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5. Division Algorithm for Polynomial
For any two polynomials p(x) and g(x); g(x) 0
We can find two polynomials q(x) and r(x) such that p(x) =g(x) x q(x) + r(x)
Where r(x) = 0 or degree of r(x) < degree of g(x)
Chapter 2
1 Mark Questions
1. Verify that one and two are the zeroes of the polynomial p(x) = (x-1) (x
2)
2. Show that 3 is a zero of the polynomial x3 - 8x2 + 8x + 21
3. Verify whether
3
1=x
is a zero of the polynomial p(x) = 3x +1
4. Find the zeroes of the polynomial x2-3
5. Find the zeroes of the quadratic polynomial 4u2 + 8u
2 Mark Questions
Example : Find a quadratic polynomial in which sum of zero = -1 and product of zeros
is4
1
Let the quadratic polynomial be ax2+bx+c with zeroes and
According to given 111 ==
=+a
b
a
b
4
1
4
1==
a
c
Take a = L.C.M. of (1,4) = 4
b=4 and c=
14
4=
so required polynomial is 4x2 + 4x +1
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6. Find a quadratic polynomial each with the given numbers as the sum and
product of its zeroes
503
12 ,)ii(,)i(
7. Prove that x2+6x+15 has no zeroes.
8. Find the zeroes of 3x2-x-4 and verify the relationship between the zeroes and
the co-efficient.
9. Divide 2x2 + 3x+1 by (x+2), find the quotient and the remainder.
10. Check whether the first polynomial is a factor of the second polynomial by division
method.
i) p(x) = x4 -5x + 6; g(x) = 2-x2
ii) p(x) = 42 - 19x -5x2 +2x3; g(x) = (x-2)
3 Marks Questions
Example : On dividing x3 -3x2 + x + 2 by polynomial g(x), the quotient and the remainder
were (x-2) and -2x+4 respectively. Find g(x)
Here p(x) = x3 - 3x2 +x + 2, q (x) = x-2, r(x) = -2x +4 By using division alogorithm.
g(x) x (x-2) + (-2x+4) = x3 - 3x2 + x + 2
g(x) x(x-2) = x3 - 3x2 + x + 2 + 2x - 4
g(x) x (x-2) = x3 - 3x2 + 3x-2
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Dividing x3-3x2+3x-2 by (x-2) we get g(x) = x2-x+1
11. Find all the zeroes of 2x4-3x3 - 3x2 + 6x -2 if two of its zeroes are 22 and
12. Find all the zeroes of the polynomial x3+6-7x if one of its zero is -3.
13. Divide x3-3x2-x+3 by (x+1) and verify the division algorithm.
14. If 5 is a zero of the polynomial x3-6x2+3x+10, find the other two zeroes.
Answers :
Q.4 33 , Q. 5 0, - 2; Q.6 i) 1233 2 + xx ii) 52 +x
Q.8 13
4, Q.9 Quotient 2x-1, Remainder 3
Q.10 (i) No (ii) Yes Q.11 and 1 Q.12 1,2 Q. 14 -1,2,5
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CHAPTER - 2
POLYNOMIAL
2 Marks Questions
Q.1 P(x) = (x-1) (x-2)
One or two are the Zeroes of the polynomial
P(1) = (1-1) (1-2)
= 2121 /+///
= 0
P(2) = (2-1) (2-2)
=2244 /+///
= 0
Q.2 Let x3
- 8x2
+ 8x + 21 = P(n)
Put x=3 in this Equation
P(3)=(3)3 - 8(3)2 + 8 (3) + 21
= 27 - 8 x 9 + 24 + 21
= 27 - 72 + 24 + 21
= 72 - 72
= 0
3 is zero of the polynomial x3 - 8x2 + 8x + 21
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Q.3 P(x) = 3x + 1
Put
3
1=x
in this Equation
13
13
3
1+
=
P
= -1 + 1
= 0
Yes x=3
1 is zero of P(x)
Hence Proved
Q.4 x2 - 3 = 0
x2 = 3
3=x
(Zeroes of the Polynomial)
Q.5 4u2 + 8u = 0
4u ( u+2) = 0
4u= 0 or u + 2 = 0
u = 0 or u = -2
(Zeroes of the quadratic Polynomial)
Q.6 (i)
3
1
2,
Let the quadratic polynomial be ax2 + bx + c with zeroes and
x2 - (sum of roots) x + Product of roots = 0
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22
==+a
b
3
1
3
1==
ac
If a = 3,
3
32x
a
b =
a=3, 23=b and c=1
So required polynomial is 1233 2 +x
ii) 0, 5
Let the quadratic polynomial be ax2+bx+c with zero and
).........(b
11
0
10
=
=+
).........(a
c2
1
55 ==+
By taking equation (1) and (2)
a=1, b=-0, c= 5
Sorequired polynomial is 52 +x
Q.8 3x2 - x - 4
3x2 + 3x - 4x - 4
3x (x+1) - 4 (x+1)
(3x -4) (x+1)
So, the value of 3x2 - x - 4 is zero when x+1 = 0 or 3x - 4 = 0 i.e. when x = -1 or
x3
4=
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therefore, the zeros of 3x2 - x - 4 are -1 and3
4
Sum of zeros =3
1
3
43
3
41 =+=++
23
1
3
1
xofCofficient
)xofCofficient(
a
b==
=
Product of Zeroes =a
cxx =
=
=
3
4
3
41
23
4
xofCofficient
termttanCons=
Q.9 Divide 2x2+3x+1 by (x+2)
1213222 +++ x(xxx
2x2
4x--------------------
- x + 1
x
2------------------
3------------------
quotient = 2x - 1
Remainder = 3
Q.10 P(x)= 42 - 19x - 5x2 + 2x3, g(x) = x-2
(xxxx 4219522 23
2x2-x-21
2x3
4x2
-------------------x2 - 19x
x2
2x---------------------
- 21x + 42
21x
42----------------
0---------------- Yex, g(x) is a factor of P(x)
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Q.11 Since two zeroes are
( ) 22222 2=+ xx)x(,and
is a factor of given
polynomial
13226332222342 ++ xx(xxxxx
2x4
4x2
---------------------------------3x3 + x2 + 6x - 2
3x3+
6x-----------------------------
x2 - 2
x2
2-----------------
0-----------------
So, 2x2 - 3x +1
by splitting the middle term
11222 + xxx
)x()x(x 1112
)X)(x( 112
so, its zeroes are given by
12
1== xandx
Therefore, the Zeroes of the given polynomial are 12
122 and,,
Q.12 x3 + 6 - 7x
if one of its Zero is -3
3
2
1 xofCofficient)xofCofficient(o
ab ===++
31
6
xofCofficient
termttanCons
a
dxx =
=
=
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0=++
03 =++ )..(.......... 13=+ 6= xx
63 +=+ )(
3
6
/
/=
2=
)(.......... 22
=
Put the value of in equation (1)
=+=
+ 3232 2
=+ 322
termmiddletheSplitingBy,0232 =+
022
2
=++( ) ( ) 0212 =
( ) ( )21
=0
So, the value of
232 +
is zero when 01=
or
2
i.e. when
21 == or
Put the value of in Eq. (2)
21
2==
Zeroes of the polynomial are 3= ,
2
1
=
=
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Q.13 Divide
x3 - 3x2 - x + 3 by (x+1)
34331223 +++ xx(xxxx
23 xx
-----------------------
xx
xx
44
34
2
2
+
---------------------
33
33
+
x
xquotiend = x2 -4x+3
----------------- Remainder = 0
0
-----------------
By division algorithm,
Dividend = Divisior x quotient + Remainder
(x3-3x2-x+3) = (x+1) (x2-4x+3) + 0
(x3-3x2-x+3) = x3-4x2+3x+x2-4x+3+0
(x3-3x2-x+3) = x3-4x2+x2+3x-4x+3+0
(x3-3x2-x+3) = x3-3x2-x+3
Hence, the division algorithm is verified
Q.14 x3-6x2+3x+10
If 5 is a zero of the polynomial
3
2
1
6
1
6
xofCofficient
)xofCofficient()(
a
b ==
=
=++
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31
10
xofCofficient
termCofficient
a
dxx =
=
=
65
6
=++
=++
)..(..........
xx
)(..........
22
2
5
10
105
10
11
56
=
=
=
=
=
=+
=+Put the value of
in Eq. (1)
12
=
+
1
22
=
022 =
022 = , by spliting the middle term
02122 =+
0212 =+ )()(
)()( 22 + = 0
So, the value of
22
is Zero when 02 = i.e. when
21 == or
Put the value of 2.Eqin
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21
2=
/
/=
Zero of Polynomial are 5= ,
B = -1
2=
Chapter 3
One Mark Questions
Q.1 If x=-1, y=5 is a solution of the equation 4x-3y = 11
Q.2 Write the condition for the following system of linear equations to have a unique
solution.
ax+by = c; px+qy=r
Q.3 Write T or F
(i) -3x+4y = 5;
02
156
2
9=+ y
x
having no solution.
Q.4 Fill up the blank spaces
(i) When l1and l
2are parallel lines then the system has ......................
(ii) When l1and l
2are coincident then the system has .........................
Q.5 (i) Write the standard form of linear equation in two variables.
Ans. : (1) No (2)q
b
p
a (3) True (4) (i) no solution 4(ii) infinite number of solutions
(5) ax+by =c
SOLVED EXAMPLES
Ex. 1 Find the value of K for which the system of equations 3x+5y=0, kx+10y=0 has
a non-zero solution.
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Solution Here,10
53
2
1
2
1 ==b
b,
ka
a
for a non-zero Solution, we have
30510
53
2
1
2
1 kor,k
.e.i,b
b
a
a
6k
Ex.2 Solve the following pair of linear equation :
3x-y=3, 9x-3y=9
Solution 3x-y=3 .........(1)
9x-3y= 9 .......(2)
from (1)
y = 3x -3 ........ (3)
put (3) in (2)
9x - 3(3x-3) = 9
9x - 9x + 9 = 9
9 = 9, which is true hence, the given system of equations has
infinitely many solutions.
Two marks question
Q.1 On comparing the ratios2
1
2
1
b
b,
a
aand
2
1
c
c, find out whether the following pair of
linear equations intersect at a point or parallel or coincident.
2x+3y = 4
3x+5y = 5
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Q.2 Solve the following system of linear equations by substitution method.
5x+2y = 2 and 2x+3y = -8
Q.3 Solve the following pair of linear equations by elimination method.
5x + 10 y = 28
15x=20y -121
Q.4 Check whether the following system of equations has unique solution, no
solution or infinitely many soultion.
4x-3y=1
x - 2y = 4
Ans. (1) consistent at a point
(2) x = 2, y = 4
(3)10
41
5
13 == yx
(4) unique solution
Three Marks questions
Q.1 Draw the graph of the equations 4x - y = 4 and 4x + y=12
Determine the vertices of the triangle formed by the lines, representing these
equations, and the x-axis. Shade the triangular region so formed.
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Q.2 For what value of k, will the system of equations :
x+2y = 5
and, 3x+ky -15=0 has (i) a unique solution (ii) no solution.
Q.3 Find two numbers whose sum is 28 and seven times their difference is equal to
four times of their sum.
Q.4 Solve : 5x + 2y = 2 and 2x+3y = -8 and hence find the value of m for which
y=mx+4
Ans. (1) (1,0), (2,4) and (3,0)
(2) (i) 6k (ii) no value of k
(3) 22 and 6(4) x=2, y = -4, m= -4
Six Marks Questions
Q.1 Solve the following pair of equations by reducing them to a pair of linear
equations :
6
13
2
2
3
12
3
1
2
1=+=+
yx;
yx
Q.2 A bag contains 94 coins of 50 paise and 25 paise denominations. If the total
worth of these coins be Rs. 29.75. Find the number of coins in each kind.
Q.3 A boat goes 24km. upstream and 28km. downstream in 6hrs. It goes 30 km
upstream and 21km. downstream in 6 hrs. Find the speed of the boat in still
water and also speed of stream.
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Q.4 Solve the following systems of equations :
62 ==+xy
yx,xy
yx
Ans. (1)3
1
2
1== y,x
(2) No. of 50 paise coins = 25
No. of 25 paise coins = 69
(3) Speed of stream = 4km/hr
Speed of boat = 10 km/hr
(4)4
1
2
1== y;x
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Chapter 3
Linear Equations in two variables
Q.1 4x - 3y = 11
x = -1, y = 5
4(-1) - 3(5) = 11
-4 - 15 = 11
-19 11
x = -1, y = 5 is not a soltuion of the equation
Q.2 ax + by = c
a1
= a, b1
= b, c1
= c
px + qy = n
a2
= p b2=q c
2=n
r
c
q
b
p
a
c
c
b
b
a
a
2
1
2
1
2
1
Q.3 -3x + 4y = 5;2
15
2
9+byx
having no solution = True
Q.4 i) When l1 and l2 are parallel lines the system has no solution.
ii) When l1and l
2are coincident then the system has infinte many solution.
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Q.5 a1x + b
1y +c
1=0
a2x + b2y+c2=0
Q.1 2x + 3y = 4
a1
= 2 b1
= 3 c1=4
3x + 5y = 5
a2
= 3 b2
= 5 c2=5
5
4
5
3
3
2
2
1
2
1
2
1
=== c
c
b
b
a
a
2
1
2
1
2
1
c
c
b
b
a
a
linear equations intersect at a point.
Q.2 By substitution method
5x + 2y = 2
2x + 3y = -8
5x + 2y = 2
5x = 2 - 2y
5
22 yx
=
2x + 3 y = -8
835
22
2 =+
y
y
835
44=+
y
y
5
401544 =+ yy
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11y = -40 - 4
11y = -44
411
44=
=y
5
22 yx
=
5
422 )(x
=
5
82+=x
25
10==x
Q.3 By elimination method
5x + 10 y = 28 (1), 15 x = 20y -121
15x - 20y = -121 ........... (2)
15 x 5x + 10y = 28
5 x 15x -20y = -121
--------------------------------------
75x + 150y = 420
75x
100y =
605
--------------------------------------
250y = 1025
y =
10
41
250
1025=
75x + 150y = 420
75x + 150
10
41=420
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75x + 615 = 420
75x = 420 - 615
75x = -195
75
195=x
5
13=x
Q.4 4x - 3y = 1
a1 = 4 b1 = -3 c1 = 1
x -2y = 4
a2
= 1 b2
= -2 c2=4
4
1
2
3
1
4
2
1
2
1
2
1 =
==
c
c
b
b
a
a
2
1
2
1
2
1
c
c
b
b
a
a
Unique Solution
Q.2 1) x + 2y = 5
3x + ky = -15
a1
= 1 b1
= 2 c1=5
a2
=3 b2=k c
2=-15
For unique solution
2
1
2
1
2
1
c
c
b
b
a
a
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15
52
3
1
k
k
2
3
1
6k
For no solution
x + 2y = 5
a1
= 1 b1= 2 c
1=5
3x + ky = -15
a2
= 3 b2
= k c2= -15
2
1
2
1
2
1
c
c
b
b
a
a=
15
52
3
1
=
k
k
2
3
1=
6=k
15
52
k
305 k
65
305 =
k
6k
Q.4 5x + 2y = 2
2x + 3y = -8
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By Substitution method
5x = 2 - 2y
5
22 yx
=
835
222 =+
y
y
835
44=+
y
y
4 - 4y + 15y = -40
11y = -40-4
11y = -44
411
44=
=y
5
22 yx
=
5
422 )(x
=
5
82 +=x
25
10==x
y = mx + 4
-4 = m(2) + 4
-4 -4 = 2m
-8 = 2m
42
8==
=m
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Q.1 23
1
2
1=+
yx
Let by
,ax
==11
232=+
ba
3a + 2b = 12 ............ (1)
by
ax
Let ==11
6
13
2
2
3=+
ba
2a + 3b = 13 .............(2)
Multiply 1 by 2 and 2 by 3.
2 x (3a + 2b = 12)
3 x (2a + 3b = 13)
-------------------------
6a + 4b = 24
6a
9b =
39-------------------------
-5b = - 15
35
15==b
6a + 4 (3) = 246a = 24 - 12
26
12==a
111==
ya
x
1
31
1
21==
yx
2x = 1 3y = 1
3
1
2
1== yx
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Q.4 2=+
xy
yx
x + y = 2xy.............(1)
6=
xy
yx
x-y = 6xy ...............(2)
But Elimination Method :
x + y = 2xy
x - y = 6xy
----------------2x = 8xy
2 = 8y
4
1
8
2==y
4
12
4
1==+ xxx
xx2
1
4
1=+
4
1
2
1 = xx
4
1
2
=
x
24 = x
2
1= x
Q.3 Let Ist no. = n
Another no = 28-n
7 [x-(28-x)] = 4(28)
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14x - 196 = 112
14x = 196 + 112
14x = 308
14
308=x
x = 22
1st no. = x = 22
Another no = 28 - x = 28 - 22 = 6
Q.1 4x - y = 4 4x + y = 12
4
4 yx
+=
4
12 yx
=
x 1 2 3 x 3 2 1
y 0 4 8 y 0 4 8
Ans. vertices = (1,0), (2,4) & (3,0)
8 (1,8) (3,8)
6
4 (2,4)
2
(1,0) (3,0)
0 2 4 6 8 x axis
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Q.2 Let 50 paise coins = x
Let 25 paise coins = y
Acc to Ques. x+y = 94 ................ (1)
Total 50 paise coins =2100
50 x.Rs
x=
Total 25 paise coins =
4100
25 y.Rs
y=
Acc. to Ques.
752942
.y
.x
=+
2x + y = 119 .......................... (2)
By Elimination method
2 x ( x + y = 94)
1 x ( 2x + y = 119)
-----------------------------------------
2x + 2y = 118
2xy = 119
-----------------------------------------y = 69
x + y = 94
x + 69 = 94
x = 94 - 69
x = 25
No. of 50 paise coins = x = 25
No. of 25 paise coins = y = 69
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Q.3 Let speed of boat = x km/hr
Let speed of stream = y km/hr
Speed of boat down stream = x+y
Speed of boat upstream = x - y
Case I
Time of boat in downstream =
yx +
28
Time of boat in upstream =yx
24
Acc. to Question 62428
=
++ yxyx
Let byx
ayx
=
=+
11
28a + 24b = 6 ............................(1)
Case II Time of boat in downstream yx +=
21
Time of boat in upstream =yx
=30
Acc. to Qus.2
16
3021=
+
+ yxyx
Let byx
ayx
=
=+
11
21a + 30b = 2
13
42a + 60b = 13 .................. (2)
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By Elimination method
60 x (28a + 24b = 6)24 x (42a + 60b = 13)
------------------------------------
1680a + 144b = 360
1008a
144b =
312
------------------------------------672a = 48
42
3
672
48==a
42
3=a
28a + 24b = 6
62442
328 =+
b
62421
42=+ b
21
42624 =b
21
4212624
=b
24
1
21
84xb=
21
7=b
byx
ayx
=
=+
11
21
71
42
31=
=
+ yxyx
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3x + 3y = 42...(3) 7x - 7y = 21 ......(4)
By Elimination method
7 x (3x + 3y = 42)
3 x (7x - 7y = 21)
------------------------------------
21x + 21y = 294
21x
21y =
126
------------------------------------42y = 168
442
168==y
3x + 3y = 42
3x + 3(4) = 42
3x+12 = 42
3x = 42-12
3
30=x
x=10
Speed of boat = x = 10km/h
Speed of Stream = y = 4km/h
0