
Ans:
i) p(1) = 2 × 1 + 5 = 7
p(10) = 2 × 10 + 5 = 25

ii) p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(10) = 3 × 10² + 6 × 10 + 1 = 361

iii) p(1) = 4 × 1³ + 2 × 1² + 3 × 1 + 7 = 16
p(10) = 4 × 10³ + 2 × 10² + 3 × 10 + 7 = 4237

Question 5.
For each of the polynomial p(x) given below, compute p(0), p(1) and p(-1)
i) p(x) = 3x + 5
ii) p(x) = 5x – 8
iii) p(x) = 3x² + 6x + 1
iv) p(x) = 2x² – 5x + 3
v) p(x) = 4x³ + 2x² + 3x + 7
vi) p(x) = ax³ + bx² + cx + d
Answer:
i) p(0) = 3 × 0 + 5 = 5
p(1) = 3 × 1 + 5 = 8
p(-1) = 3 × -1 + 5 = 2

ii) p(0) = 5 × 0 – 8 = -8
p(1) = 5 × 1 – 8 = -3
p(-1)= 5 × -1 – 8 = -13

iii) p(0) = 3 × 0 + 6 × 0 + 1 = 1
p(1) = 3 × 1² + 6 × 1 + 1 = 10
p(-1)= 3 × (-1)² + 6 × – 1 + 1 = -2

iv) p(0) = 2 × 0 – 5 × 0 + 3 = 3
p(1) = 2 × 1² – 5 × 1 + 3 = 0
p(-1)= 2 × (-1)² – 5 × -1 + 3 = 10

v) p(0) = 4 × O + 2 × O + 3 × O + 7 = 7
p(1) = 4 × 1³ + 2 × 1² + 3 × 1 +7 = 16
p(-1)= 4 × (-1)³ + 2 × (-1)² + 3 × -1 + 7 = 2

vi) p(0) = a × 0 + b × 0 + c × 0 + d = d
p(1) = a × 1³ + b × 1² + c × 1 + d = a + b + c + d
p(-1)= a × (-1)³ + b × (-1)² + c × -1 + d = -a + b- c + d

Polynomials Class 9 Extra Questions and Answers Kerala Syllabus

Question 1.
In a polynomial p(x) = 2x³ + ax² – 7x + b, p (1) = 3, p (2) = 19. Then find the value of ‘a’ and ’b’?
Answer:
p (x) = 2x³ + ax² – 7x + b
We have p(1) = 3
p(1) = 2(1)³ + a(1)² – 7(1) + b
3 = 2 + a – 7 + b
a + b = 8 ………..(1)
p(2) = 19
p (2) = 2(2)³ + a(2)² – 7(2) + b
19= 16 + 4a- 14 + b
4a + b = 17 ……..(2)
Subtracting equation (1) from (2)
we get 3a = 9, a = 3

Put a = 3 in equation (1)
we get b = 8 – 3 = 5

Question 2.
In a polynomial p(x) = 2x³ + 9x² + kx + 3, p (- 2) = p (- 3) Find the value of k.
Answer:
p (x) = 2x³ + 9x² + kx + 3

p (-2) = 2(-2)³ + 9(-2)² + k (-2) + 3
= – 16 + 36 – 2k + 3
= 23-2k

p (-3) = 2(-3)3 + 9(-3)2 + k (-3) + 3
= -54 + 81 – 3k + 3
= 30 – 3k

We have p (-2) = p (-3)
23 – 2k = 30 – 3k
3k – 2k = 30 – 23
k = 7

Question 3.
From the polynomial p(x) = 2x² – 3x + 1 find p (0), p (1) and p (-1).
Answer:
p(x) = 2x² – 3x + 1
p (0) = 2(0)² – 3(0) + 1 = 1
p(1) = 2(1)² – 3(1) + 1 = 0
p (-1) = 2(- 1)² – 3(-1) + 1 = 2 + 3 + 1 = 6

Question 4.
In a polynomial p(x) = 2x³ – 7x² + kx + 20, p(2) = p(3)
i) Find the value of k.
ii) Using the value of k, write the polynomial.
iii) Find p (1).
Answer:
i) p(2) = 2(2)³ – 7(2)² + k(2) + 20
= 16 – 28 + 2k + 20
= 8 + 2k

p (3) = 2(3)3 – 7(3)2 + k(3) + 20
= 54 – 63 + 3k + 20
= 11 + 3k

P(2) = P(3)
8 + 2k = 11 + 3k
3k – 2k = 8 – 11
k = -3

ii) p(x) = 2x³ – 7x² – 3x + 20

iii) p(1) = 2(1)³ – 7(1)² – 3(1) + 20
= 2 – 7 – 3 + 20
= 12

Question 5.
In the polynomial p(x) = 3x² – ax + 1 Find ’a’ satisfying p(2) = 2
Answer:
p(x) = 3x² – ax + 1

p (1) = 3x² – ax + 1
= 3 – a + 1
= 4 – a
Given p(1) = 2, 4-a = 2, a = 4- 2 = 2

Question 6.
i) p(x) = 4x – 5, find p(2)
ii) Write the first degree polynomial q(x) with q (2) = 4 and q (1) = 0
Answer:
i) p(2) = 4(2) – 5 = 8 – 5 = 3
ii) Let q(x) = ax + b
q (2) = 2a + b
4 = 2a + b ….(1)

q (1) = a +b
0 = a + b ….(2)
Solving equations (1) and (2) we get a = 4 and b = -4
Thus q(x) = 4x – 4

Question 7.
p(x) = 2x² + 3x + 5
i) Find p (1), p (0)
ii) Write the second degree polynomial p(x) with p (0) = 2 and p (1) = 5
Answer:
i) p(1) = 2 × 1² + 3 × 1 + 5
= 2 + 3 + 5
= 10

p(0) = 2 × 0² + 3 × 0 + 5 = 5

ii) p(x) = ax³ + bx + c
p(0) = a × 0² + b × 0 + c = c
Given p(0) = 2 c = 2

p(1) = a × 1² + b × 1 + c
= a + b + c

Given p(1) = 5
a + b + c = 5
a + b = 3
Therefore,p(x) = 2x³ + x + 2
The equation a + b = 3 is satisfied for any pair of values (a, b), provided a is non-zero.