Polynomials
Class 9th Mathematics Part 2 Kerala Board Solution
- In rectangles with one side 1 centimetre shorter than the other, take the length of the…
- From the four corners of a rectangle, small squares are cut off and the sides are…
- Consider all rectangles that can be made with a 1 metre long rope. Take one of its…
- An 1 metre wide path goes around a square ground. The relation between the length of a…
- A liquid contains 7 litres of water and 18 litres of acid. More acid is added to it.…
- Two poles of heights 3 metres and 4 metres are erected upright on the ground, 5 metres…
- Write each of the operations below as an algebraic expression, find out which are…
- First degree polynomial with p(1) = 1 and p(2) = 3. Find polynomials p(x) satisfying…
- First degree polynomial with p(1) = -1 and p(-2) = 3. Find polynomials p(x) satisfying…
- Second degree polynomial with p(0) = 0, p(1) = 2 and p(2) = 6. Find polynomials p(x)…
- Three different second degree polynomials with p(0) = 0 and p(1) = 2. Find polynomials…
- Taking p(x) = 2x^2 + 3x + 5, q(x) = x^2 + 4x + 1 and s(x) = p(x) + q(x), calculate…
- What polynomial added to x^2 + 4x - 5 gives 2x^2 - 3x + 1?
- What polynomial subtracted from x^2 + 4x - 5 gives 2x^2 - 3x + 1?…
- Find polynomials p(x) and q(x) such that p(x) + q(x) = x^2 - 4x + 1 and p(x) - q(x) =…
- Taking p(x) = 3x^2 - 2x + 4, write the following as polynomials: i) (x + 1) p(x) + (x -…
Questions Pg-147
Question 1.In rectangles with one side 1 centimetre shorter than the other, take the length of the shorter side as x centimetres.
i) Taking their perimeters as p(x) centimetres, write the relation between p(x) and x as an equation.
ii) Taking their area as a(x) square centimetres, write the relation between a(x) and x as an equation.
iii) Calculate p(1), p(2), p(3), p(4), p(5). Do you see any pattern?
iv) Calculate a(1), a(2), a(3), a(4), a(5). Do you see any pattern?
Answer:
Given one side is smaller than the other by 1 cm.
∴ The two adjacent sides of the triangle are x, x + 1.
(i) perimeter p(x) = 2 × [(x) + (x + 1)]
⇒ p(x) = 4x + 2 – (1)
(ii) Area a(x) = (x) × (x + 1)
⇒ a(x) = x2 + x – (2)
(iii) by (1)
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
Here the difference between a number and its successor is always 4.
We can see that they are in Arithmetic Progression (AP) with a common difference of 4.
(iv) by (2)
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
Here the difference between a number and its previous number is increasing by 2.
a(2) – a(1) = 4
a(3) – a(2) = 6
a(4) – a(3) = 8
a(5) – a(4) = 10
∴ Their common difference is in AP.
Question 2.
From the four corners of a rectangle, small squares are cut off and the sides are folded up to make a box, as shown below:
i) Taking side of the square as x centimetres, write the dimensions of the box in terms of x.
ii) Taking the volume of the box as v(x) cubic centimetres, write the relation between v(x) and x as an equation.
iii) Calculate 
Answer:
(i) The side of the square = x
⇒ The dimensions of the box are
L = 7 – 2x
B = 5 – 2x
H = x
(ii) v(x) = L × B × H
⇒ v(x) = (7 - 2x) × (5 – 2x) × (x)
⇒ v(x) = (35 + 4x2 -24x) × (x)
⇒ v(x) = 4x3 – 24x2 + 35x – (1)
(iii) by (1)
⇒ 
⇒ 
Similarly,
v(1) = 4(1)3 – 24(1)2 + 35(1)
⇒ v(1) = 15
And
⇒ 
Question 3.
Consider all rectangles that can be made with a 1 metre long rope. Take one of its sides as x centimetres and the area enclosed as a(x) square centimetres.
i) Write the relation between a(x) and x as an equation.
ii) Why are the numbers a(10) and a(40) equal?
iii) To get the same numbers as a(x), for two different numbers as x, what must be the relation between the numbers?
Answer:
According to the question the perimeter of the rectangle must be always equal to 100 cm.
Now, let one side be x cm and other adjacent side be y cm.
⇒ P = 2 × [x + y]
⇒ 100 = 2x + 2y
⇒ y = 50 – x – (1)
(i) a(x) = (50-x) × (x) – (by (1))
a(x) = 50x – x2
(ii) a(10) = 50 × (10) – (10)2 = 400
a(40) = 50 × (40) – (40)2 = 400
Now if a side a is 10 cm, then the other side
P = (10 + y) × 2
P = 100
⇒ 50 – 10 = y
⇒ y = 40 cm
∴ the in case of x = 10 cm y = 40 cm, and in case of x = 40 cm y = 10 cm. So, in both cases we have same rectangle hence the area is same.
Questions Pg-150
Question 1.Write each of the relations below in algebra and see if it gives a polynomial. Give reasons for your conclusion also.
An 1 metre wide path goes around a square ground. The relation between the length of a side of the ground and the area of the path.
Answer:
Let the side of the ground be a.
The width of the path is 1 m.
⇒ Length of one side of the path is a + 2 m.
⇒ Area of the path = (a + 2)2 – (a)2
Now, Area of the path is a is defined p(a)
⇒ p(a) = (a + 2)2 – (a)2
⇒ p(a) = 4 + 4 × a
Question 2.
Write each of the relations below in algebra and see if it gives a polynomial. Give reasons for your conclusion also.
A liquid contains 7 litres of water and 18 litres of acid. More acid is added to it. The relation between the amount of acid added and the change in the percentage of acid in the liquid.
Answer:
A liquid is 7 litres water and 18 litres acid.
Let x litres acid is added to this liquid.
Now,
The liquid is 7 litres water and (18 + x) litres acid.
Now,
Initial percentage of acid = 
⇒ Initial percentage of acid = 72% -- (1)
Final percentage of acid = 
– (2)
Let change in percentage of the acid be R
⇒ 
Question 3.
Write each of the relations below in algebra and see if it gives a polynomial. Give reasons for your conclusion also.
Two poles of heights 3 metres and 4 metres are erected upright on the ground, 5 metres apart. A rope is to be stretched from the top of one pole to some point on the ground and from there to the top of the other pole:
The relation between the distance of the point on the ground from the foot of a pole and the total length of the rope.
Answer:
Let the distance of the point on ground be x m from the 3 m pole.
⇒ This distance of the point on ground from 4 m pole is (5-x) m. (∵ the distance between the poles is 5 m.)
Now the length of the rope is AB + BC.
By Pythagoras Theorem
AB2 = AD2 + DB2
⇒ AB2 = 32 + x2
⇒ 
Similarly,
Now, let the length of rope be a R
⇒ R = AB + BC
⇒ 
Question 4.
Write each of the operations below as an algebraic expression, find out which are polynomials and explain why.
i) Sum of a number and its reciprocal.
ii) Sum of a number and its square root.
iii)Product of the sum of difference of a number and its square root.
Answer:
(i) Let the number be x.
So its reciprocal is 
⇒ 
It is not a polynomial as it has a negative power of x, but for a polynomial all the powers should be positive integers.
(ii) Let the number be x.
So its square root is √x
Sum = x + √x
It is not a polynomial as it has square root which is fractional exponent, but to be a polynomial it should only have positive integer exponents.
(iii) Let sum of number and its square root be a.
⇒ a = x + √x
Let difference be b.
⇒ b = x - √x
Product of a and b.
⇒ a × b = (x + √x) × ( x - √x)
⇒ a × b = x2 – x
It’s a Polynomial as all the exponents of x are positive integer.
Question 5.
Find polynomials p(x) satisfying each set of conditions below.
First degree polynomial with p(1) = 1 and p(2) = 3.
Answer:
let the polynomial be p(x)
p(x) = ax + b (where a and b are constants)
p(1) = a + b = 1 – (1) (given in the question)
p(2) = 2a + b = 3 – (2) (given in the question)
by (1) and (2)
a = 2, b = -1
⇒ p(x) = 2x – 1
Question 6.
Find polynomials p(x) satisfying each set of conditions below.
First degree polynomial with p(1) = –1 and p(–2) = 3.
Answer:
Let the polynomial be p(x)
⇒ p(x) = ax + b
⇒ p(-1) = -a + b = -1 – (1)
⇒ p(-2) = -2a + b = 3 – (2)
By (1) and (2)
a = -4, b = -5
⇒ p(x) = -4x -5
Question 7.
Find polynomials p(x) satisfying each set of conditions below.
Second degree polynomial with p(0) = 0, p(1) = 2 and p(2) = 6.
Answer:
Let the polynomial be p(x)
p(x) = ax2 + bx + c (a,b,c are constants)
⇒ p(0) = c = 0
⇒ p(1) = a + b = 2 – (1)(∵ c = 0)
⇒ p(2) = 4a + 2b = 6 – (2)
By (1) and (2)
a = 1, b = 1
⇒ p(x) = x2 + x
Question 8.
Find polynomials p(x) satisfying each set of conditions below.
Three different second degree polynomials with p(0) = 0 and p(1) = 2.
Answer:
Let an arbitrary polynomial be p(x)
p(x) = ax2 + bx + c
⇒ p(0) = c = 0
⇒ p(1) = a + b = 1
⇒ p(x) = ax2 + bx (where a + b = 1)
There could be infinite number of such polynomials but we have to find only 3.
They are any three possible polynomials.
Questions Pg-153
Question 1.Taking p(x) = 2x2 + 3x + 5, q(x) = x2 + 4x + 1 and s(x) = p(x) + q(x), calculate p(10), q(10), s(10), p(10) + q(10).
Answer:
(i) p(10)
p(x) = 2x2 + 3x + 5
p(10) = 2(10)2 + 3(10) + 5
p(10) = 235 – (1)
(ii) q(10)
q(x) = x2 + 4x + 1
q(10) = (10)2 + 4(10) + 1
q(10) = 141 – (1)
(iii) s(x) = p(x) + q(x)
s(x) = 2x2 + 3x + 5 + x2 + 4x + 1
s(x) = 3x2 + 7x + 6
s(10) = 3(10)2 + 7(10) + 6
s(10) = 376
(iv) p(10) + q(10)
by (1) and (2)
p(10) + q(10) = 235 + 141
p(10) + q(10) = 376
This is the same result as (iii)
⇒ Polynomials have Commutative Property of Addition.
Question 2.
What polynomial added to x2 + 4x – 5 gives 2x2 – 3x + 1?
Answer:
Let the required polynomial p(x)
Now, according to the question
x2 + 4x – 5 + p(x) = 2x2 – 3x + 1
⇒ p(x) = 2x2 – 3x + 1 - x2 - 4x + 5
⇒ p(x) = x2 - 7x + 6
Question 3.
What polynomial subtracted from x2 + 4x – 5 gives 2x2 – 3x + 1?
Answer:
Let the required polynomial p(x)
Now, according to the question
x2 + 4x – 5 – p(x) = 2x2 – 3x + 1
⇒ p(x) = x2 + 4x – 5 - 2x2 + 3x – 1
⇒ p(x) = -x2 + 7x – 6
Question 4.
Find polynomials p(x) and q(x) such that p(x) + q(x) = x2 – 4x + 1 and p(x) – q(x) = x2 + 5x –2.
Answer:
p(x) + q(x) = x2 – 4x + 1 – (1)
p(x) – q(x) = x2 + 5x –2 – (2)
by adding (1) and (2)
2 × p(x) = 2x2 + x –1
⇒ 
by (1)
q(x) = x2 – 4x + 1 – p(x)
⇒ 
⇒ 
Question 5.
Taking p(x) = 3x2 – 2x + 4, write the following as polynomials:
i) (x + 1) p(x) + (x – 1) p(x)
ii) (x + 1) p(x) – (x – 1) p(x)
iii) 
Answer:
(i) p(x) = 3x2 – 2x + 4
Now,
(x + 1) p(x) + (x – 1) p(x)
taking p(x) common
⇒ p(x)[x + 1 + x-1]
⇒ 2x × p(x)
⇒ 2x × (3x2 – 2x + 4)
⇒ 6x3 – 4x2 + 8x
(ii) p(x) = 3x2 – 2x + 4
Now,
(x + 1) p(x) - (x – 1) p(x)
taking p(x) common
⇒ p(x)[x + 1-x + 1]
⇒ 2 × p(x)
⇒ 2 × (3x2 – 2x + 4)
⇒ 6x2 – 4x + 8
(iii) p(x) = 3x2 – 2x + 4
Now,
taking p(x) common
⇒ 
⇒ p(x)
⇒ 3x2 – 2x + 4