In the previous section we saw some solved examples in which we were asked to write expressions, and then determine whether they are polynomials or not. In this section we will see Polynomial operations.
■ Consider an object stationed at O in the fig.20.8(a) below.
It starts from rest, and move towards the
left with an uniform velocity of 1 metre/sec. What will be it's distance
from O, 1 second after the beginning of the journey?
■ Consider an object stationed at O in the fig.20.8(a) below.
 |
| Fig.20.8 |
Ans: It is given that the
object starts from rest, and that the speed is a uniform velocity of
1 metre/sec.
From this uniform velocity, it
is clear that, the object travels a distance of 1 m in every second.
So, at the end of 1 s, the object will be at a distance of 1 m from O
towards the left side.
• What will be it's distance
from O, 2 seconds after the beginning of the journey?
Ans: The object travels 1 m in
every second. So after 2 seconds, it will be 2 m from O, towards the
left side.
• What will be it's distance
from O, 5 seconds after the beginning of the journey?
Ans: The object travels 1 m in
every second. So after 5 seconds, it will be 5 m from O, towards the
left side.
• We can write an algebraic
expression for the distance from O. If 'l' is the distance from O
after 't' seconds, then we can write: l = t sec × 1 metre/sec. That is., l = t
metre
• But l is not a constant. It
depends on 't'. So we write:
l(t) = t metre
■ Consider another object
stationed at O. It starts from rest, and move towards the right with
an uniform velocity of 2 m/s. What will be it's distance from O, 1
second after the beginning of the journey?
Ans: The object travels 2 m in
every second. So after 1 second, it will be 2 m from O, towards the
right side.
• What will be it's distance
from O, 2 seconds after the beginning of the journey?
Ans: The object travels 2 m in
every second. So after 2 seconds, it will be 4 m from O, towards the
left side.
• What will be it's distance
from O, 5 seconds after the beginning of the journey?
Ans: The object travels 2 m in
every second. So after 5 seconds, it will be 10 m from O, towards the
left side.
• We can write an algebraic
expression for the distance from O. If for the second object, 'd' is
the distance from O after 't' seconds, then we can write: d = t sec × 2 metre/sec. That is., d = 2t metre
• But d is not a constant. It
depends on 't'. So we write:
d(t) = 2t metre
Now consider a more advanced
case:
■ The first object, which
travels towards the left, begins it's journey from a point P, which
is 2 m to the left of O. This is shown in fig.b. It's speed is the same 1 metre/sec
■ At the same instant, the
second object which travels towards the right, begins it's journey
from a point Q, which is 1 m to the right of O. It's speed is the
same 2 metre/sec
■ Consider the first object.
After t seconds, it will be at a distance of t metres from P. So it's
total distance from O will be (t+2) m
• We can write an algebraic
expression for the distance from O. If 'l' is the distance from O
after 't' seconds, then we can write: l(x) = t+2
■ Consider the second object.
After t seconds, it will be at a distance of 2t metres from P. So
it's total distance from O will be (2t+1) m
• We can write an algebraic
expression for the distance from O. If 'd' is the distance from O
after 't' seconds, then we can write: d(x) = 2t+1
■ So we have derived two
polynomials:
• l(x) = t+2
• d(x) = 2t+1
■ Now comes a practical
application:
• The two objects begin their
journeys at the same instant.
♦ The first object, from P,
towards the left.
♦ The second object, from Q, towards the
right.
• After 7 seconds What is the
distance between the two objects?
Solution:
• After 7 seconds, the distance
of the first object from O = t+2 = 7 + 2 = 9 m
• After 7 seconds, the distance
of the second object from O = 2t+1 = 2×7 + 1 = 14 +1 = 15 m
• So total distance between the
two objects = 9 + 15 = 24 m
We got the answer. But is
there any other method to find it? Let us try:
• The distance for the first
object from O is given by t+2
• The distance for the second
object from O is given by 2t+1
• So total distance = (t+2) +
(2t+1) = 3t+3
• After 7 seconds, the distance
= 3×7 + 3 = 21 + 3 = 24 m
■ We get the same answer. And we
see that the second method is easier. In this method, the two polynomials were added.
■ We can add any two polynomials
like this. For example, if
• p(x) = x2 – 2x + 3
• q(x) = 2x2 + 5x - 1
• We can write: p(x) + q(x) = (x2 – 2x + 3) + (2x2 + 5x - 1)
= (x2+2x2) + (-2x+5x) + (3-1)
= 3x2 + 3x + 2
■ We can do subtraction also:
p(x) - q(x) = (x2 – 2x + 3) - (2x2 + 5x - 1)
= (x2 – 2x + 3) - 2x2 - 5x + 1
= (x2-2x2) + (-2x-5x) + (3+1)
= -x2 -7x + 4
■ And multiplication too:
p(x)q(x) = (x2 – 2x + 3)(2x2 + 5x - 1)
= [(x2 × 2x2) + (-2x × 2x2 ) + (3 × 2x2)] + [(x2 × 5x ) + (-2x × 5x )+ (3 × 5x )] + [ (x2 × -1) + (-2x × -1 ) + (3 × -1 )]
= [2x4 + -4x3 + 6x2] + [5x3 + -10x2 + 15x] + [-x2 + 2x + -3]
= (2x4) + (-4x3 + 5x3) + (6x2 + -10x2 + -x2) + (15x + 2x) + (-3)
= 2x4 + x3 - 5x2 + 17x -3
We will now see some solved examples
Solved example 20.10
Given that:
• p(x) = 2x2 + 3x + 5
• q(x) = x2 + 4x + 1
• s(x) = p(x) + q(x)
Calculate p(10), q(10), s(10) and [p(10) + q(10)]
Solution:
1. p(10) = 2×102 + 3×10 + 5 = 200 + 30 + 5 = 235
2. q(10) = 102 + 4×10 + 1 = 100 + 40 + 1 = 141
3. s(x) = p(x) + q(x) = [2x2 + 3x + 5] + [x2 + 4x + 1] = 3x2 + 7x + 6
So s(10) = 3×102 + 7×10 + 6 = 300 + 70 + 6 = 376
4. p(10) + q(10) = 235 + 141 = 376
• We find that the results in (3) and (4) are the same. That means s(10) = p(10) + q(10)
Solved example 20.11
What polynomial added to x2 + 4x - 5 gives 2x2 - 3x + 1
Solution:
Let the required polynomial be p(x). When this p(x) is added to [x2 + 4x - 5] , we will get [2x2 - 3x + 1]
That is., [x2 + 4x - 5] + p(x) = [2x2 - 3x + 1]
⇒ p(x) = [2x2 - 3x + 1] - [x2 + 4x - 5] = [2x2 - 3x + 1 - x2 - 4x +5] = x2 - 7x + 6
Solved example 20.12
What polynomial subtracted from x2 + 4x - 5 gives 2x2 - 3x + 1
Solution:
Let the required polynomial be q(x). When this q(x) is subtracted from [x2 + 4x - 5] , we will get [2x2 - 3x + 1]
That is., [x2 + 4x - 5] - p(x) = [2x2 - 3x + 1]
⇒ p(x) = [x2 + 4x - 5] - [2x2 - 3x + 1] = [x2 + 4x - 5 - 2x2 + 3x -1] = -x2 + 7x - 6
• It may be noted that, if we multiply the above polynomial by '-1', we will get the result in the previous example.
Solved example 20.13
Find polynomials p(x) and q(x) such that:
• p(x) + q(x) = x2 - 4x + 1
• p(x) - q(x) = x2 + 5x - 2
Solution:
1. We have two sums. Let us add them together:
[p(x) + q(x)] + [p(x) - q(x)] = [x2 - 4x + 1] + [x2 + 5x - 2]
⇒ p(x) + q(x) + p(x) - q(x) = 2x2 + x - 1
⇒ 2p(x) = 2x2 + x - 1
⇒ p(x) = x2 + (1⁄2)x - 1⁄2
2. Substitute this value of p(x) in any one of the given equations. let us take the first:
• p(x) + q(x) = x2 - 4x + 1
⇒[x2 + (1⁄2)x - 1⁄2 ] + q(x) = [x2 - 4x + 1]
⇒q(x) = [x2 - 4x + 1] - [ x2 + (1⁄2)x - 1⁄2]
⇒q(x) = x2 - 4x + 1 - x2 - (1⁄2)x + 1⁄2
⇒q(x) = -(41⁄2)x + 11⁄2 .
Solved example 20.14
Given that p(x) = 3x2 - 2x + 4. Calculate each of the following:
(i) [(x+1)p(x)] + [(x-1)p(x)]
(ii) [(x+1)p(x)] - [(x-1)p(x)]
(iii) [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)]
Solution:
■ part (i):
1. (x+1)p(x) = (x + 1) × (3x2 - 2x + 4)
= [(x × 3x2 ) + (x × -2x )+ (x × 4)] + [ (1 × 3x2 ) + (1 × -2x ) + (1 × 4)]
= [3x3 -2x2 + 4x] + [3x2 -2x + 4]
= 3x3 + x2 + 2x + 4
2. (x-1)p(x) = (x - 1) × (3x2 - 2x + 4)
= [(x × 3x2 ) + (x × -2x )+ (x × 4)] + [ (-1 × 3x2 ) + (-1 × -2x ) + (-1 × 4)]
= [3x3 -2x2 + 4x] + [-3x2 +2x - 4]
= 3x3 - 5 x2 + 6x - 4
3. [(x+1)p(x)] + [(x-1)p(x)] = (1) + (2) = (3x3 + x2 + 2x + 4) + (3x3 - 5 x2 + 6x - 4)
= 6x3 - 4x2 + 8x
This is the solution of part (i). The same result can be obtained by another method:
• [(x+1)p(x)] + [(x-1)p(x)] = p(x)[(x+1) + (x-1)] = p(x)[2x]
• p(x)[2x] = (3x2 - 2x + 4)(2x)
= [(3x2 × 2x ) + (-2x × 2x )+ (4 × 2x)]
= 6x3 - 4x2 + 8x
■ We will use this method for part (ii):
• [(x+1)p(x)] - [(x-1)p(x)] = p(x)[(x+1) -(x-1)] = p(x)[2]
• p(x)[2] = (3x2 - 2x + 4)(2)
= [(3x2 × 2 ) + (-2x × 2 )+ (4 × 2)]
= 6x2 - 4x + 8
■ We will use the same method for part (iii):
• [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)] = (1⁄2)p(x)[(x+1) -(x-1)] = (1⁄2)p(x)[2] = p(x)
• So the answer to the given question is p(x) itself. That is:
• [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)] = p(x) = 3x2 - 2x + 4
Solved example 20.15
Given that
p(x) = 3x2 - 2x + 4
q(x) = 5x + 1
Calculate each of the following:
(i) p(x) + q(x)
(ii) p(x) - q(x)
(iii) p(x)q(x)
Solution:
(i) p(x) + q(x) = [3x2 - 2x + 4] + [5x + 1 ] = 3x2 + 3x + 5
(ii) p(x) - q(x) = [3x2 - 2x + 4] - [5x + 1 ] = 3x2 - 7x + 3
(iii) p(x)q(x) = [3x2 - 2x + 4] × [5x + 1 ]
= [(3x2 × 5x ) + (-2x × 5x )+ (4 × 5x)] + [(3x2 × 1 ) + (-2x × 1 ) + (4 × 1)]
= [(15x3) + (-10x2 )+ (20x)] + [(3x2) + (-2x ) + (4)]
= 15x3 - 7x2 + 18x + 4
■ And multiplication too:
p(x)q(x) = (x2 – 2x + 3)(2x2 + 5x - 1)
= [(x2 × 2x2) + (-2x × 2x2 ) + (3 × 2x2)] + [(x2 × 5x ) + (-2x × 5x )
= [2x4 + -4x3 + 6x2] + [5x3 + -10x2 + 15x] + [-x2 + 2x + -3]
= (2x4) + (-4x3 + 5x3) + (6x2 + -10x2 + -x2) + (15x + 2x) + (-3)
= 2x4 + x3 - 5x2 + 17x -3
We will now see some solved examples
Solved example 20.10
Given that:
• p(x) = 2x2 + 3x + 5
• q(x) = x2 + 4x + 1
• s(x) = p(x) + q(x)
Calculate p(10), q(10), s(10) and [p(10) + q(10)]
Solution:
1. p(10) = 2×102 + 3×10 + 5 = 200 + 30 + 5 = 235
2. q(10) = 102 + 4×10 + 1 = 100 + 40 + 1 = 141
3. s(x) = p(x) + q(x) = [2x2 + 3x + 5] + [x2 + 4x + 1] = 3x2 + 7x + 6
So s(10) = 3×102 + 7×10 + 6 = 300 + 70 + 6 = 376
4. p(10) + q(10) = 235 + 141 = 376
• We find that the results in (3) and (4) are the same. That means s(10) = p(10) + q(10)
Solved example 20.11
What polynomial added to x2 + 4x - 5 gives 2x2 - 3x + 1
Solution:
Let the required polynomial be p(x). When this p(x) is added to [x2 + 4x - 5] , we will get [2x2 - 3x + 1]
That is., [x2 + 4x - 5] + p(x) = [2x2 - 3x + 1]
⇒ p(x) = [2x2 - 3x + 1] - [x2 + 4x - 5] = [2x2 - 3x + 1 - x2 - 4x +5] = x2 - 7x + 6
Solved example 20.12
What polynomial subtracted from x2 + 4x - 5 gives 2x2 - 3x + 1
Solution:
Let the required polynomial be q(x). When this q(x) is subtracted from [x2 + 4x - 5] , we will get [2x2 - 3x + 1]
That is., [x2 + 4x - 5] - p(x) = [2x2 - 3x + 1]
⇒ p(x) = [x2 + 4x - 5] - [2x2 - 3x + 1] = [x2 + 4x - 5 - 2x2 + 3x -1] = -x2 + 7x - 6
• It may be noted that, if we multiply the above polynomial by '-1', we will get the result in the previous example.
Solved example 20.13
Find polynomials p(x) and q(x) such that:
• p(x) + q(x) = x2 - 4x + 1
• p(x) - q(x) = x2 + 5x - 2
Solution:
1. We have two sums. Let us add them together:
[p(x) + q(x)] + [p(x) - q(x)] = [x2 - 4x + 1] + [x2 + 5x - 2]
⇒ p(x) + q(x) + p(x) - q(x) = 2x2 + x - 1
⇒ 2p(x) = 2x2 + x - 1
⇒ p(x) = x2 + (1⁄2)x - 1⁄2
2. Substitute this value of p(x) in any one of the given equations. let us take the first:
• p(x) + q(x) = x2 - 4x + 1
⇒[x2 + (1⁄2)x - 1⁄2 ] + q(x) = [x2 - 4x + 1]
⇒q(x) = [x2 - 4x + 1] - [ x2 + (1⁄2)x - 1⁄2]
⇒q(x) = x2 - 4x + 1 - x2 - (1⁄2)x + 1⁄2
⇒q(x) = -(41⁄2)x + 11⁄2 .
Solved example 20.14
Given that p(x) = 3x2 - 2x + 4. Calculate each of the following:
(i) [(x+1)p(x)] + [(x-1)p(x)]
(ii) [(x+1)p(x)] - [(x-1)p(x)]
(iii) [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)]
Solution:
■ part (i):
1. (x+1)p(x) = (x + 1) × (3x2 - 2x + 4)
= [(x × 3x2 ) + (x × -2x )
= [3x3 -2x2 + 4x] + [3x2 -2x + 4]
= 3x3 + x2 + 2x + 4
2. (x-1)p(x) = (x - 1) × (3x2 - 2x + 4)
= [(x × 3x2 ) + (x × -2x )
= [3x3 -2x2 + 4x] + [-3x2 +2x - 4]
= 3x3 - 5 x2 + 6x - 4
3. [(x+1)p(x)] + [(x-1)p(x)] = (1) + (2) = (3x3 + x2 + 2x + 4) + (3x3 - 5 x2 + 6x - 4)
= 6x3 - 4x2 + 8x
This is the solution of part (i). The same result can be obtained by another method:
• [(x+1)p(x)] + [(x-1)p(x)] = p(x)[(x+1) + (x-1)] = p(x)[2x]
• p(x)[2x] = (3x2 - 2x + 4)(2x)
= [(3x2 × 2x ) + (-2x × 2x )
= 6x3 - 4x2 + 8x
■ We will use this method for part (ii):
• [(x+1)p(x)] - [(x-1)p(x)] = p(x)[(x+1) -(x-1)] = p(x)[2]
• p(x)[2] = (3x2 - 2x + 4)(2)
= [(3x2 × 2 ) + (-2x × 2 )
= 6x2 - 4x + 8
■ We will use the same method for part (iii):
• [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)] = (1⁄2)p(x)[(x+1) -(x-1)] = (1⁄2)p(x)[2] = p(x)
• So the answer to the given question is p(x) itself. That is:
• [1⁄2(x+1)p(x)] - [1⁄2(x-1)p(x)] = p(x) = 3x2 - 2x + 4
Solved example 20.15
Given that
p(x) = 3x2 - 2x + 4
q(x) = 5x + 1
Calculate each of the following:
(i) p(x) + q(x)
(ii) p(x) - q(x)
(iii) p(x)q(x)
Solution:
(i) p(x) + q(x) = [3x2 - 2x + 4] + [5x + 1 ] = 3x2 + 3x + 5
(ii) p(x) - q(x) = [3x2 - 2x + 4] - [5x + 1 ] = 3x2 - 7x + 3
(iii) p(x)q(x) = [3x2 - 2x + 4] × [5x + 1 ]
= [(3x2 × 5x ) + (-2x × 5x )
= [(15x3) + (-10x2 )
= 15x3 - 7x2 + 18x + 4
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