Chapter 20 - Polynomials

In the previous section we completed the discussion on Similar triangles. In this section we will learn about Polynomials.

Consider a rectangle shown in fig.20.1.1(a). It has a length (l) = 3 cm and width (b) = 2 cm.

Fig.20.1

• We can easily calculate it's perimeter.

p = 2(l+b) = 2(3+2) = 2 × 5 = 10 cm

1. Suppose we increase the length by 1 cm, and width also by the same 1 cm, the new length = 4 cm, and new width = 3 cm

• Then new perimeter = 2(4+3) = 14 cm. This is shown in fig.20.1.1(c)

2. Suppose we increase the original length by 2 cm, and original width also by the same 2 cm, the new length = 3 +2 =5 cm, and new width = 2 + 2 = 4 cm

• Then new perimeter = 2(5+4) = 18 cm. This is shown in fig.20.1.2(c)

3. Suppose we increase the original length by 1¹⁄₂ cm, and original width also by the same 1¹⁄₂ cm, the new length = 3 + 1¹⁄₂ = 4¹⁄₂ cm, and new width = 2 + 1¹⁄₂ = 3¹⁄₂ cm

• Then new perimeter = 2(4¹⁄₂ + 3¹⁄₂) = 16 cm

■ So we find that, when the dimensions of the rectangle changes, the perimeter also changes. We can write a general form (see fig.20.1.3):

• Let the increase in length be 'x'. Then the new length = (3+x)

• Let the increase in width also be 'x'. Then the new width = (2+x)

• Then, the new perimeter = 2 [(3+x) + (2+x)] = 2 [5+2x] = 10+4x = 4x +10 

We can say this:

• We have a rectangle with length 3 cm and width 2 cm

• If the length and width are increased by the same amount 'x', the new perimeter will be given by 4x+10

• So this is an easy method to calculate new perimeter. Let  us put some values for x. The new perimeters are tabulated below:

Value of perimeter for different values of 'x':

x (cm)	1	11⁄2	2	21⁄4	3
p (cm)	14	16	18	19	22

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Now we will consider area. 
The original area = l × b = 3 × 2 = 6 cm (see fig.20.2.1 below)

Fig.20.2

1. When the length and width are increased by 1 cm, new length = 4 cm, and new width = 3 cm. So new area = 4 × 3 = 12 cm²

• In the fig.20.2.1.c, the additional areas are marked by green rectangles. We can write:

• Original area = 6, additional area = 3 +1 +2 = 6. So total area = 6 +6 = 12

2. When the original length and width are increased by 2 cm, new length = 5 cm, and new width = 4 cm. So new area = 5 × 4 = 20. This is shown in fig.20.2.2.c

• In the fig.20.2.2.c, the additional areas are marked by green rectangles. We can write:

• Original area = 6, additional area = 6 +4 +4 = 14. So total area = 6 +14 = 20
■ So we find that, when the dimensions of the rectangle changes, the area also changes. We can write a general form (see fig.20.2.3):

• Let the increase in length be 'x'. Then the new length = (3+x)

• Let the increase in width also be 'x'. Then the new width = (2+x)

• Then new area = [(3+x)(2+x)]. We have learned how to multiply them, when we saw identities. We get: [(3+x)(2+x)] = 6 + 2x + 3x + x² = x²+ 5x + 6

[The value '6' in the above result is the area of the red rectangle in fig.c. The others 2x, 3x and x² are the areas of additional green rectangles]

• We can write: New area = x²+ 5x + 6

• So this is an easy method to calculate new areas. Let us put some values for x. The new areas are tabulated below:

Value of area for different values of 'x':

x (cm)	1	11⁄2	2	3	4
a (cm)	12	153⁄4	20	30	42

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So we have seen two equations:

■ New perimeter = 4x+10

■ New area = x²+ 5x + 6

Apart from making our calculations easier, such equations have wider applications.

• Instead of 'New perimeter', let us write 'p(x)'. Then the equation will become: p(x) = 10 + 4x

• 'p(x)' indicates that we must not put a 'permanent value' for perimeter.
• It is kind of a 'warning', which tells us that, in this particular problem, the perimeter, (which is denoted by 'p') will change based on the value of 'x'

• The manner in which the change takes place, will be given on the right side of the '=' sign.

• Similarly, in the second equation, instead of 'New area', let us write 'a(x)'. Then the equation will become: a(x) = x²+ 5x + 6

• 'a(x)' indicates that we must not put a 'permanent value' for area.
• It is kind of a 'warning', which tells us that, in this particular problem, the area, (which is denoted by 'a') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign.

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We have seen perimeter and area. Now let us see volume. 

1. In fig.20.3(a) below, we have a red box of length 3 cm, width 2 cm, and height 1 cm. So it's volume = 3 × 2 × 1 = 6 cm³ 

Fig.20.3

2. Its length and width are increased by 'x' cm as indicated by the green rectangles in fig.a. It's height is increased by the same 'x' cm, as indicated by the yellow rectangle in fig.a.

3. If we give a height of 1 cm for the green rectangles, they will reach the top level of the red box. This is shown in fig.b. Then, the additional volumes will be: 
♦ 3 ×x ×1 = 2x cm³ 
♦ 2 ×x ×1 = 3x cm³
♦ x ×x ×1 = x² cm³  

• So the total volume in fig.b = 6 + 2x + 3x + x² = x²+ 5x + 6

4. Now, in fig.c, the additional height x cm is also given volume. The volume of this yellow box = [(2+x)x(3+x)] = [(6+5x+x²)x] = x³+ 5x² + 6x

5. Thus, the total volume in fig.c = x²+ 5x + 6 + x³+ 5x² + 6x 
= x³+ 6x² + 11x + 6

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The same result can be obtained by multiplying the new length, width and height together. This is shown below:

(3+x)(2+x)(1+x) = (6+5x+x²)(1+x) 
= 6+5x+x²+6x+5x²+x³ = x³+6x²+11x+6

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We can write: v(x) = x³+6x²+11x+6

• 'v(x)' indicates that we must not put a 'permanent value' for volume.
• It is kind of a 'warning', which tells us that, in this particular problem, the volume, (which is denoted by 'v') will change based on the value of 'x'
• The manner in which the change takes place, will be given on the right side of the '=' sign

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In the next section, we will see some solved examples demonstrating the creation of such expression in real life situations.

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Chapter 15.3 - Equations in two variables - Identities

In the previous sections we saw some advanced cases in the solution of 'two equations with two variables'. We also saw some solved examples. In this section, we will see another type of equations.

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We will be using some identities that we have learned before:
1. (x+y)² = x² + 2xy + y²
2. (x-y)² = x² - 2xy + y²
3. (x+y)² - (x-y)² = 4xy
4. (x+y)² =  (x-y)² + 4xy
5. x² - y² = (x+y)(x-y)

• (1) and (2) are basic identities
• (3) is simply (1) - (2)
• (4) is obtained by rearranging (3)
• (5) is easily obtained by expanding it's right side

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Consider the problem: There are two squares of different sizes. The sides of the larger square is 5 cm greater than the sides of the smaller square. The area of the larger square is 55 sq.cm more than the smaller square. What is the length of the sides of each square?
Solution:
1. Let the length of the larger square = x. So area of the larger square = x²
2. Let the length of the smaller square = y. So area of the smaller square = y² 
3. Use condition 1: Larger side is 5 cm greater: x = y+5
4. Use condition 2: Area is 55 greater: x² = y² + 55
5. From (3) we get x-y = 5
6. From (4) we get: x²-y² = 55
7. Now we use an identity that we have learned before: x² - y² = (x+y)(x-y)
8. From (7) we get x+y = ^(x2-y2)⁄_(x-y)
9. So we can write: x+y = 55/5 = 11
10. Isolate x from (9). We get: x = 11-y
11. Substitute this value of x in (5). We get: (11-y)-y = 5 ⇒ 11-2y = 5 ⇒ 2y = 11-5 ⇒ 2y = 6
∴ y = 6/2 = 3
12. Substitute this value of y in (3). We get: x = 3 + 5 = 8
13. So we can write: Side of the larger square = x = 8 cm, and, the side of the smaller square = 3 cm
14. Check: From (4) we get: 82 = 32 + 55 ⇒ 64 = 9+55 ⇒ 64 = 64

Another problem:
The perimeter of a rectangle is 10 m and it's area is 5¹⁄₄ sq.m. What are the lengths of the sides?
Solution:
1. Let the length of the rectangle = x
2. Let the breadth of the rectangle = y
3. Use condition 1: perimeter = 10: 2(x+y) = 10 ⇒ x+y = 10/2 ⇒ x+y = 5
4. Use condition 2: Area = 5¹⁄₄ : xy = 5¹⁄₄
5. Now we use an identity that we have learned before: (x+y)² -(x-y)² = 4xy
6. In (5), we have the value of (x+y). This we have from (3)
7. In (5), we have the value of xy. This we have from (4)
8. What we do not have in (5) is (x-y). So isolate it: (x-y)² = (x+y)²-4xy
9. Substitute the values of (x+y) and xy in (8). We get:
10. (x-y)² = 5²- 4 × 5¹⁄₄ ⇒ (x-y)² = 25 - 21 ⇒ (x-y)² = 4 ⇒ x-y = 2
11. Now, from (3), we have x+y, and from (10), we have x-y 
12. In (10), isolate x. We get: x = 2+y. 
13. Substitute this x in (3). We get: (2+y)+y = 5 ⇒ 2 +2y = 5 ⇒ 2y = 3 
∴ y = 3/2 = 1¹⁄₂ m
14. Substitute this value of y in (3). We get: x+ 3/2 = 5 ⇒ x = 5 - 3/2 = 7/2 = 3¹⁄₂ m
15. So we can write: The length of the rectangle = x = 3¹⁄₂ m, and breadth = y = 1¹⁄₂ m
16. Check: use condition 2: Area = xy = 7/2 × 3/2 = 21/4 = 5¹⁄₄ sq.m

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Now we will see some solved examples:
Solved example 14.11
A 10 m long rope is to be cut into two pieces, and a square is to be made using each. The difference in the areas enclosed must be 1¹⁄₄ sq.m. How should it be cut?
Solution:
1. Let the length of one part = x
2. Let length of the remaining = y
3. Use condition 1: Total length of rope = 10 m: x+y = 10
4. Use condition 2:
• x is made into a square. Side of that square will be ^x⁄₄. So area of that square will be ^x2⁄₁₆ 
• y is made into a square. Side of that square will be ^y⁄₄. So area of that square will be ^y2⁄₁₆ 
• Difference in areas enclosed = ^x2⁄₁₆ - ^y2⁄₁₆ = ^(x2-y2)⁄₁₆
• Condition 2 states that this difference must be 1¹⁄₄sq.mt. So we can write: ^(x2-y2)⁄₁₆ = 1¹⁄₄
⇒ ^(x2-y2)⁄₁₆ = ⁵⁄₄ ⇒ x² - y² =  ^(16×5)⁄₄ ⇒ x² - y² = 20   
5. Use the identity: x² - y² = (x+y)(x-y)
6. Rearranging (5) we get: x-y = ^(x2-y2)⁄_(x+y)   ⇒ x-y = ²⁰⁄₁₀  ⇒ x-y = 2
7. Isolate x from (6). We get: x = y+2
8. Substitute this value of x in (3). We get: (y+2)+y = 10 ⇒ 2y+2 = 10 ⇒ 2y = 8
∴ y = 8/2 = 4
9. Substitute this value of y in (7). We get: x = 4+2 = 6
10. So we can write: The rope should be cut in such a way that, one piece has a length = x = 6m, and the remaining piece has a length = y = 4 m
11. Check: Use condition 2: 
• Area of first square = ^x2⁄₁₆ = ⁶²⁄₁₆ = ³⁶⁄₁₆ = ⁹⁄₄ = 2¹⁄₄
• Area of second square = ^y2⁄₁₆ = ⁴²⁄₁₆ = ¹⁶⁄₁₆ = 1
• Difference in area = 2¹⁄₄ - 1 = 1¹⁄₄

Solved example 14.12
The length of a rectangle is 1 m more than it's breadth. It's area is 3³⁄₄ sq.m. What are it's length and breadth?
Solution:
1. Let the length of the rectangle = x
2. Let the breadth of the rectangle = y
3. Use condition 1: Length more than breadth by 1 m: x = y+1 ⇒ x-y=1
4. Use condition 2: Area = 3³⁄₄ : xy = 3³⁄₄
5. Use the identity: (x+y)² =  (x-y)² + 4xy. We get: (x+y)² =  1² + 4 × 3³⁄₄
⇒ (x+y)² = 1 + 15 ⇒ (x+y)² =16 ⇒ (x+y) = √16 ⇒ (x+y)= 4
6. Isolate x from (5). We get: x = 4-y
7. Substitute this value of x in (3). We get: (4-y)-y = 1 ⇒ 4-2y = 1 ⇒ 2y = 3 ⇒ y = 3/2 = 1¹⁄₂
8. Substitute this value of y in (6). We get: x = 4 - 3/2 = 5/2 = 2¹⁄₂ 
9. So we can write: Length of the rectangle = x = 2¹⁄₂ m, and breadth = y = 1¹⁄₂ m
10. check: Area = xy = ³⁄₂ × ⁵⁄₂ = ¹⁵⁄₄ = 3³⁄₄

Solved example 14.13
The hypotenuse of a right triangle is 6¹⁄₂ cm, and it's area is 7¹⁄₂ sq.cm. Calculate the length of it's perpendicular sides
Solution:
1. Let the length of one perpendicular side = x
2. Let the length of the other perpendicular side = y
3. Use condition 1: Hypotenuse = 6¹⁄₂
• Using Pythagoras theorem, x² + y² = [ 6¹⁄₂ ]² ⇒ x² + y² = [ ¹³⁄₂ ]²  ⇒ x² + y² = ¹⁶⁹⁄₄ 
4. Use condition 2: Area = 7¹⁄₂
• In a right triangle, one perpendicular side can be taken as the base, and the other perpendicular side can be taken as the height. Let x be the base, and y, the height
• So area = ¹⁄₂ × base × height = ¹⁄₂ × xy = ^xy⁄₂ 
Thus we can write: ^xy⁄₂ = 7¹⁄₂ ⇒ ^xy⁄₂ = ¹⁵⁄₂ ⇒ xy = 15 ⇒ 2xy = 30
5. Use identity: (x+y)² = x² + 2xy + y² ⇒ (x+y)² = (x² + y²)+ 2xy 
6. We have all values on the right side. Substituting those values we get: (x+y)² = (¹⁶⁹⁄₄) + 30 ⇒ (x+y)² = ^(169+120)⁄₄ 
⇒ (x+y)² = ²⁸⁹⁄₄ ⇒ x+y = √(²⁸⁹⁄₄) ⇒ x+y = ¹⁷⁄₂
7. Use identity: (x-y)² = x² - 2xy + y² ⇒ (x-y)² = (x² + y²) - 2xy 
8. We have all values on the right side. Substituting those values we get: (x-y)² = (¹⁶⁹⁄₄) - 30 ⇒ (x-y)² = ^(169-120)⁄₄ 
⇒ (x-y)² = ⁴⁹⁄₄ ⇒ x-y = √(⁴⁹⁄₄) ⇒ x-y = ⁷⁄₂
9. Isolate x from (8). We get: x = ⁷⁄₂ + y
10. Substitute this value of x in (6). We get: ⁷⁄₂ + y + y = ¹⁷⁄₂  ⇒ ⁷⁄₂ + 2y = ¹⁷⁄₂ 
⇒  2y = ¹⁷⁄₂ - ⁷⁄₂  = ¹⁰⁄₂ = 5 ⇒ 2y = 5 
∴ y = ⁵⁄₂
11. Substitute this value of y in (6). We get: x + ⁵⁄₂ = ¹⁷⁄₂ ⇒ x = ¹⁷⁄₂ - ⁵⁄₂ = ¹²⁄₂ = 6
12. So we can write: One perpendicular side = x = 6 cm, and the other perpendicular side = y = ⁵⁄₂ cm
13. Check: Area =  ¹⁄₂ × 6 × ⁵⁄₂ = ³⁰⁄₄ = ¹⁵⁄₂ = 7¹⁄₂ sq.cm

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We have completed the discussion on Equations in two variables. In the next section we will see irrational numbers.

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