CHAPTER – 2 UNIT – 1
BANKING
POINTS TO REMEMBER






Banks accepts deposits and lend money.
You will understand the terms drawee and drawer.
Cheques or withdraw forms are used to draw money from the bank.
Bank draft is bankers cheque.
In S.B. any number of deposits are allowed but withdrawals are restricted.
Interest is calculated on the basis of daily balance.
UNIT – 1 EXERCISE 2.1.6
1. Enter the following particulars into a saving bank pass book.
(a) Balance brought forward: 2250
(b) By cash: 3500 on 06-06-2012
(b) To self: 1250 on 11-06-2012
(d) By bank draft:4800 on 23- 06-2012
(e) By cheque no. 323263: 750 on 30-06-2012
Solution:
Date
Particulars
01.06.2012
06.06.2012
11.06.2012
23.06.2012
30.06.2012
Balance B/F
By Cash
To Self
By Bank Draft
By Cheque
Cheque
no.
Debit
Credit
3,500.00
1,250.00
323263
4,800.00
750.00
Balance
2,250.00
5,750.00
4,500.00
9,300.00
10,050.00
2. The particulars of a SB account is given here. Calculate the monthly interest on
daily product basis.
Date
Particulars
Cheque no. Debit
Credit
Balance
01.05.2010 Balance B/F
2,842.00
04.05.2010 By Cash
4,600.00
7,442.00
12.05.2010 To Geetha
843655
2,500.00
4,942.00
21.05.2010 By Self
800.00
4,142.00
560090
7,500.00 11,642.00
30.05.2010 By Cheque
Solution: The daily product for the month is calculated as follows:
Daily product for 3 days (from 1st to 3rd )
= 2,842×3= 8,526
Daily product for the W and t 8 days (from 4 th to 11th ) = 7,442×8= 59,536
Daily for the next 9 days (from 12 th to 20th )
= 4,942×9= 44,478
Daily for the 9 days (from 21s t to 29th)
= 4,142×9= 37,278
Daily for one day (to 30 th )
= 11642×1= 11,642
Daily product for may 2010
=1,61,460
Interest for the month at the rate of 4% PTR = 161460× 4 × 1 = 17.69=18.00
100
100 365
3. Pranava frequently uses his debit card for online shopping. In a month, he had
an in initial amount of 9,800 in a bank. He credited 11,000 to his account on
07.12.2011. He used his card for online shopping and an amount of 6,125 was
debited from his account. He credited 3,400 on 19.12.2011 by cheque no
565677. Enter these particulars in to SB account and find interest at 4% per
annum on daily product basis.
Solution:
Date
01.12.2011
07.12.2011
Particulars Cheque No.
Debit
Credit
Balance
Balance B/F
9,800.00
By Cash
11,000.00
20,800.00
To
6,125.00
14,675.00
19.12.2011 By Cheque
787865
3,400.00
18,075.00
24.12.2011 To Cheque
123432
4,800.00
13,275.00
To Cheque
565677
750.00
12,525.00
4. The entries in an SB account pass book are given below. Calculate the interest
at 4% per annum for the month on daily product basis.
Date
01.01.2009
05.01.2009
10.01.2009
21.01.2009
Particulars
Balance B/F
By Cash
To Cheque
By Cash
Cheque No.
Debit
Credit
2,200.00
945800
1,250.00
7,800.00
Balance
6,400.00
8,600.00
7,350.00
15,150.00
27.01.2009
30.01.2009
Solution:
By Cheque
By Cheque
663119
124322
750.00
2,100.00
15,900.00
13,800.00
The daily product for the month is calculated as follows:
6,400 × 4
25,600.00
8,600 × 5
43,000.00
7,350 × 11
80.850.00
15,150 × 8
90,900.00
15,900 × 3
47,470.00
13,800 × 2
27,600.00
3,15,650.00
Interest for the month =
PTR = 315650×1×4 = Rs. 34.59 = 35.00
100
365×100
5. Kruthi opens SB account in a bank on 4.6.2010 with an initial deposit of 1,500.
She credits 4,00 to her account on 09-06-2010. She receives a banker cheque
in her name for 10,200 on her name and she eposits it to her accunt on
20.06.2010. she issues a cheque (no. 2290898) for her brother for 4,500 on
21.06.2010. she deposits 2,500 on 24-06-2010 and withdraws 1,000 on
30-6-2010. Enter these into SB account pass book and find the interest she
gets for the month on daily product basis.
Solution:
Date
04.06.2010
09.06.2010
20.06.2010
21.06.2010
24.06.2010
30.06.2010
Particulars
Open SB A/c
By Cheque
To Cheque
By Cash
To Self
Cheque No.
Debit
Credit
4,000.00
10,200.00
2290898
4,500.00
2,500.00
1,000.00
Balance
1,500.00
5,500.00
15,700.00
11,200.00
13,700.00
12,700.00
The daily product for the month is calculated as follows:
1,500 × 5
7,500.00
5,500 × 11
60,500.00
15,700 × 1
15,700.00
11,200 × 3
33,600.00
13,700 × 6
82,200.00
12,700 × 1
12,700.00
2,12,200.00
Interest for the month = PTR = 212200×1×4 = Rs. 23.25 = 23
100
365×100
CHAPTER – 3
UNIT- 4 DIVISION
POINTS TO REMEMBER
 Addition, substraction and multiplication of two polynomials always give a
polynomial.
 We generally get two polynomials quotient and remainder when a
polynomial is divided by another.
 Complete divisibility occur when the remainder is o.
 Dividend:
(Divisor × Quotient) + remainder .
EXERCISE 3.4.2
1. Find the quotient in each case:
(i)
X8
= x8-3 = x5
X3
(ii)
M8
M3
= m8-3 = m5
(iii)
8p14
=
P8
8p14-8 8p6
5
(iv)
30k9
6k2
=
(v)
(-15x8)
5x5
-9a10
-3a9
= -15 X
5
= -9a10-9
-3
-9y 6
9y2
= -9y6-2
9
(vi)
(vii)
30
6
k
9-2
= 5k7
8-5
=-3x3
= 3a1 = 3a
= -1×y4 = -y4
(viii) 2x2y
Xy
2
(i x)
(x)
2x2y×2 = 4x2-1y1-1 = 4x1y0 = 4x -1 = 4x
(xy)
=
-18h4h5
-62h2
= -18 h 4-2 k 5-2 = 3h2 k3
-6
1 x6
8
1 x3
4
= 1 × 4 x6-3
8 1
= 1 x3
2
2. Area of a rectangle = length × breadth = 800x2
Length
= 10x
Breadth
= Area
Length
= 800x2
10x
= 800x2-1 = 80x
10
3. An isosceles right triangle has its area 20x2. What is the area of the
equilateral triangle constructed on one of its legs?
[Hint: area of an equilateral triangle with side length a is Ṿ3.a2 ]
4
Solution: in isosceles right triangle ABC
A
AB = BC = a cm,
Area of ABC = 1 ×AB×BC
2
2
20X = 1 ×a×a
2
2
20x = a2
1
2
a
B
A2 = 40X2…………..(1)
The area of an equilateral
le
a
C
With AB or BC as side
Area of = V3 a2
4
[a2 = 40×2 from (1)]
= Ṿ3 ×40×x2
4
= 10Ṿ3 x2
4. The area of a rectangle is 36x4 and one of its sides is 4x. A square is
constructed on the other side. What is the radio of the area of the square
to that of the rectangle?
Area of a rectangle
A1 = 36x4
Length
L1 = 4x
Area of the square of length 4x ,
A2 = (4x)2 = 16x2
A1
A2
=
36x4 = 36
=9
2
4-2
16x
16 x
4 x2
5. The area of a square is 64x2 . an equilateral triangle is constructed on one of
its sides. What is the altitude of this triangle?
Area of the square = 64x2
Length of the sides = V64x2 = 8x
Altitude of the equilateral
with side 8x is
(8x)2 = (4x)2 + h2
64x2 = 16x2 + h2
H2 = 64x2 – 16x2
H2 = 48x2
H = v48x2
= v16×3x = 4v3x
EXERCISE 3.4.3
1. Arrange the following in descending powers of x:
(i)
X2 + x5+ 2x3 + x2 – 2 + 3x
X5 + 2x3 + x2 + 3x-2
(ii)
X8 + x9 + x12 – 3x7 + x2 + 1
X12 + x9 + x8 – 3x7 + x +1
(iii)
X2y6 + x5y2 + x6y4 + 3x2y8
X6y4 + x5y2 + x2y6 + 3x2y8
2. Perform the following divisions
(i)
3x2 + 4x – 4 by x+2
X+2 ) 3x2+4x -4 (3x-2
3x2 + 6x
(-) (-)
-2x-4
-2x-4
0
(ii)
2x2 – 9x + 9 by x-3
x-3)2x2 – 9x+9 (2x-3
2x2 – 6x
(-) (+)
-3x + 9
-3x + 9
0
(iii)
2x2 + 2x+11 by x+3
X+3)3x2 = 13x+11(2x-4
2x2 + 6x
(-) (-)
-4x + 11
-4x – 12
-1
(iv)
3x2=11 – 13x by x-3
x-3)3x2-13x + 11 (3x-4
3x2-9x
(-) (+)
-4x + 11
-4x + 12
-1
(V)
x 3 – 5x2 + 8x – 4 by x-2
x-2) x3-5x2 + 8x-4 (x2-3x+2
x3 – 2x2
(-) (+)
-3x2 + 8x
-3x2 + 6x
2x-4
2x-4
0
(vi)
6x3+x2-23x +12by 2x - 3
2x-3)6x3+x2-23x + 12 (3x2 = 5x-4
6x3-9x2
(-) (+)
10x2-23x
10x2-15x
-8x + 12
-8x + 12
0
(vii)
2x2-7x+6 by x-2
x-2)2x2-7x+6(2x-3
2x2-4x
-3x+4
-3x+6
0 0
(viii)
x4-0.x3 – 3x2 + 0.x – 4 by x+2
X+2)x4+0.x3-3x2+0.x-4 (x3-2x2+x-2
X4+2x3
(-) (-)
-2x3-3x2
-2x3-4x2
(+) (+)
X2+0.x
X2+2x
-2x-4
-2x-4
0
(ix)
3x2+x3 + 4 by x+2
X+2)x3+3x2+0.x+4(x3+x-2
X3+2x2
X2+0.2
X2+2.x
-2x+4
-2x-4
8
(xi)
X4-16 by x-2
x-2 )x4+0.x3+0.x2+0.x-16(x3+2x2+4x+8
x4-2x2
2x3+0.x2
2x3-4x2
4x2+0.x
4x2-8x
8x-16
8x-16
0
(xii)
X3-1 by x-1
x-1 )x3+0.x3+0.x-1(x3+x+1
x3-x3
x2 +0.x
x2- x
x–1
x–1
0
(xiii)
8x3- 27 by 2x-3
2x-3)8x3+0.x2+0.x-27 (4x2+6x+9
8x3- 12x3
12x2+0.x
12x2-18x
18x –27
18x- 27
0
3.
Divide
(i)
X5+a5 by x+a
X+a)x5+0.x4+0.x3a2+0.x2a3+0.xa4+a5(x4-x3a+x2a2-xa3+a4
X5+x4a
-x4a+0.x3a2
-x4a-x3a2
X3a2+0.x2a3
X3a2+x2a3
-x2a3+0.xa4
-x2a3- xa4
Xa4+a5
Xa4+a5
0
(ii)
X7-y7 by x-y
X6+y+x4y2+x2y2+x3y3+x2y4+xy5+y
x-y)x7+0.x6y+0.x5y2+0.x4y3+0.x3y4+0.x2y5+0.xy6-y7
x 7-x2y
x6y+0.x5y2
x5y2- x4y5
x6y2+0.x4y3
x5y2- x4y3
x4y3+0.x3y4
x4y3 – x 3y4
x3y4+0.x2y5
x3y4 – x 2y5
x2y6+0.xy6
x2y5- xy6
xy8-y7
xy6-y7
0
(iii)
X9+y9 by x3+y3
X6-x3y3+y6
X3+y3)x9+0.x6y3+0.x3y6+y9
X9+ x6y3
(-)
-x6y3+0.x3y6
-x6y3-x3y6
X3y6+y9
X3y6+y9
0
4.
Divide a+b by a1/3 – b1/3
A1/3-a1/3+b2/3
A1/3+b1/3)a+0.a2/3b1/3+0.a1/3 b1/3+b
a+a2/3 b1/3
(-)
-a2/2 b1/3 +0.a1/3 b2/3
-a2/3 b1/3- a1/3 b2/3
A1/3b2/3 +b
A1/3 b2/3+b
0
5. Divide a2-b2 by a1/2– b1/2
A3/2+ ab1/2 +a1/2 b+b3/2
A1/2+b1/2) a2+0.a 3/2 b 1/2 +0.ab – b2
A2 – a3/2 b1/2
A3/2+0.ab
A3/2-ab
Ab+0.a½ b3/2
Ab- a½ b3/2
A½ b3/2-b2
A½ b /2-b2
0
6. Which of the following are visible by x+a? (that is the division leaves 0
remainder?
(i)
X3-a3
X2-xa+a2
X+a)x +0.x2a+0.xa2-a3
X3+x 2a
-x2a+0.xa2
-a2a-xa2
-xa2-a3
-xa2+a3
-2a3
3
(x+a) is not a divisible by x3-a3 since it leaves a remainder = -2a3.
(ii)
X4-a4
X3-x2a+xa2-a3
X+a)x4+0.x3a+0.x2a2+0.xa3-a4
X4+x3a
-x3a+0.x2a2
-x3a-x2a2
X2a2+0.xa3
X2a2+xa3
-xa3-a3
-xa3-a4
0
(x+a) is divisible by x4-a4
(iii)
X7 + a7
X6-x5+x4a2-x3a3+x2a4-xa5+a6
X+a)x7+0.x6a+0.x5a2+0.x4a3+0.x3a4+0.x2a5+0.xa6+a7
X6+x6a
-x6a+0.x5a2
-x6a-x5a2
X6a2+0.x4a3
X5a2+x4a3
-x4a3+0.x3a4
-x4a3-x3a4
X3a4+0.x2a5
X3a4-x2a5
X2a5+0.xa6
X2a5-xa5
Xa6-a7
Xa6-a7
0
Thus (x+a) divides x7+a7
(iv)
X8 + a8
X7-x8a+x 5a2-x4a3+x3a4-x2a5+xa6-a7
X+a)x8+0.x7a+0.x6a2+0.x5a3+0.x4a4+0.x3a5+0.x2a6+0.xa7+a3
X8+x7a
-x7a+0.x6a2
-x7a-x6a2
+x6a2 +0.x6a3
+x6a2 +x5a3
-x5a3+0.x4a4
-x5a3-x4a4
X4a4+0.x3a5
X4a4- x3a5
-x3a5+0.x2a4
-x3a5- x2a6
X2a6+0.a7
X2a6+xa7
-xa7+a8
-xa7+a8
2a8
(x+a) is not divisible by x8+a8 since it leaves a remainder of 2a8 .
1. Divide
(i)
4x5+7x4+14x2+3x2-8x+6 by x2+3x+2
X2+3x+2) 4x5+7x4+14x3+3x2-8x+6(4x3-5x2+21x-50
4x5+12x4+8x3
4
-5x + 6x3+ 3x2
-5x4- 15x3-10x2
+
2
-21x +13x3-8x
-21x3+63x3+42x
2
-50x – 50x+ 6
-50x2-150x-100
100x+106
X3+5x2+4x – 4 by x2+3x- 2
(ii)
X+2
X +3x-2)x +5x2+4x – 4
X3+3x3-2x
2x2+6x – 4
2x2+6x – 4
0
4
4
2
(iii) A +4b by a +2ab+b2
2
3
A2 – 2ab+3b2
A2+2ab+b2)a4+0.a3b+0.a2b2+0.ab3+4b4
A4 +2a3+a2b2
-2a3b-a2b2+0.ab3
-2a3b-4a2b2-2ab3
+3a2b2+2ab3+4b4
+3a2b2+6ab3+3b4
-4ab3+b4
x4-4x2+12x+9 by x2+2x-3
(iv)
X3-2x+3
X +2x-3)x4+0.x3- 4x2+ 12x+9
X4 +2x3-3x2
-2x3-x2 + 12x
-2x3-4x2+6x
3x2+6x+9
3x2+6x-9
18
2
(iv)
2x5-7x4-2x3+18x2-3x-8 by x3-2x2+1
2x2- 3x – 8
X3-2x2+1)2x5- 7x4 – 2x3 + 18x2-3x-8
2x5-4x4+0+2x2
-3x4- 2x3+16x2-3x-8
-3x4+ 6x3+0 – 3x
-8x3 + 16x2 – 8
-8x3 + 16x2 – 8
0
7x4+4x2+3x – 5 by 2x2+3x-22
(v)
7 x2 -21 x +107
2
4
8
2
4
3
2
2x +3x-2)7x +0.x +4x +3x-5
7x4+21x3 – 7x2
2
-21x3+11x2+3x-5
2
-21 x3 63 x2+21 x
2
4
2
107 x2 - 15 x -5
4
2
107 x2 + 321 x -107
4
8
4
-381 x - 87
8
4
X6 – 1 by x3-2x+2x – 1
(vi)
X3+2x2+2x+1
X -2x +2x-1)x6+0.x5+0.x4+0.x3+0.x2+0.x-1
X6-2x5+2x4-x3
2x5-2x4+x3+0.x2+0.x -1
2x5-4x4+4x3 -2x2
2x4-3x3 + 2x2 + 0.x -1
2x4-4x3+4x2 -2x
X3-2x2+2x -1
X3-2x2+2x -1
0
3
2
2. Find the remainder when x2+px2+qx+1
Is divided by x2+px+q
X
X +px+q)x +px2+qx+1
X3+px2 +qx
2
3
1
3. What should be added to x5 -1 to be completely divisible by x2 +3x -1?
The remainder is 1
4. What should be added to x6-64 to be completely divisible by a4-16
A2
A4 – 16) a6 – 0.a2-64
A6- 16a2
16a2 – 64
16a2-64 should be added to a6-64 to be completely divisible by a4 – 16.
ADDITIONAL PROBLEMS ON DIVISION
1. Divide
(i)
4x5 by 8x3
4x5
8x3
(ii)
(iii)
= 4 x5-3 = 1 x2
8
2
3x3y2 by 1 xy
3
3 2
3x y
1xy
= 3x 3 x3-1 y2-1 = 9x2y
3
1
4x6+3x5 by 2x2
4x6+3x5
2x2
= 4x6 + 3x5
2x2
2x2
= 4 x6-2 + 3 x5-2
2
2
(iii)
= 2x4+3 x3
2
12
6
3
4πa +8a by 2πa
4πa12+8a6
2πa3
(iv)
= 4πa12 + 8a6
2πa3
2πa3
= 2a12-3 + 4a6-3
Π
9
= 2a + 4 a3
Π
X12+x10+x8+x6 by x3
X3)x12+x10+x8+x6 (x9 +x7+x5 +x3
X12
0+x10+x8+x6
X10
0+x8 +x6
X8
0+x6
X6
0
2. Divide
(i)
2x4+x2+1 by 2x+1
3
2
X -1 x + 3 x – 3
4
4
2x+1)2x +0.x +x +0.x+1
2x4+x3
-x3 +x2+0.x+1
4
2
3
2
X3 - 1 x2
2
3 x2 + 0.x+1
2
3 x2 +3 x
1 4
-3 x +1
4
- 3 x -3
4
4
7
4
(ii)
3x6+4x3+x2+1 by 3x+4.
X5-x4+x3+1 x2 – 2 x+ 2
3
3
3
6
5
4
3 2
3x+3)3x +0.x +0.x +4x -x +0.x+2
3x6+3x5
-3x5+0.x4
-3x5 – 3x4
3x4+4x3
3x4+3x3
X3 – x 2
X3 + x 2
-2x2+0.x
-2x2-2x
2x+2
2x+2
0
(iii)
X5 – a5 by x-a.
X4+x3a+x2a2+xa3+a 4
x-a)x5+0.x4a+0.x3a2+0.x2a3+0.xa4-a5
x5- x4a
x4a+0.x3a2
x4a-x3a2
+x3a2+0.x2a3
+x3a2- x2a3
X2a3+0.xa4
X2a3- xa4
Xa4- a5
Xa4- a5
0
(iv)
X5 – a5 by x+a
X4-x3a+x2a2-xa3+a4
X+a)x5+0.x4a+0.x3a2+0.x2a3+0.xa4-a5
X5+ x4a
-x4a+0.x3a2
-x4a-x3a2
X3a2+0.x2a3
X3a2+x2a3
-x2a3+0.xa4
-x2a3- xa4
Xa4- a5
Xa4+ a5
-2a5
(v)
32x5 -16x4+8x3 – 2 by 2x+1.
16x4-16x3+12x2-6x+3
2x+1)32x5 – 16x4+8x3+0.x2+0.x -2
32x5 + 16x4
-32x4+ 8x3
-32x4- 16x3
24x3+0.x2
24x3+12x2
-12x2+0.x
-12x2-6x
6x- 2
6x+ 3
-5
(vi)
a3+b3-3ab by a+b
A2-ab+b2
A+b)a3-0.a2b+0.ab2+b3- 3ab
A3+a2b
-a2b+ 0.ab2
-a2b-ab2
Ab2+b3 – 3ab
Ab2+b3
-3ab
3. Divide:
(i)
2x4- 3x2 -3x -2 by x2+1
2x2- 3x – 2
X2+1)2x4 – 3x3 -0.x2 – 3x – 2
2x4 0.x + 2x2
-3x3 – 2x2 – 3x – 2
-3x2 + 0 – 3x
-2x2 – 2
-2x2 – 2
0
(ii)
-2x3 by x2+1-2x.
X4+0.2x3+3x2=2x+1
X2-2x+1)x6+0.x5+0.x4-2x3+0.x2+0.x+1
X6+2x5+x 4
2x5- x4 -2x3
2x5-4x4+2x3
3x4- 4x3+0.x2
3x4- 6x3+3x2
(+)
2x3- 3x2+0.x
2x3- 4x2+2x
X2 – 2x+1
X2 – 2x+1
0
(iii)
A4+4a2 +16 by a2+2a+4.
A2 – 2a+4
A2+2a+4) a4 + o.a3 +4a2 +0.a+16
A4 +2a3 +4a2
-2a3- 0+0.a+16
-2a3 – 4a2 – 8a
4a2+8a+16
4a2+8a+16
0
(iv)
Y6+x6 – 2x3y3 by x2+y2 -2xy.
X4+2x3y+3x2y2+2xy3+y4
X2-2xy+y2)x6+0.x5y+0.x4y2 – 2x3y3+0.x2y4+0.xy5+y6
X6- 2x5y+x4y2
2x5y – x4y2 – 2x3y3
2x5y – 4x4y2+ 2x3y3
3x4y2-4x3y3+0.x2y4
3x4y2-6x3y3+3x2y4
2x3y3- 3x2y4+0.xy5
2x3y3- 4x2y4+2xy5
X2y4- 2xy5+y6
X2y4- 2xy5+y6
0
4. Divide: x5-px4+qx3- px-1 by x-1.
X4+(1-p)x3+(1-p+q)x2+(1-p)x+1
x-1)x5-px4+qx3-qx2+px-1
x5-x4
(1-p)x4+qx3
(1-p)x4- (1-p)x3
(1-p+q)x3- qx2
(1-p+q)x3- (1-p+q)x2
(1-p)x2+px
(1-p)x2-(1-p)x
x-1
x-1
0
5. Divide: x3(y-z) +x3(z-x)+x3(x-y) by y2-xz-z2+xy [write in descending powers of
(y).
Solution: [write in descending powers of y]
Dividend=x3y-x3z+y3z-xy3+z3x+z3y
Descending powers of y
Dividend =y3z-yz3+z3x-x3y+x3y-y3x
Divisor =y 2 + xy – xz –z2
Y2+xy-xz-z2) y2z- yz3 +z3x –x3z+ x3y- y3x(yz-xz-xy+x2
Y2z- yz3+xy2z-xyz2
(-) (+) (-) (+)
Z3x +xyz2-xyz2- x3z+x3y- y3x
Z2x +x2z2-xy2z- x2yz
(-) (-) (+) (+)
Xyz2- x3z+ x3y –y3x –x2y2 + x2yz
Xyz2- x2z2
-y3x
+ x2yz
(-) (+)
(+)
(-)
3
2 2
3
2 2
-x z + x y + x y – x z
-x3z + x2y2 + x3y – x2z2
(+) (-)
(-) (+)
0
2
Q =yz-xz-xy+x
Q =z(y-x) – x(y-x) = (y-x) (z-x)
6. Divide: a-b by a1/4 – b1/4
A3/4+a1/2b1/4+a1/4b1/2+b3/4
A1/4-b1/4)a+0.a3/4b1/4b1/4 +0.a1/2b1/2+0.a1/4b3/4-b
a- A3/4b1/4
A3/4b1/4+0.a1/2b1/2
A3/4b1/4- a1/2b1/2
A1/2b1/2+0.a1/4b1/4
A1/2b1/2 – a1/4b1/4
A1/4b3/4- b
A1/4b3/4- b
0
7. Divide: a3 (b2- c2)+b3 +b3(c2-a2)+c3(a2-b2) by ab+bc+ca
A3(b2 – c2)+b3(c2 –a2)+c3(a2-b2)
=a3b2-a3c2+b3c2-b3a2+c3a2-c3b2
A2b-ab2+b2c – bc2+ac2 – a2c
Ab+bc+ca)a3b2-a2b3+b3c2-b2c3+c3a2-a3c3
A3b2 + a2b2c+a3bc
-a2b3-a2b2c – a3bc+b3c2
-a2b3-ab3c – a2b2c
-a3bc+ab3c+b3c2-a3c2
-a3bc-a2c2b-a3c2
-ab3c+ac2b+b3c2+a2c3
-ab3c+b3c2+ab2c2
-a2c2b-ab2c2+a2c3-b2c3
-a2c2b+abc3+a2c3
-ab3c2-abc3-b2c3
-ab2c2+b2c3-abc3
0
Q =a2b-ab2+b2c-bc2+ac2-a2c
8. What is to be added to x6+4x3+7 in order to be completely divisible by
x2+2x+2?
X2+2x+2)x6+0x5+0x4+4x3+0x2+0x+7
X6+2x5+2x4
-2x5-2x4-4x3
-2x5-4x4-4x3
2x4+8x3+0x2
2x4+4x3+4x2
4x3-4x2+0x
4x3+8x2+8x
-12x2-8x+7
-12x2-8x-24
16x+31
6x-31 should be added.
9. How much one should add to x8+a8 so that the resulting expression is
completely divisible by x2+a2?
X6-x4a2+x2a4-a6
X2+a2)x8+0.x6a2+0.x4a4+0.x2a6+a8
X8 +x6a2
-x6a2+0.x4a4
-x6a2-x4a4
X4a4+0.x2a6
X4a4+0.x2a6
-x2a6 +a8
-x2a6-a8
2a8
2a8 should be added to x8+a8 to be completely divisible by x2+a2.
10. A triangle has a area x3 +7x2+14x+6 and one of its sides is x2+4x+2. Find the
corresponding altitude.
Area of triangle A= 1 bh
2
3
2
H= 2a = (x +7x +14x+6)
B
x 2+4x+2
X +3
X +4x+2)x +7x2+14x+6
X3 +4x2+2x
3x2+12x+6
3x2+12x+6
0
2
3
Altitude = 2(x+3).
NON-TEXTUAL QUESTIONS
1. Find the quotient in the following:
(i)
17
= a17-7 =a10
A
7
A
(ii)
4
Y = y4-4 = y0 =1
4
Y
3 2
(iii)
X y z =x3-1 y2-1z= x2yz
xy
(iv)
Xb
Xc
a
- xc
xa
b
- xa
xb
= (xb-c)a (xc-a)b (xa-b)c
= xa (b-c) xb(c-a) xc(a -b)
= xa b-a c-ab+ca-bc
= x0
=1
(v)
A4/3
A-2/3
2. Divide
4 + 2 = a2
a3 3
c
(i)
2x2-x-15 by (x-3)
2x + 15
x-3)2x2- x -15
2x2- 6x
5x-15
5x-15
0
(ii)
X3+8 by x+2
X2- 2x+4
X+2)x3+0.x2+0.x+8
X3+2x2
-2x2+0.x
-2x2- 4x
4x+8
4x+8
0
(iii)
X6-y6 by x-y
X5+x4y+x3y2+x2y3+xy4+y5
x-y)x6+0.x5y+0.x4y2+0.x3y3+0.x2y4+0.xy5-y6
x 6 – x5y
x5y+0.x4y2
x5y-x4y2
x4y2+0.x3y3
x4y2- x3y3
x3y3+0.x2y4
x3y3- x2y4
x2y4+0.xy5
x 2y4- xy5
xy5- y6
xy5- y6
0
3. What should be added to x4-1 to be completely divisible by
x2+x-1?
X2-x+1
X2+x-1)x4+0.x3 +0.x2+0.x-1
X4 + x3 – x2
-x3+ x2+ 0x
-x3- x2+ x
(+) (-)
2x2- x -1
2x2+2x-2
-3x+1
-3x+1 should be added to x4-1 to be completely divisible by
x2+x-1.
4.
divide: x6+3x5+2x4+x2+2x+1 by x4+x3+4x2+1
X2+2x-4
X4+x3 +4x2+1)x6+3x5+2x4+0.x3+x2+2x+1
X6+x 5+4x4
+x2
2x5-2x4+0.x3+2x+1
2x5+2x4-8x3+2x
-4x4+ 8x3
+1
4
3
2
-4x - 4x -16x -4
-4x3+16x2+5