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Question-1

Find the remainder when y3 + y2 - 2y + 5 is divided by y - 5.

Solution:

Again, we should evaluate p(5)

Let p(y) = y3 + y2 - 2y + 5
 ... p(5) = 53 + 52 - 2 x 5 + 5
            = 125 + 25 - 10 + 5
            = 145

Thus, we find that p(5) is the remainder when

 p(y) is divided by y - 5.

Question-2

Determine the remainder when polynomial p(x) is divided by x – 2.

Solution:
p(x) = x4 - 3x2 + 2x - 5

According to remainder theorem, the required remainder will be = p(2)


    p(x) = x4 - 3x2 + 2x - 5

... p(2) = 24 - 3(2)2 + 2(2) - 5

           = 16 - 12 + 4 - 5 = 3

Question-3

Show that x + a is a factor of xn +an for any odd +ve integer n.
Solution:

Let f(x) = xn + an

 x + a will be the factor of xn +an if f(-a) = 0 

Now f(- a) = (- a)n+ an 0    (since n is a odd +ve integer)

Thus (x + a) is a factor of xn + an.

Question-4

When f(x) = x4 - 2x3 + 3x2 - ax is divided by x + 1 and x - 1, we get remainders as 19 and 5 respectively.
Find the remainder if f(x) is divided by x - 3.

Solution:
When f(x) is divided by (x + 1) and (x - 1), the remainders are 19 and 5 respectively.
... f(-1) = 19 and f(1) = 5
⇒ (- 1)4 - 2(- 1)3 + 3(- 1)2 - a(- 1) + b = 19
⇒1+ 2 + 3 + a + b = 19
... a + b = 13 ------------ (i)

Again, f(1) = 5
⇒14 - 2
× 13 + 3 × 12 - a ×1 b = 5
⇒1 - 2 + 3 - a + b = 5
... b - a = 3 ------------- (ii) 

Solving eqn (i) and (ii), we get
a = 5 and b = 8
Now substituting the values of a and b in f(x), we get
... f(x) = x4 - 2x3 + 3x2 - 5x +8

Now f(x) is divided by (x - 3) so remainder will be f(3)
... f(x) = x4 - 2x3 + 3x2 - 5x +8
⇒f(3) = 34 - 2
× 33 + 3 × 32 - 5 × 3 + 8
           = 81 - 54 + 27 - 15 + 8 = 47

Question-5

What must be subtracted from 4x4 - 2x3 - 6x2 + x - 5 so that the result is exactly divisible by 2x2 + 3x - 2 ?

Solution:
Since 2x2 + 3x - 2 is of degree 2, so when

p(x) = 4x4 - 2x3 - 6x2 + x - 5 is divided by

q(x) = 2x2 + 3x - 2, the remainder should be a linear expression (degree of remainder < degree of divisor).

Let the remainder r(x) = ax + b for exact division this remainder should be subtracted from p(x) Now let
   f(x) = p(x) - r(x)
         = 4x4 - 2x3 - 6x2 + x - 5 - (ax + b)
   f(x) = 4x4 - 2x3 - 6x2 + (1 - a) x - 5 - b

Again, we have
  q(x) = 2x2 + 3x - 2
         = 2x2 + 4x - x - 2
         = 2x (x + 2) - 1(x + 2)
         = (x + 2) (2x - 1)


Since x + 2 and 2x - 1 are factors of q(x) and f(x) is exactly divisible by q(x), hence (x + 2) and (2x - 1) are also factors of f(x), i.e.
 f(-2) = 0 and f( ) = 0
... f(-2)=4(- 2)4 - 2(- 2)3 - 6(- 2)2 + (1 - a) (- 2) - b - 5 = 0

64 + 16 - 24 - 2 + 2a - b - 5 = 0
  2a - b = - 49 …… (i)

Again f( ) = 0

4 ()4- 2( )3 - 6 ()2 + (1 - a) x  - b - 5 = 0
- - + - -b - 5 = 0
- - b - 6 = 0
a + 2b = - 12 …… (ii)

Solving eqns (i) and (ii), we get a = - 22 and b = 5.
... r(x) = - 22x + 5 should be subtracted.

Question-6

If both (x + 1) and (x - 1) are factors of ax3 + x2 - 2x + b, find a and b.

Solution:
Let p(x) = ax3 + x2 - 2x + b

Since (x + 1) and (x - 1) are the factors of p(x),
... p(-1) = 0 and p(1) = 0
... p(- 1) = a(-1)3 + (-1)2 - 2(-1) + b = 0
⇒ - a
+ 1 + 2 + b = 0
⇒ a - b =
3 …… (i)

Again, p(1) = a(1)3 + (1)2 - 2(1) + b = 0
⇒ a + 1 - 2 + b = 0
⇒ a + b = 1 …… (ii)
Solving equations (i) and (ii) we get a = 2 and b= -1

Question-7

If x2 - 1 is a factor of ax4 + bx3 + cx2 + dx + e, show that a + c + e = b + d = 0.

Solution:
Since x2 - 1 = (x + 1)(x - 1) is a factor of
          p(x) = ax4 + bx3 + cx2 + dx + e
... p
(x) is divisible by (x + 1) and (x - 1) separately
⇒ p(1) = 0 and p(- 1) = 0
    p(1) = a(1)4 + b(1)3 + c(1)2 + d(1) + e = 0
⇒ a + b + c + d + e = 0 ……(i)
      

Similarly, p(- 1) = a(- 1)4 + b(- 1)3 + c(- 1)2+ d(- 1) + e = 0
  ⇒ a - b + c - d + e = 0
            ⇒ a + c + e = b + d
……(ii)

Putting the value of a + c + e in eqn (i), we get
    a + b + c + d + e = 0
⇒ a + c + e + b + d = 0
      ⇒ b + d + b + d = 0
              ⇒ 2(b + d) = 0
                  ⇒ b + d = 0
……(iii)

Comparing equations (ii) and (iii), we get
     a + c + e = b + d = 0

Question-8

Factorise each of the following expressions:
(i) 48x3 - 36x2
(ii) 5x2 - 15xy
(iii) 15x3y2z - 25xy2z3

Solution:
(i) 48x3 - 36x2
    = 12.4.x2.x - 12.3.x2
    = 12x2(4x - 3)

(ii) 5x2 - 15xy = 5x
(x - 3y)

(iii) 15x3y2z - 25xy2z3 = 5xy2z (3x2 - 5z2)

Question-9

Factorise
(i) 2x2(x + y) - 3(x + y) 
(ii) 5xy(5x + y) - 5y(5x + y)
(iii) x(x2 + y2 - z2) + y(x2 + y2 - z2) + z(x2 + y2- z2)
(iv) ab(a2 + b2 - c2) + bc(a2 + b2 - c2) + ca(a2+b2-c2)

Solution:
(i) 2x2(x + y) - 3(x + y)
    = (x + y) (2x2 - 3)

(ii) 5xy(5x + y) - 5y(5x + y)
     = 5y(5x + y) (x - 1)

(iii) x(x2 + y2 - z2) + y(x2 + y2 - z2) + z(x2 + y2 - z2)
     = (x2 + y2 - z2) (x + y + z)

(iv) ab(a2 + b2 - c2) + bc(a2+b2-c2) + ca(a2+b2 - c2)
     = (a2 + b2 - c2) (ab + bc + ca)

Question-10

Factorise
(i) 6ab - b2 + 12ac - 2bc
(ii) x2 + y - xy - x
(iii) x3 - 2x2y + 3xy2 - 6y3
(iv) a(a + b - c) – bc

Solution:
(i) 6ab - b2 + 12ac - 2bc
    = (6ab - b2) + (12ac - 2bc)
    = b(6a - b) + 2c(6a - b)
    = (6a - b) (b + 2c)

(ii) x2 + y - xy - x
    = x2 - xy - x + y
    = x(x - y) - (x - y)
    = (x - y) (x - 1)

(iii) x3 - 2x2y + 3xy2 - 6y3
     = (x3 - 2x2y) + (3xy2 - 6y3)
     = x2(x - 2y) + 3y2(x - 2y)
     = (x - 2y) (x2 + 3y2)

(iv) a(a + b - c) - bc
     = a2 + ab - ac - bc
     = a(a + b) - c(a + b)
     = (a + b) (a - c)

Question-11

Factorise each of the following expressions.
(i) 25x2y2 - 20xy2z + 4y2z2
(ii) 4x2 - 4x + 7
(iii) 
(iv) 4a2 + 12ab + 9b2 - 8a - 12b

Solution:
(i) 25x2y2 - 20xy2z + 4y2z2
     = (5xy)2 - 2
×5xy × 2yz + (2yz)2
     = (5xy + 2yz)2
     = y2 (5x + 2z)2

(ii) 4x2 - 4x + 7
      = (2x)2 + 2
×2x ×+()2 = (2x +)2 

(iii) 
     = 

(iv) 4a2 + 12ab + 9b2 - 8a - 12b
      = (4a2 + 12ab + 9b2) - 4(2a + 3b)
      = (2a)2 + 2
× 2a × 3b + (3b)2 - 4(2a + 3b)
      = (2a + 3b)2 - 4(2a + 3b)
      = (2a + 3b) (2a + 3b - 4)

Question-12

Factorise
(i) 25x2 - 36y2
(ii) 2ab - a2 - b2 + 1
(iii) 36a2 - 12a + 1 - 25b2
(iv) a4 - 81b4
(v) a12b4 - a4b12
(vi) 4x2 - 9y2 - 2x - 3y
  

Solution:
(i) 25x2 - 36y2
    = (5x)2 - (6y)2
    = (5x + 6y) (5x - 6y)


(ii) 2ab - a2 - b2 + 1
    = 1 - (a2 + b2 - 2ab)
    = 1 - (a - b)2
    = (1 + a - b) (1 - a + b)


(iii) 36a2 - 12a + 1 - 25b2
     = (6a)2 - 2 x 6a x 1 + 12 - (5b)2
     = (6a - 1)2 - (5b)2
     = (6a - 1 + 5b) (6a - 1 - 5b)
     = (6a + 5b - 1) (6a - 5b - 1)

(iv) a4 - 81b4
     = (a2)2 - (9b2)2
     = (a2 + 9b2) (a2 - 9b2)
     = (a2 + 9b2) (a + 3b) (a - 3b)

(v) a12 b4 - a4 b12
     = a4b4 (a8 - b8)
     = a4b4 [(a4)2 - (b4)2]
     = a4b4 (a4 + b4) (a4 - b4)
     = a4b4 (a4 + b4) (a2 + b2) (a2 - b2)
     = a4b4 (a4 + b4) (a2 + b2) (a + b) (a - b)

(vi) 4x2 - 9y2 - 2x - 3y
     = (2x)2 - (3y)2 - (2x + 3y)
     = (2x + 3y) (2x - 3y) - (2x + 3y)
     = (2x + 3y) (2x - 3y - 1)

Question-13

Factorise by completing the squares.
(i) a4 + a2 + 1
(ii) y4 + 5y2 + 9
(iii) x4 + 4
(iv) x4 + 4x2 + 3

Solution:
(i) a4 + a2 + 1
    = a4 + a2 + 1 + a2 - a2
    = a4 + 2a2 + 1 - a2
    = (a2)2 + 2a2 + 1 - a2
    = (a2 + 1)2 - a2
    = (a2 + 1 + a) (a2 + 1 - a)
    = (a2 + a + 1) (a2 - a + 1)

(ii) y4 + 5y2 + 9
    = (y2)2 + 5y2 + (3)2 + y2 - y2
    = (y2)2 + 6y2 + (3)2 - y2
    = (y2 + 3)2 - y2
    = (y2 + y + 3) (y2 - y + 3)

(iii) x4 + 4
    = x4 + 4 + 4x2 - 4x2
    = (x2)2 + (2)2 + (2x)2 - 4x2
    = (x2 + 2)2 - (2x)2
    = (x2 + 2x + 2) (x2 - 2x + 2)
    
(iv) x4 + 4x2 + 3
     = x4 + 4x2 + 4 - 1
     = x4 + 4x2 + 3 + 1 - 1
    = (x2)2 + 2 × 2x2 + (2)2 - 1
    = (x2 + 2)2 - 1
    = (x2 + 2 + 1) (x2 + 2 - 1) = (x2 + 3) (x2 + 1)

Question-14

Factorise the following expressions:
(i) a3 - 27
(ii) 1 - 27x3
(iii) 8x3 - (2x - 3y)3
(iv) a8 - a2b6
(v) a3 - 5b3

Solution:
(i) a3 - 27
    = a3 - (3)3
    = (a - 3) (a2 + 3a + 9)

(ii) 1 - 27x3 = 13 - (3x)3
                 = (1 - 3x) [12 + 1 × 3x + (3x)2]
                 = (1 - 3x) (1 + 3x + 9x2)

(iii) 8x3 - (2x - 3y)3 = (2x)3 - (2x - 3y)3
     = [2x - (2x - 3y)]
[(2x)2 + 2x(2x - 3y) + (2x - 3y)2]
     = (2x-2x+ 3y) (4x2 + 4x2- 6xy + 4x2+ 9y2-12xy)
     = 3y(12x2 + 9y2 - 18xy)
     = 3y×3(4x2 + 3y2 - 6xy)
     = 9y(4x2 + 3y2 - 6xy)

(iv) a8 - a2b6 = a2(a6 - b6)
      = a2[(a3)2 - (b3)2]
      = a2 (a3 + b3) (a3 - b3)
      = a2 (a + b) (a2 - ab + b2) (a - b) (a2 + ab + b2)
      = a2 (a + b) (a - b) (a2 - ab + b2) (a2 + ab + b2)

(v)  a3 - 5b3 = a3 - (b)3
      = (a- b)(a2+ ab+5b2)

Question-15

Factorise the following
(i) 16p3q3 + 54r3
(ii) + 8b3
(iii) 2 a3 + 3 b3
(iv) 8a4b + ab4
(v) a7 - 64a

Solution:
(i) 16p3q3 + 54r3
    = 2(8p3q3 + 27r3)
    = 2[(2pq)3 + (3r)3]
    = 2(2pq + 3r) [(2pq)2 - 2pq . 3r + (3r)2]
    = 2 (2pq + 3r) (4p2q2 - 6pqr + 9r2)

(ii) + 8b3 = ()3 + (2b)3
                 = ( +2b)( -ab+4b2)

(iii) 2 a3 + 3 b3
     = ( a)3+( b)3
     = ( a + b)(2a2 - ab + 3b2)

(iv) 8a4b + ab4 = ab(8a3+ )
                           = ab[ ]
                           = ab (2a+ )

(v)  a7 - 64a
      = a(a6 - 64)
      = a(a6 - 26)
      = a[(a3)2 - (23)2]
      = a(a3 + 23) (a3 - 23)
      = a(a + 2) (a2 - 2a + 4) (a - 2) (a2 + 2a + 4)
      = a(a + 2) (a - 2) (a2 - 2a + 4) (a2 + 2a + 4)

Question-16

Factorise
(i) x3 + 9x2+ 27x + 27
(ii) x3 – 9x2y + 27xy2 – 27y
3

Solution:
(i) x3+9x2+ 27x + 27
             = (x)3+ 3(x)2(3)+3(x)(3)2 + (3)3
             = (2x + 3)3

(ii) x3–9x2y + 27xy2 – 27y3
                = x3–3(x)2(3y)+3(x)(3y)2 – (3y)3
                = (x – 3y)3

Question-17

Using identities, find the value of
(i) 1012
(ii) 982
(iii) (0.98)2 
(iv) 11 × 9
(v) 190 × 190 - 10 × 10

Solution:
(i) 1012 = (100 + 1)2
           = 1002 + 2 × 100 × 1 + 12
           = 10000 + 200 + 1 = 10201

(ii) 982 = (100 - 2)2
           = 1002 - 2 × 100 × 2 + 22
           = 10000 - 400 + 4 = 9604

(iii) 0.982 = (1 - 0.02)2
              = 12 - 2 × 1 × 0.02 + (0.02)2
              = 1 - 0.04 + 0.0004
              = 0.9604

(iv) 11 × 9 = (10+1) (10-1)
                = 102 - 12
                = 99
       
(v) 190 × 190 - 10 × 10 = 1902 - 102
                                  = (190 + 10) (190 - 10)
                                  = 200 × 180
                                  = 36000

Question-18

If x + y = 5 and xy = 4, find x - y, using identities.

Solution:
    (x - y) 2 = x2 - 2xy + y2
                  = x2 + 2xy + y2 -
4xy
                  =
(x + y)2 - 4xy

                  = 52 - 4 × 4
               = 25 - 16 = 9

   ∴  x – y = 9
              = 3

Question-19

Expand
(i) (x + 5y + 6z)2
(ii) (2a - 3b + 4c)2
(iii) (-a + 6b + 5c)2                 
(iv) (-p +4q - 3r)2

Solution:
Using the identity
(x + y + z)2 = x2 + y2 + 2xy + 2yz + 2xz, we have
(i) (x + 5y + 6z)2    
= (x)2 + (5y)2 + (6z)2 + 2(x)(5y) + 2(5y)(6z) + 2(x)(6z)
= x2 + 25y2 + 36z2 + 10xy + 60yz + 12 xz

(ii) (2a - 3b + 4c)2
= (2a)2+(-3b)2+(4c)2+2(2a)(-3b)+2(-3b)(4c)+2(2a)(4c)
= 4a2+ 9b2 + 16c2 - 12ab - 24bc + 16ac

(iii) (-a + 6b + 5c)2
= (-a)2+(6b)2+(5c)2+2(-a)(6b)+2(6b)(5c)+2(-a)(5c)
= a2 + 36b2 + 25c2 - 12ab + 60bc - 10ac

(iv) (-p + 4q - 3r)2
= (-p)2+(4q)2+(-3r)2+2(-p)(4q)+2(4q)(-3r)+2(-p)(-3r)
= p2 + 16q2 + 9r2 - 8pq - 24qr + 6pr

Question-20

Expand
(i) (2x + 5y)3
(ii) (5p – 3q)3
(iii) (-a + 2b)3 

Solution:
(i) Using the identity
   (x + y)3 = x3 + 3x2y + 3xy2 + y3 we get
(2x + 5y)3 = (2x)3 + 3(2x)2 (5y) + 3(2x)(5y)2 + (5y)3
               = 8x3 + 60x2y + 150xy2 + 25y3

(ii)
Using the identity
    (x - y)3 = x3 - 3x2y + 3xy2 - y3   we get
 (5p - 3q)3 = (5p)3-3(5p)2(3q)+3(5p)(3q)2-(3q)3
              = 125p3 - 225p2q + 135pq2 - 27q3

(iii)
Using the identity for (x + y)3 we get
(-a + 2b)3 = (-a)3+3(-a)2(2b)+3(-a)(2b)2+(2b)3
               = a3 + 6a2b - 12ab2 + 8b3

Question-21

Evaluate, using identities
(i) 1023
(ii) 993

Solution:
(i) 1023 = (100 + 2)3
           = (100)3 + 3(100)2 (2) + 3(100)(22) + 23
           = 10,00,000 + 60,000 + 1200 + 8  
            = 10,61,208

(ii) 993 = (100 - 1)3
           = 1003 - 3(100)2(1) + 3(100)(12) - 13
           = 10,00,000 - 30,000 + 300 - 1
           =  9,70,299

Question-22

Simplify (2a + b)3 + (2a- b)3

Solution:
(2a + b)3 + (2a - b)3 = (8a3 + 12a2b + 3ab2 + b3)+(8a3 - 12a2b + 3ab2 - b3)
                              = 16a3 + 6ab2

Question-23

Evaluate, using the identities
(i) 505 x 503,
(ii) 37 x 26

Solution:
(i) 505 x 503 = (500 + 5) (500 + 3)
                     = 5002 + (5 + 3) (500) + 5 x 3
                     = 250000 + 4000 + 15 = 254015
 
(ii) 37 x 26 = (30 + 7) (30 - 4)
                  = 302 + (7 - 4) (30) - 7 x 4
                  = 900 + 90 - 28 = 962

Question-24

Find the products   
(i) (2a + 5b)(4a2 - 10ab + 25b2)
(ii) (2a - 5b)(4a2 + 10ab + 25b2)

Solution:
(i) (2a + 5b)(4a2 - 10ab + 25b2)
             = (2a + 5b)[(2a)2 - (2a)(5b) + (5b)2]
             = (2a)3 + (5b)3 (from identity 10 above)
             = 8a3 + 125b3

(ii) (2a - 5b)(4a2 + 10ab + 25b2)
              = (2a - 5b)[(2a)2 + (2a)(5b) + (5b)2]
              = (2a)3 - (5b)3   (from identity 11 above)
              = 8a3 - 125b3

Question-25

Factorise
(i) 30x3y + 24x2y2 – 6xy
(ii) 5x(a – b) + 6y(a –b)
 

Solution:
(i) The greatest monomial that is a common factor of the three terms is 6xy.
...    30x3y + 24x2y2 - 6xy = 6xy(5x2 + 4xy – 1)


(ii) Here the polynomial (a – b) is a common factor.
...5x(a – b) + 6y(a – b) = (a – b)(5x + 6y)

Question-26

Factorise
(i) 7a3 + 7a – 2a2 – 2
(ii) 4ax + 3by – 3ay –4bx
(iii) x3 + y3 + x2y + xy2

Solution:
(i) 7a3 + 7a – 2a2 – 2 = 7a(a2 + 1) – 2(a2 + 1)
                               = (a2 + 1)(7a – 2)


(ii)
The terms of 4ax + 3by – 3ay – 4bx can be rearranged and factorized.
4ax – 4bx – 3ay + 3by = 4x(a – b) - 3y(a - b)
                                = (a – b)(4x – 3y)


(iii) x3 + y3 + x2y + xy2 = x3 + x2y + xy2 + y3 
                                  = x2(x + y) + y2 (x + y)
                                  = (x + y) (x2 + y2)

Question-27

Factorise
(i) 9x2 + 30xy + 25y2
(ii) 9x2 – 30xy + 25y2

Solution:
(i) 9x2 + 30xy + 25y2   
     = (3x)2 + 2(3x)(5y) +(5y)2  
     = (3x +5y)2  


(ii) 9x2 – 30xy + 25y2  
    = (3x)2 – 2(3x)(5y) + (5y)2
    
= (3x – 5y)2

Question-28

Factorize
(i) 9x2 – y2
(ii) (3 – x)2 – 36x2
(iii) (2x – 3y)2 – (3y + 4y)2
(iv) 16x4 – y4

Solution:
(i) 9x2 – y2 = (3x)2 – (y)2
                = (3x + y)(3x – y)


(ii) (3 – x)2 – 36x2 = (3 – x)2 – (6x)2
                              
= (3 - x + 6x) (3 - x - 6x)
                          = (3 + 5x)(3 – 7x)


(iii) (2x – 3y)2 – (3x + 4y)2 = (2x – 3y + 3x + 4y)(2x – 3y – 3x – 4y)
                                      = (5x + y)(- x – 7y) 
                                      = -(5x + y) (x + 7y)


(iv) 16x4 – y4 = (4x2)2 – (y2)2
                       
= (4x2 + y2)(4x2 –y2)
                    = (4x2 + y2)(2x + y) (2x - y)

Question-29

Factorise
(i) x2 + y2 + 9z2 + 2xy + 6yz + 6xz, 
(ii) x2 + 4y2 + 9z2 – 4xy – 12yz + 6xz

Solution:
(i) x2 + y2 + 9z2 + 2xy + 6yz + 6xz
              = x2+(y)2+(3z)2+ 2(x)(y)+2(y)(3z)+2(x)(3z)  
              = (x + y + 3z)2


(ii) x2 + 4y2 + 9z2 _ 4xy – 12yz + 6xz
                   = x2 + (2y)2 + (3z)2 - 2(x)(2y) - 2(2y)(3z)+ 2(x)(3z)    
                   = x2 + (2y)2 + (3z)2 + 2(x)(-2y) + 2(-2y)(3z) + 2(x)(3z)
                   = (x - 2y + 3z)2

Question-30

Factorise x6 – y6

Solution:
x6 – y6 = (x2)3 – (y2)3
           = (x2 – y2)[(x2)2 + x2y2 +(y2)2]
           = (x + y)(x – y)(x4 + x2y2 + y4)


(Here the third factor can be further factorised as shown)

          = (x + y)(x – y)(x4 + 2x2y2 + y4 – x2y2)
          = (x + y)(x – y)[(x2 + y2)2 – x2y2]
          = (x + y)(x – y) (x2 + y2 + xy)(x2 + y2 – xy)




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