POLYNOMIALS
A polynomial in x is an expression of the form
p(x) = a0 + a1x + a2x2 + ………………. + anxn
Where a0, a1, a2…………. an are real numbers and n is a non-negative integer and an  0.
A polynomial having only one term is called monomial.
A polynomial having two terms is called binominal.
A polynomial having three terms is called trinomial.
Degree of polynomial: The highest power of x in a polynomial is called degree of a polynomial.
A polynomial is said to be linear if it is of degree 1.
A polynomial is said to be quadratic if it is of degree 2.
A polynomial is said to be cubic if it is of degree 3.
A polynomial is said to be biquadratic if it is of degree 4.
Sum and difference of two polynomial
Let p(x) and q(x) be two polynomials
p(x) = x4 + 3x3 – 2x2 + 6x + 9
q(x) = 2x3 + 3x2 + 6x + 3
The sum of two polynomial gives a polynomial
Here p(x) + q(x)=x4 + 5 x3 + x2 + 12x + 12
Similarly we can find difference of two polynomials
p(x) – q(x)= x4 + x3 – 5x2 + 6
Multiplication of two polynomial
p(x)q(x)= (x4 + 3x3 – 2x2 + 6x + 9)( 2x3 + 3x2 + 6x + 3)
Division of a polynomial by a polynomial- if we divide p(x) by g(x) and we get quotient q(x) and remainder
r(x) then we can write
p(x) = g(x) q(x) + r(x) degree r(x) < degree g(x)
Illustration 1: Divide p(x) by g(x), where p(x) = x4 + 1 and g(x) = x + 1
Solution:
x3 – x2 + x – 1
x+1
x4 + 1
x4 + x3
–x3 + 1
–x3 – x2
x2 + 1
x2 + x
–x + 1
–x – 1
2
Here, the quotient q(x) = x3 – x2 + x – 1 and the remainder r(x) = 2
We write x4 + 1 = (x3 – x2 + x – 1) (x + 1) + 2.
Note: Notice that the degree of g(x) is less than the degree of p(x). Therefore it is always
possible to divide a polynomial of higher degree by a polynomial of lower degree (in the same
variable). The process above stops as soon as the remainder is zero or the degree of the
remainder becomes smaller than that of the divisor.
Illustration 2: Let p(x) = x4 + 2x3 – 3x2 + x – 1. Find the remainder when p(x) is divided by x – 2.
Solution:
x3 + 4x2 + 5x + 11
x–2
x4 + 2x3 – 3x2 + x – 1
x4 – 2x3
4x3 – 3x2 + x – 1
4x3 – 8x2
5x2 + x – 1
5x2 –10x
11x – 1
11x – 22
21
Remainder is 21.
In the above example let us evaluate p(2). We get
p(2) = 24 + 2(23) – 3x2 + 2 – 1
= 16 + 16 – 12 + 2 – 1 = 21.
Thus we find that p(2) is equal to the remainder when p(x), is divided by x – 2.
Remainder Theorem: Let p(x) be any polynomial of degree  1, and a any real number. If p(x) is divided by
x – a. then the remainder is p(a).
Proof: Let us suppose that, when p(x) is divided by x – a, the quotient is q(x) and remainder is r(x). So we
have
p(x) = (x – a) q(x) + r(x), where r(x) = 0 or degree r(x) < degree (x – a)
Since degree of (x – a) is 1, either r(x) = 0 or degree of r(x) = 0 (<1).
So r(x) is a constant, say r.
Thus for all values of x, p(x) = (x – a) q (x) + r where r is a constant. In particular, for x = a,
p(a) = 0. q(a) + r = 0 + r = r. This proves the theorem.
Illustration 3: What would be the remainder when (4x3 + 7x2 – 5x + 3) is divided by (x + 2).
Solution:
Let f(x) = 4x3 + 7x2 – 5x + 3.
We may write, (x + 2) = [x – (–2)]
Thus, when f(x) is divided by [x – (–2)], then by remainder theorem, remainder = f(–2).
Now, f(–2)= [4x (–2)3 + 7 x (–2)2 – 5x(–2) + 3]
= [4x (–8) + 7 x 4 + 10 + 3] = (–32 + 28 + 10 + 3) = 9.
Required remainder = 9
Illustration 4: Using remainder theorem, find the remainder when (4x3 – 12x2 + 15x – 3) is divided by (2x – 1).
Solution:
Let f(x) = 4x3 – 12x2 + 15x – 3.
1

We may write, (2x – 1) = 2  x   .
2


1
1 
 1  1
Now f      4 x 12 x  15 x  3 
4
2 
 2  8
 Required remainder = 2.
15
1

  3  3 2
2
2

Synthetic Division
The method of synthetic division can be illustrated from the following example:
Find the quotient and remainder when 2x4 + 5x3 + 3x2 + 7x – 3 is divided by (x – 2)
x – 2)2x4 + 5x3 –3x2 + 7x – 3(2x3 + 9x2 + 15x + 37
2x4 – 4x3
9x3 – 3x2
9x3 – 18x2
15x2 + 7x
15x2 – 30x
37x – 3
37x – 74
71
Quotient is 2x3 + 9x2 + 15x + 37. Remainder is 71.
The divided is 2x4 + 5x3 – 3x2 + 7x – 3
Divisor is x – 2. The quotient will be of third degree.
2 = 1st coefficient in quotient = 2 = 1st coefficient in divided.
9 = 2nd coefficient in quotient = 2(2) + 5 = 2 (1st coefficient in quotient) + 2nd coefficient in dividend.
15 = 3rd coefficient in quotient = 2(9) + (–3) = 2(2nd coefficient in quotient) + 3rd term in dividend
37 = 4th coefficient in quotient = 2(15) + 7 = 2(3rd coefficient in quotient) + 4th term in dividend.
71 = Remainder = 2(37) + (–3) = 2(4th term in quotient) + 5th term in dividend.
The above process can be illustrated as follows:
Write down the coefficient only of the dividend.
2 5
3
7
3
 2(2)  2(9)
 2(15)  74
2
2  9  15
 37
 71
3
2
Hence quotient is 2x + 9x + 15x + 37. Remainder is 71.
Synthetic division to find quotient and remainder is also called the method of find quotient and remainder by
the method of detached coefficients.
Illustration 5: Divide 2x4 + 5x3 – 3x2 + 7x – 3 by (x + 2) the method of detached coefficients.
2
5
3
7
3
 2( 2)  1( 2)  ( 5)( 2)  17( 2)
2
1
2
Solution:
5
3
 17
 37
2
Quotient is 2x + x – 5x + 17. Remainder is –37.
If p(x) is a polynomial of degree n > 0, then it follows the remainder theorem that p(x) = (x – a). q(x) + p(a),
where q(x) is a polynomial of degree n – 1. If p(a) = 0, then p(x) = (x – a), q(x) + p(a), where q(x) is a
polynomial of degree n – 1. If p(a) = 0, then p(x) = (x – a). q(x). we say that (x – a) is factor of p(x)
Factor Theorem : Let p(x) be polynomial of degree n > 0. If p(a) = 0 for a real number a, then (x – a) is a
factor of p(x). Conversely, if (x – a) is a factor of p(x), the remainder p (a) must be zero.
Illustration 6: Show that (x + 3) as well as (3x + 2) is a factor of the polynomial (3x3 + 8x2 – 5x – 6).
Solution :
Let f(x) = 3x3 + 8x2 – 5x – 6. By factor theorem, (x + 3) will be a factor of f(x), if f(–3) = 0.
Now, f(–3) = 3(–3)3 + 8 (–3)2 – 5(–3) – 6 = [3 x (–27) + 8 x 9 + 15 – 6] = (–81 + 72 + 15 – 6) = 0.
 (x + 3) is a factor of f(x)

2 

 2 
Again, (3x + 2) = 3  x    3  x      .
3 

 3 

 2
 By factor theorem, (3x + 2) will be factor of f(x), if f     0.
 3
3
2
 2   2
 2
 2 
Now f     3 x     8 x     5 x     6 
 3    3 
 3
 3  
  8 
  8 32 10
 4  10

= 3 x    8 x     6   
   6   0.
27
9
3
9
9
3
 

  
 
Illustration 7: Given that (x – 1) and (x + 2) are factors of the polynomial (x3 + ax2 + bx – 8). Find the values of a
and b. With these values of a and b. Factorize the given polynomial.
Solution :
Let f(x) = (x3 + ax2 + bx – 8). Then
(x – 1) is a factor of f(x).
 f(1) = 0
[ By factor theorem ]
 (13 + a x 12 + b x 1 – 8) = 0
[ f(1) = (13 + a x 12 + b x 1 – 8)]
1+a+b–8=0
a+b=7
…..(i)
Again, (x + 2) is a factor of f(x)
 f(–2) = 0
[ By factor theorem ]
 (–2)3 + a x (–2)2 + b x (–2) – 8 = 0
[ f(–2) = (–2)3 + a x (–2)2 + b x (–2) – 8 ]
 –8 + 4a – 2b – 8 = 0
 4a – 2b = 16
 2a – b = 8……..(ii)
Adding (i) and (ii), we get 3a = 15 or a = 5.
Substituting a = 5 in (i), we get 5 + b = 7, i.e., b = 2
With these values of a and b, we have f(x) = x3 + 3x2 + 2x – 8.
Now, (x – 1) and (x + 2) are factors of f(x)  (x2 + x – 2) is a factor of f(x)
x+4
_________________________
2
x + x – 2 ) x3 + 5x2 + 2x – 8 (
x3 + x2 – 2x
______________
4x2 + 4x – 8
4x2 + 4x – 8
______________
0
_______________
 On dividing f(x) = x3 + 5x2 + 2x – 8 by (x2 + x – 2), we get
Quotient = (x + 4)
 f (x) = x3 + 5x2 + 2x – 8
= (x2 + x – 2) (x + 4)
= (x –1) (x + 2) (x + 4)
Hence, (x3 + 5x2 + 2x – 8) = (x –1) (x + 2) (x + 4).
Illustration 8: A Polynomial f(x) with rational coefficients leaves remainder 15 when divided by (x – 3) and
remainder 3 when divided by (x – 1). Find the remainder when the polynomial is divided by (x – 1)
and (x – 3)
Solution:
f(x) = ax2 + bx + c
f(x) = Q(x) (x – 3) + 15
f(x) = Q(x) (x – 1) + 3.
f(x) = Q(x) (x – 3)(x – 1) + (ax + b)
f(3) = 0 + 3a + b = 15
f(1) = 0 + a + b = 3
2a = 12
a=6
b = 3 – 6 = –3
 Remainder is 6x – 3.
Exercise 1: If f(x) is a polynomial in x, 4 and 10 are the remainders when f(x) is divided by (x – 1) and (x – 2).
Then find the remainder when f(x) is divided by (x – 1)(x – 2).
A.S.Rao, 2006
SQUARE ROOT:
If y 
x or y2 = x, then y is called the square root of x.
As per the above definition 4 is the square root of 16 since 42 = 16. But also (–4)2 = 16  16 =4.
The positive square root of 16 is called the principal square root of 16. In general
root we mean only the principal square root.
Hence
x 2 =  x. By square
x 2 = x when x > 0
x 2 = –x when x < 0
x 2 = 0 when x = 0. In short x 2 = |x|
In what follows we consider only the non–negative square roots. Modulus sign is to be put whenever it is
needed. For instance a2  2ab  b2  (a  b)2  (a  b)  (a  b) if a  b) .
Illustration 9: Find the square root of a8b12c6
1
Solution:
a8b12c 6  (a8b12c 6 ) 2  a4b6c 3
25a 4
4b 2
Illustration 10: Find the square root of
Solution:
25a 4
=
4 b2
Illustration11: Find
5
 
2
2
 a2

 b

2

5 a2
 =
2 b

121a5
9 ab 6
2
2
 11   a2 
11a 2
Solution:
=
=



3 
3b 3
9 ab6
 3   b 
Square root by inspection:
a2 + 2ab + b2 = (a + b)2
a2 – 2ab + b2 = (a – b)2
121a5
 a2  2ab  b2 = (a + b)
a2  2ab  b2 = (a – b)
If the given expression is reduced to one of the above forms, the square root can be written down by
inspection.
Extraction of square root by division method:
Principle behind finding the square root is given below.
Let us find the square root of 25x2 – 30x + 9
Let 25x2 – 30x = (5x + a)2
25x2 – 30x + 9 = 25x2 + 10ax + a2
 –30x + 9 = 10ax + a2
Equating coefficient of x on both sides
 10a = –30 or a = –3
We can show this briefly in the division method
5x – 3
5x
25x2 – 30x + 9
25x2
5(x)2
10x – 3
–30x + 9
(10x – 3) (–3)
–30x + 9
0
Procedure:
(1) The square root of 25x2 is 5x. This is the first term of the square root. Multiply (5x) by (5x) which gives
25x2. Substract 25x2 from the dividend. We get –30x + 9 as the next two terms.
30 x
(2) Multiply 5x by 2 which gives 10x.
gives –3. This becomes the second term of the square root.
10 x
Adding –3 to 10x, we get 10x – 3. Multiplying this by –3. This becomes the second term of the square root.
Additing –3. This becomes the second term of the square root. Adding –3 to 10x, we get 10x – 3. Multiplying
this by –3 we get –30x + 9. The remainder becomes zero.
Square root is (5x – 3)
Illustration 12:
Solution:
Find the square root of 4x4 + 12x3y + 13x2y2 + 6xy3 + y4
2x2 + 3xy + y2
4x +12x3 y + 13 x2 y2 + 6xy3 + y4
4x4
12x3 y + 13 x2 y2
4
2x2
4x2 + 3xy
12x3 y + 9x2 y2
4x2 + 6xy + y2
4x2 y2 + 6xy3 + y4
4x2 y2 + 6xy3 + y4
(2x2)2 = 4x4
(4x2 + 3x) 3xy


12x3 y
 3xy 
 since
2
4x


(4x2 + 6xy + y2) y2


4x 2 y 2
 y2 
 since
2
4x


0
Hence the square root is 2x2 + 3xy + y2
Square root by the method of Indeterminate coefficients
Illustration 13: Find the square root of x4 + 6x3 + 17x2 + 24x + 16
Solution:
The expression is of the fourth degree in x. Hence the square root will be an expression of
second degree in x.
x4 + 6x3 + 17x2 + 24x + 16 = (x2 + ax + b)2
Equating coefficients of the like powers of x on both sides, we get
2a = 6  a = 3
a2 + 2b = 17  2b = 17 – a2 = 17 – 9 = 8  2b = 8 or b = 4.
Hence the required square root is x2 + 3x + 4.
Illustration 14:
Solution:
If ax2 + bx + c is a perfect square, then prove that b2 = 4ac.
Illustration 15:
Solution:
Show that (x + 1) (x + 4) (x + 3)(x + 2) + 1 is a perfect square
Let ax2 + bx + c = (px + q)2
ax2 + bx + c = (px + q)2
Equating coefficients of like powers x
p2 = a
2pq = b 4p2q2 = b2
and q2 = c 4ac = b2
Hence b2 = 4ac
The given expression = (x + 1) (x + 4) (x + 2) (x + 3) + 1
= (x2 + 5x + 4) (x2 + 5x + 6) + 1
2
Put y = x + 5x
The expression reduces to (y + 4)(y + 6) + 1 = y2 + 10y + 25 = (y + 5)2 = (x2 + 5x + 5)2
Which is a perfect square.
Exercise 2 : Find the square root of 1 – 4x + 10x2 – 20x3 + 25x4 – 24x5 + 16x6
Exercise 3 : Find the square root of x 2 +
Exercise 4 : Find the square root of
x2
y
2
+
Exercise 5 : Find the square root of x 4 +


1
x
- 2 x +
2
y2
4x
2
-
x
y
+
1
x 
y
2x
+3
-
3
4
1
1 

 4  x2 + 2 + 6
x4
x


Exercise 6 : Find the square root using undetermined coefficient method
X4 + 6x3 + 17x2 + 24x + 16
Exercise 7 : Find the square root using undetermined coefficient method
X6 – 4x4 – 6x3 + 4x2 + 12x + 9
Exercise 8 : If x4 + ax3 + bx2 + cx + d = (x2 + px + q)2, then show that 2q = b –
a 2 2c
.
=
4
a
Homogeneous function
An integral function is said to be homogeneous, if each of its term is of the same degree with respect to any
set of variables.
eg. 2x2 + 4y2 – 5xy is a homogeneous expression of the second degree in x and y.
The product of two homogeneous functions of degrees m and n respectively is a homogeneous function of
degree m + n.
eg. (2 + 5y) (4x2 + 5y2) = 8x3 + 10xy2 + 20x2y + 25y3
2x + 5y is of first degree in x and y.
4x2 + 5y2 is of second degree in x and y and both are homogeneous in x and y.
Their product is homogenous and of third degree in x and y.
General forms of homogeneous integral functions of different degrees
Degree
Variables
1
x, y
Forms of homogeneous functions
ax + by
2
x, y
ax + bxy + cy2
3
x, y
ax3 + bx2y + cxy2 + dy3
1
x, y, z
2
x, y, z
2
ax + by + cz
2
2
ax + by + cz2 + fyz + gzx + hxy
Symmetric functions
A function is said to be symmetric with respect to any set of variables if the interchange of any pairs of the
set of variables does not alter the value of the function.
eg. 5 + 2x + 2y does not change in value if the variables x and y are interchanged. The function 5x + y + z is
symmetric with respect to y and z, but it is not symmetric with respect to x and y.
The expression xy + yz + zx + x + y + z is symmetric with respect to x, y, z.
pqr
p q r
and 2 2 2 2 2 2 are symmetric with respect to p, q, r.
 
qr rp pq
p q  q r r p
Hence the necessary and sufficient condition that an integral function to be symmetrical is that all the terms
of any one type shall have the same coefficient.
If the variables x and y are interchanged ax2 + bxy + cy2 then the resulting expression is ay2 + byx + cx2 then
the resulting expression is ay2 + byx + cx2
In order that ax2 + bxy + cy2 is symmetric, we must have a = c.
Similarly
Forms of symmetric integral functions:
Degree
1
2
1
2
Variables
x, y
x, y
x,y,z
x,y,z
Functions
a(x + y) + b
a (x2 + y2) + bxy + c (x +y) + d
a (x + y + z) + b
a (x2 + y2 + z2) + b(xy + yz + zx) + c(x + y + z) + d
Homogeneous symmetric integral function
Degree
1
2
1
2
Variables
x, y
x, y
x,y,z
x,y,z
Functions
a(x + y)
a (x2 + y2) + bxy
a (x + y + z)
a (x2 + y2 + z2) + b(xy + yz + zx)
Cyclic symmetry
In the expression 2y2z + 2z2x +2x2 y which consists of three terms
(i) 2z2x can be obtained from 2y2z if we replace y by z and z by x.
(ii) 2x2y can be obtained from 2z2x if we replace z by x and x by y.
(iii) 2y2z can be obtained from 2x2y if we replace x by y and y by z.
The selected term from which the other forms are obtained may be called the typical term. The function can
evidently be determined when a typical term is given.
Hence we can say that the function 2y2 z+ 2z2x + 2x2y possesses cyclic symmetry.
Again the expression a2(b – c) + b2 (c – a) + c2 (a – b) possesses cyclic symmetry.
We can get the second term the first by replacing a by b, b by c and c by a. We can get the third term from
the second if we replace b by c, c by a and a by b.
y 2 z2 x 2
y2
It can also be seen that
possesses cyclic symmetry. The typical term can be taken as
.


z
x
y
z
Standard Formulae:
1.
2.
3.
4.
5.
6.
7.
8.
9.
10.
11.
12.
13.
14.
15.
(a + b)2 = a2 + 2ab + b2
(a – b)2 = a2 – 2ab + b2
(a + b) (a – b) = a2 – b2
(a + b)2 + (a – b)2 = 2(a2 + b2)
(a + b)2 – (a – b)2 = 4ab
(a + b + c)2 = a2 + b2 + c2 + 2bc + 2ca + 2ab =  a2 + 2  bc
(a + b)3 = a3 + b3 + 3ab ( a + b) = a3 + 3a2b + 3ab2 + b3
(a – b)3 = a3 – b3 –3ab (a – b) = a3 – 3a2b + 3ab2 – b3
(x + a) (x + b) = x2 + (a + b) x + ab
(x + a) (x + b) (x + c) = x3 + x2 (a + b + c) + x (ab + bc + ca) + abc
a3 + b3 = (a + b) (a2 – ab + b2)
a3 – b3 = (a – b) (a2 + ab + b2)
a3 + b3 + c3 – 3abc = (a + b + c) (a2 + b2 + c2 – ab – bc – ca)
If a + b + c = 0, then a3 + b3 + c3 – 3abc = 0 or a3 + b3 + c3 = 3abc
1
a2 + b2 + c2 – ab – bc – ca = [(a – b)2 + (b – c)2 + (c – a)2]
2
Illustration 16: Factorise x3 + y3 + z3 – 3xyz and hence prove that x2 + y3 + z3 = 3xyz if x + y + z = 0
Solution:
x3 + y3 + z3 – 3xyz = [(x + y)3 + z3] – 3xy(x + y) – 3xy (x + y + z)
= (x + y + z) [(x + y)2 – (x + y)z +z2] – 3xy (x + y + z)
= (x + y + z) [(x + y)2 – (x + y)z + z2 – 3xy
= (x + y + z) (x2 + y2 + z2 – xy – yz – zx)
Illustration 17: Show that if x2 + y2 + z2 = xy + yz + zx then x = y = z
Solution:
x2 + y2 + z2 = xy + yz + zx  x2 + y2 + z2 – xy – yz – zx
1 2
[2x + 2y2 + 2z2 – 2xy – 2yz – 2zx] = 0
2
1
i.e [(x – y)2 + (y – z)2 + (z – x)2] =0
2

The sum of 3 perfect squares cannot be equal to zero unless each quantity of equal to zero.
(x – y) = 0  x = y
(y – z) = 0  y = z
 x = y = z.
Exercise 9 : Resolve into factors a2(b + c) + b2(c + a) + c2(a + b) + 2abc
3

 
3
Exercise 10 : Resolve into factors  a  -  a  .
 abc   abc 

Exercise 11 : Resolve into factors

(a + b) (a - b) .
2
abc
Exercise 12 : Resolve into factors a4(b – c) + b4(c – a) + c4(a – b)
Exercise 13 : Resolve into factors (a + b + c)5 – a5 – b5 – c5
Answers to exercises:
1. 6a – 2
5. x2 
1
x2
2
1
x
3. x   1
4.
6. x2 + 3x + 4
7. x3 – 2x – 3
9. (a + b) (b + c) (c + a)
10. 3(a + b) (b + c) (c + a)
11. –(a – b) (b – c) (c – a)
12. –(a – b) (b – c) (c – a) (a2 + b2 + c2 + ab + bc + ca)
13.
x y 1


y 2x 4
2. 4x3 – 3x2 + 2x – 1
1
(a + b) (b + c) (c + a) [–83(a2 + b2 + c2) + 145(ab + bc + ca)]
5
ASSIGNMENTS
SUBJECTIVE
LEVEL – I
1.
On dividing (kx3 + 9x2 + 4x – 10) by (x – 3), we get 5 as remainder. Find the value of k.
2.
The polynomials f(x) = ax3 + 3x2 – 3 and g(x) = 2x3 – 5x + a, when divided by (x – 4), leave the same
remainder in each case. Find the value of a.
3.
Using factor theorem, factorize the polynomial 2x3 + 3x2 – 17x – 30 and hence solve the equation
2x3 + 3x2 – 17x – 30=0.
4.
5.
If (x + a) is a common factor of the polynomials f(x) = x2 + mx + n and g(x) = x2 + px + q, show that
 qn 
a =
.
pm
Show that (1+3x) is a factor of the polynomial f(x) = (3x3 + 7x2 – 43x – 15) and hence factorize f(x)
6.
Prove (a2 + b2) (x2 + y2) = (ax – by)2 + (bx + ay)2
7.
Factorise (x + y + z)3 + (x – y – z)3 – 8x3
8.
A polynomial leaves remainders 0, 1 and 2 when divided by x, (x – 1), (x – 2) and respectively. Find
the remainder when polynomial is divided by x (x – 1) (x – 2).
9.
Find the remainder when (a – b) x2 + (b – c) x + (c – a) is divided by x – 1.
10.
If a + b + c = abc then the value of a  b  b  c  c  a + abc is ………..
1 ab 1 bc 1 ca
LEVEL – II
1.
A polynomial f(x) leaves remainders 2 and 3 when divided by x +1 and (x – 3) respectively. Find the
remainder when f(x) is divided by (x + 1) (x – 3).
2.
The remainder when x303 – x302 + x301 – x300 …….. + x – 1 is divided by x3 – x is.
3.
Given that x3 + 2x – 1 = (x + a) (x + b) (x + c). Find the value of
(a) a2 + b2 + c2
(b) a3 + b3 + c3
4.
Find the remainder and quotient when 2x3 – 5x2 – x + 3 is divided by (1 – 3x).
5.
Factorise
(i)
1 – 3a + 3a2 + 26a3
(ii)
2b2 c2 + 2c2 a2 + 2a2 b2 – a4 – b4 – c4
OBJECTIVE
LEVEL – I
Select the correct alternative (A), (B), (C),(D)from each of the following,Indicate your choice by writing the
appropriate letter only.
1.
If (x – a)2 be a factor of x3 + px + q then
(A) p = 2a2; q = 3a2
(B) p = –3a2; q =2a3
(C) p =3a2; q=2a3
(D) None of these
2.
The homogeneous function of the second degree in x and y having 2x – y as a factor.
A) 2x2 + xy – y2
(B) 3x2 + 2xy + y2
(C) x2 + xy + 2y2
(D) None of these
3.
The factors of the polynomial expression 15 – x – 6x2 are
(A) (3x + 5) and (2x + 3)
(B) (5 – 3x) and (2x + 3)
(C) (3 – 2x) and (3x + 5)
(D) None of these
4.
If x –
1
1
= 7 then the value of x3 – 3 is
x
x
(A) 333
(B) 343
(C) 364
(D) None of these
5.
If x2 – 3x + 2 is a factor of the expression x4 + ax2 + b, then the values of a and b are given by
(A) a = –5; b = 4
(B) a = 4; b = –5
(C) a = 5; b = –4
(D) None of these
6.
If a + b + c=6; bc + ca + ab = 11; abc = 6, then the value of (1 – a)(1 – b)(1 – c) is
(A) 1
(B) –1
(C) 0
(D) None of these
7.
If x =
a
b
c
,y=
,z=
, then the value of xy + yz +zx+ 2xyz is
b +c
c +a
a +b
(A) 1
(B) 2
(C) 3
(D) None of these
Determine whether each of the following statements is true or false.
8.
If a + b + c = 0, then a4 + b4 + c4 = 2(b2 c2 + c2 a2 + a2 b2).
9
The expression (ab + cd)2 – (ad + bc)2 is divisible by (a – c) (b – d).
10.
A certain algebraic expression is exactly divisible by x – 2, the quotient being x2 – x – 6. Then the
expression is also divisible by x + 2.
LEVEL – II
1.
If (2a – 3b) = 5 then 8a3 – 27b3 – 90ab is
(A) 125
(B) 91
(C) –111
(D) None of these
2.
The H.C.F. of the functions x3 + (a + b)x2 + (ab + 1) x + a and bx3 + (ab + 1)x2 + (a + b) x + 1 is
(A) x2 + ax + 1
(B) x2 + bx + 1
(C) x2 + x + a
(D) None of these
3.
If a + b = 9, x = 5 and a – b – x = 2 then value of (a – b) [x3 – 2ax2 + a2x – (a+b)b2] is
(A) 445
(B) 252
(C) 376
(D) None of these
4.
x4 – 3x3 – x – 5 = (x + 1) (......)
(A) x3 – 4x2 + 4x – 5
(B) x3 + 4x2 – 4x + 5
5.
(C) x3 – x2 – 5
4 4
If abx2 = (a – b)2 (x + 1) then the value of 1 +  2 is
x x
2
2
2

 a 
 a b 
ab 
(A) 
(B) 
(C) 


 a  b 
 ab 
 ab 

(D) None of these
(D) None of these
6.
For what value of x will 4x4 + 12x3 – 11x2 – 15x – 5 be a perfect square?
(A) 3
(B) –2
(C) 2
(D) None of these
7.
The square root of 9a2 – 24ab + 16b2 is
(A) 4b – 3a
(B) 3a + 4b
(C) –3a – 4b
The L.C.M. of 6x2 –19x + 10 and 3x2 + x – 2 is
(A) (2x – 5) (x + 1) (3x + 2)
(C) (3x – 2) (2x – 5) (x + 1)
(B) (x + 1) (2x + 5) (3x – 2)
(D) None of the above
8.
(D) None of these
9.
The value of p for which the function 4x4 – 12x3 + 17x2  12x + p is a perfect square is
(A) 4
(B) –4
(C) –2
(D) None of these
10.
What must be added to (x2 + 7x + 4) (x2 + 7x – 2) to make it a perfect square?
(A) –9
(B) 9
(C) 1
(D) None of these
ANSWERS
SUBJECTIVE
LEVEL – I
8
9
5
3. 3,  , –2
2
7. 6x(x + y + z) (y + z  x)
1. k = 
2. a = 1
5. (x – 3) (x + 5) (3x + 1)
8. x
9. 0
10. 0
LEVEL – II
1.
1
(x + 9)
4
2.
151x2 + 152x  1
3.
–4, –3
4.
Rem:
5.
(i) (1 + 2a) (1 – 5a + 13a2)
(ii) (a + b + c) (a + b – c) (b + c – a) (c + a – b)
22 
59  2 2 13
, 
x 
x
9
27 
27  3
OBJECTIVE
LEVEL – I
1.
B
2.
A
3.
C
4.
C
5.
A
6.
C
7.
A
8.
T
9.
T
10.
T
LEVEL – II
1.
A
2.
A
3.
B
4.
A
5.
B
6.
C
7.
A
8.
C
9.
A
10.
B