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Question-1

Which of the following expressions are polynomials in one variable and which are not? State reasons for your answer.
(i) 4x2 – 3x + 7 (ii) y2 + (iii) 3 + t
  (iv) y +  (v) x10 + y3 + t50

Solution:
(i) 4x2 – 3x + 7 – is a polynomial in one variable.

(ii) y2 +
  is a polynomial in one variable.

(iii) 3 + t
  is not a polynomial since the power of the variable is not a whole number.

(iv) y +
 is not a polynomial since the power of the variable is not a whole number.

(v) x10 + y3 + t50 is a polynomial in three variables.

Question-2

Write the co-efficients of x2 in each of the following:
(i) 2 + x2 + x  (ii) 2 – x2 + x3  (iii) 
 x2 + x  (iv) x – 1

Solution:
(i) Coefficient of x2 is 1. 

(ii) Coefficient of x2 is –1.

(iii) Coefficient of x2 is .

(iv) Coefficient of x2 is ‘0’.

Question-3

Give one example each of a binomial of degree 35, and of a monomial of degree100.

Solution:
 
3x35+5 and 4x100
 

Question-4

Write the degree of each of the following polynomials:
(i) 5x3 + 4x2 +7x (ii) 4 – y2(iii) 5t - (iv) 3.

Solution:
(i) Degree is 3.

(ii) Degree is 2.

(iii) Degree is 1.

(iv) No degree.

Question-5

Classify the following as linear, quadratic and cubic polynomials:
(i) x2 + x (ii) x – x3 (iii) y +  y2 + 4 (iv) 1 + x (v) 3t (vi) r2 (vii) 7x3

Solution:
(i) Quadratic Polynomial.

(ii) Cubic Polynomial.

(iii) Quadratic Polynomial.

(iv) Linear Polynomial.

(v) Linear Polynomial.

(vi) Quadratic Polynomial

(vii) Cubic polynomial.

Question-6

Find the value of the polynomial 5x - 4x2 + 3 at
(i) x = 0 (ii) x = -1 (iii) x = 2

Solution:
Let f(x) = 5x - 4x2 + 3

(i) when x =0

 f (0) = 5(0) – 4(0)2 + 3 = 3


(ii) When x = -1

f (-1) = 5(-1) – 4(-1)2 + 3 = -5 - 4 + 3 = -6

(iii) Value of f(x) at x = 2

       f(2) = 5(2) – 4(2)2 + 3

            = 10 – 16 + 3 = -3

Question-7

Find p(0), p(1) and p(2) for each of the following polynomials:
(i) p(y) = y2 – y + 1 (ii) p(t) = 2 + t + 2t2 – t3 (iii) p(x) = x3 (iv) p(x) = (x
  1) (x + 1)

Solution:
(i)  p(y) = y2 – y + 1 p(0) = (0)2 – (0) + 1 = 1

    p(1) = (1)2 – (1) + 1 = 1

And, p(2) = (2)2 – (2) + 1 = 4 – 2 + 1 = 3


(ii)  p(t) = 2 + t + 2t2 – t3

p(0) = 2 + 0 + 2(0)2 – (0)3 = 2

   p(1) = 2 + 1 + 2(1)2 – (1)3 = 2 + 1 – 1 = 4

and p(2) = 2 + 2 + 2(2)2 – (2)3 = 2 + 2 + 8
 8 = 4

(iii) p(x) = x3 

  p(0) = (0)3 = 0

      p(1) = (1)3 = 1

and p(2) = (2)3 = 8

(iv) p(x) = (x  1) (x + 1)

   p(0) = (0  1)(0 + 1) = (  1)(1) =  1

       p(1) = (1
 1) (1 + 1) = (0)(2) = 0

and, p(2) = (2
 1) (2 + 1) = (1)(3) = 3

Question-8

Verify whether the following are zeroes of the polynomial, indicated against them.
(i) P(x) = 3x + 1, x = - (ii) P(x) = 5x -, x =

(iii) P(x) = x2 – 1, x = 1, -1 (iv) P(x) = (x + 1) (x - 2), x = -1,2

(v) P(x) = x2, x = 0 (vi) P(x) = lx +m, x = -
 

(vii)P(x) = 3x2 – 1, x = -, (viii) P(x) = 2x + 1, x =

Solution:
(i) P(x) = 3x + 1, x = - , P = 3  +1 = -1 + 1 = 0 -   is a zero of P(x).

(ii)  P(x) = 5x -, x =   

  is not a zero of P(x).

(iii) P(x) = x2 – 1, x =1,-1

      P(1) = (1)2 - 1 = 1 - 1= 0

     P(-1) = (- 1)2 - 1 = 1 - 1 = 0

1,-1 are zeros of P(x).


(iv) P(x) = (x + 1) (x - 2), x = -1, 2 

      P(-1)= (-1 + 1)(-1 – 2)

              =(0)(-3)

              = 0

     P(2) = (2 + 1)(2 – 2) = (3)(0) = 0 -1,2 are zeros of P(x).

(v) P(x) = x2, x = 0 
 

     P(0) = (0)2 - 0

0 is a zero of P(x).


(vi) P(x) = lx +m, x = -    

 P = l  +m = - m +m = 0 - is a zero of P(x)

(vii) P(x) = 3x2 – 1, x = - ,    

   P = 3-1 = 3-1 = 1 – 1 = 0

      P(
) = 3()2 -1 = 3()-1 = 4 -1 = 3 0 - is a zero of P(x) but is not a zero of P(x)

(viii) P(x) = 2x  + 1, x  

       P = 2= 1 + 1 = 2 0

 
 is not a zero of P(x)

Question-9

Find the zero of the polynomial in each of the following cases.
(i) P(x) = x + 5 (ii) P(x) = x – 5 (iii) P(x) = 2x + 5
(iv) P(x) = 3x - 2 (v) P(x) = 3x.

Solution:
(i) P(x) = x +5
     P(x) = 0 x + 5 = 0 = - 5 -5 is a zero of the polynomial P(x)

(ii) P(x) = x –5      P(x) = 0 x + 5 = 0 x = - 5 5 is a zero of the polynomial P(x)

(iii)  P(x) = 2x + 5
      P(x) = 0 2x + 5 = 0 =2x – 5 x = - -is a zero of the polynomial P(x)

(iv) P(x) = 3x - 2
      P(x) = 0 3x - 2 = 0 3x – 5 x = is a zero of the polynomial P(x).

(v)  P(x) = 3x
     P(x) = 0     3x= 0 x= 0 0 is a zero of the polynomial P(x).

Question-10

Find the remainder when x3 + 3x2 + 3x + 1 is divided by
(i) x + 1 (ii) x - 
 (iii)x (iv) x + (v) 5 +2x

Solution:
Let p(x) = x3 + 3x2 + 3x + 1
(i) x + 1
          x + 1 = 0 x = -1 Remainder = p(-1) = (-1)3 + 3(-1)2 + 3(-1) + 1 = -1 + 3 – 3 + 1 = 0


(ii) x -
         x - = 0 x = Remainder =
                   = =


(iii) x
Remainder = (0)3 +3(0)2 + 3(0) + 1 = 1



(iv) x + π x + π = 0 x = -π ∴ Remainder = (-π )3 +3(-π )2+3(-π ) + 1 = - π 3 + 3π 2 - 3π + 1


(v) 5 + 2x
5 + 2x = 0 2x – 5 x = - Remainder =
                   = - = -

Question-11

Find the remainder when x3 – ax2 +6x –a is divided by x – a.

Solution:
Let p(x) = x3 – ax2 + 6x – a

    x – a = 0

    
x = a
\ Remainder = (a)2 – a(a2) + 6(a) – a

                          = a3 – a3 + 6a – a

                          = 5a

Question-12

Check whether 7+ 3x is a factor of 3x3 + 7x.

Solution:
7+ 3x will be a factor of 3x2 + 7x.only if 7+ 3x divides 3x2 + 7x . leaving no remainder.

Let p(x) = 3x2 + 7x

  7 + 3x = 0 ⇒  3x = - 7 only if 7 + 3x divides 3x3 + 7x  leaving no remainder

Let p(x) = 3x3 + 7x.

  7 + 3x = 0 ⇒  3x = -7 ⇒  x = -

∴Remainder = 3 = - = -

7 + 3xis not a factor of 3x3 + 7x
 

Question-13

Determine which of the following polynomials has (x + 1) as a factor:

(i) x3 + x2 + x + 1 (ii) x4 + x3 + x2 + x +1

(iii) x4 + 3x3 + 3x2 + x +1 (iv) x3 - x2 - (2 +
)x +

Solution:
(i) x3 + x2 + x + 1

Let p(x) = x3 + x2 + x + 1

The zero of x + 1 is - 1.

P(-1) = (-1)3 + (-1)2 +(-1) + 1

         = -1 + 1 - 1 + 1 = 0 By factor theorem, x + 1 is a factor of x3 + x2 + x + 1.

(ii) x4 + x3 + x2 + x + 1

Let p(x) = x4 + x3 + x2 + x+1

The zero of x + 1 is -1.

P(-1) = (-1)4+(-1)3+(-1)2+(-1) + 1

         = 1 - 1 + 1 - 1 = 1 0 By factor theorem, x + 1 is not a factor of x4 + x3 + x2 + x +1.

(iii) x4 + 3x3 + 3x2 + x + 1

Let p(x) = x4 + 3x3 + 3x2 + x + 1

The zero of x + 1 is - 1.


p(- 1) = (- I)4 + 3(- 1)3 + 3(-1)2 + (-1) + 1

          = 1 - 3 + 3 - 1 + 1 = 1 0

By factor theorem, x + 1 is not a factor of x 4 + 3x 3 + Sx2 + x + 1.


(iv) Let p(x) = x 3 - x 2 - (2 +)x +

The zero of x + 1 is - 1.


p(-1) = (-1)3 - (-1)2 - (2 + )(-1) + By factor theorem, x + 1 is not a factor of x 4 -x 3 - (2 + )x + .

Question-14

Use the factor theorem to determine whether g(x) is a factor of P(x) in each of the following cases:
(i) P(x) = 
2x3 + x2 - 2x - 1, g(x) = x + 1

(ii) P(x) = x3 + 3x2 + 3x + 1, g(x) = x + 2

(iii) P(x) = x3 - 4x2 + x + 6, g(x) = x - 3

Solution:
(i)  p(x) = 2x3 + x2 - 2x - 1, g(x) = x + 1

     g(x) = 0

⇒ x + 1 = 0 ⇒ x = - 1


∴ Zero of g(x) is - 1.

Now, p(-1) = 2(-1)3 + (-1)2 – 2(-1) - 1

                 = -2 + 1 + 2 – 1 = 0

By factor theorem, g(x) is a factor of p(x).


(ii) p(x) = x3 + 3x2 + 3x + 1,g(x) = x + 2

     g(x) = 0

⇒ x + 2 = 0 ⇒ x = -2

Zero of g(x) is -2.

Now, p(-2)
            = (-2)3 + 3(-2)2 + 3(-2) + 1

            = -8 + 12 - 6 + 1 = - 1 ≠ 0

∴ By factor theorem, g(x) is not a factor of p(x).


(iii) p(x) = x3 - 4x2 + x + 6, g(x) = x - 3

      g(x) = 0

  ⇒ x – 3 = 0 ⇒ x = 3

Zero of g(x) is 3.

Now, p(3)
             = (3)3 - 4(3)2 + 3 + 6

             = 27 - 36 + 3 + 6 = 0

By factor theorem, g(x) is a factor of p(x).

Question-15

Find the value of  k , if x - 1 is a factor of P(x) in each of the following cases:

(i) P(x) = x2 + x + k ,           (ii) P(x) = 2x2 + kx +
(iii) P(x) = kx2 -x + 1,   (iv) P(x) = kx2 - 3x + k

Solution:
(i) p(x)= x2 + x + k

If x - 1 is a factor of p(x), then p(1) = 0        | By Factor Theorem (1)2 + (1) + k = 0 1 + 1 + k = 0

2 + k = 0 k = -2


(ii) p(x) = 2x2+ kx +

If x - 1 is a factor of p(x), then p(1) = 0

2(1)2 + k(1)+ =0 2 + k + =0 k = -(2 +)


(iii) p(x) = kx2 -x + 1

If x - 1 is a factor of p(x), then p(1) = 0     | By Factor Theorem k(1)2 - (1) + 1 = 0


Þ k = - 1 

(iv) p(x) =kx2 - 3x + k

If x - 1
is a factor of p(x), then p(1) = 0

k(1)2 - 3(1) + k = 0 Þ k – 3+ k = 0

Þ  2k - 3 = 0

k = By Factor Theorem

Question-16

Factorise:(i) 12x2 – 7x + 1 (ii) 2x2 + 7x + 3(iii) 6x2 + 5x - 6 (iv) 3x2 - x - 4

Solution:
(i) 12x2 – 7x + 1

12x2 – 7x + 1 = 12x2 – 4x –3x + 1

                    = 4x(3x – 1)-1(3x – 1)

                    = (4x- 1) (3x – 1)

        Let p(x) = 12x2 – 7x + 1

      Then p(x) = 12 = 12q(x)

    Where q(x) = x2 -

By trial, we find that

q =

                               = = = 0 By Factor Theorem,

is a factor of q(x)

Similarly, by trial, we find that

q = = = 0 By Factor Theorm,

is a factor of q(x)

Therefore,

            12x2 – 7x + 1 = 12

                                = 12 = (3x – 1)(4x – 1)

(ii) 2x2 + 7x + 3

2x2 + 7x + 3 = 2x2+7x+3

                  = 2x2 + 6x + x + 3

                  = 2x(x + 3) + 1(x +3)

                  =(x + 3) (2x +1)

      Let p(x) = 2x2 + 7x + 3

    Then p(x) = = 2q(x)

  Where q(x) = x2 +

By trial, we find that

          q(-3) = (-3)2 +

                  = = 0 By Factor Theorem, x – (-3), i.e. (x + 3) is a factor of q(x)

similarly, by trial, we find that

q

                = = 0 By Factor Theorem, x - is a factor of q(x)

Therefore,
2x2 + 7x + 3 = 2(x + 3)

                  = 2(x+3)

                  = (x + 3) (2x + 1)


(iii) 6x2 + 5x - 6

6x2 + 5x - 6 = 6x2 + 9x - 4x – 6

                  = 3x(2x + 3) –2(2x + 3)

                  = (2x + 3) (3x –2)

  Let p(x) = 6x2 + 5x – 6

Then p(x) = 6 = 6q(x)

Where q(x) = x2 +

= By Factor Theorem, x - , i.e., x + is a factor of q(x)

Similarly, by trial, we find that

= By Factor Theorem, x - is a factor of q(x)

Therefore
6x2 + 5x + 6 = 6

                  = 6

                  = (2x + 3) (3x – 2)


(iv) 3x2 – x - 4

3x2 – x – 4 = 3x2 – 4x + 3x – 4

                = x(3x – 4) + 1(3x – 4)

                = (3x – 4) (x + 1)

    Let p(x) = 3x2 – x – 4

  Then p(x) = 3= 3q(x)

Where q(x) = x2 -

By trial, we find that

q

      = = By Factor theorem, x - is a factor of q(x)

Similarly, by trial, we find

q(-1) = (-1)2 - By Factor Theorem, x-(-1), i.e. (x+1) is a factor of q(x).
Therefore
3x2 – x – 4 = 3

                = 3

                = (3x – 4)(x +1)

Question-17

Factorise:
(i) x3 – 2x2 – x +2  (ii) x3 – 3x29x –5

(iii) x3 +13x2 +32x +20  (iv) 2y3 +y2 2y - 1

Solution:
(i) x3 2x2 – x + 2

Let p(x) = x32x2 – x + 2

            = x2
(x 2) – 1 (x – 2)

            =
(x21) (x – 2)

            =
(x + 1) (x – 1) (x – 2)


(ii) x3 – 3x2 –9x –5

Let p(x) = x3 – 3x2 9x – 5
Synthetic Division


x3 – 3x2 –9x –5 = (x2 – 4x – 5) (x + 1)


But x2 – 4x – 5 = x2 – 5x + x – 5

                      = x (x – 5) +1 (x – 5)

(x + 1) (x – 5)

∴ x3 – 3x2 – 9x – 5 =
(x +1)(x+1)(x-5)


(iii) x3 +13x2 +32x +20

Let P(x) = x3 + 13x2 + 32x + 20

By trial, we find that

P(-1) = (-1)3 +13(-1)2+32(-1)+20 = -1 +13 –32+20 = 0

By synthetic Division



x3 + 13x2 + 32x + 20 = (x2 +12x +20 ) (x + 1)

  
But (x2 +12x +20 ) = x2 – 10x -2x +20

                              = x (x – 10) – 2(x – 10)

                              = (x – 2) (x – 10) 


∴ x3 – 13x2 + 32x + 20

                             = (x +1) (x -2) (x -10)


(iv) 2y3 +y2 – 2y - 1


Let p(y) = 2y3 + y2 – 2y – 1

By trial, we find that

     P
(1) = 2(1)3 – 2(1) – 1 = 2 +1 – 2 – 1

            = 0

∴ 
By Factor Theorem, (y – 1) is a factor of p(y)
Now,



2y3 + y22y – 1 = (2y2 +3y +1) (y – 1)

    
  But ,2y2 +3y +1 = 2y2 +2y +y + 1

                                  =
2y(y + 1) +1(y +1)

                                  =
(y + 1)(2y +1)

Question-18

Use suitable identities to find the following products:
(i) (x + 4) (x + 10) (ii) (x + 8)(x – 10) (iii) (3x + 4) (3x – 5)

(iv) (y2
  ) (y2  ) (v) (3 – 2x) (3 + 2x)

Solution:

(i) (x + 4) (x + 10)

(x + 4)(x +10) = x2 +(4 +10)x +(4)(10) ∵ Using Identity IV

                     = x2 + 14x+40


(ii) (x + 8)(x – 10)

(x + 8)(x – 10) = (x + 8)[x +(-10)x + (8) (– 10) ∵Using Identify IV

                      = x2 – 2x – 80


(iii) (3x + 4) (3x – 5)

(3x + 4) (3x – 5) = (3x + 4)(3x + (-5)]

                                                     = (3x)2 + (4 + (-5) (3x) + (4) (-5)   ∵ Using Identity IV
                                                     = 9x2 – 3x - 20

(iv) (y2 + ) (y2 - )

(y2 + ) (y2 - ) = (z + ) (z - ),   Where y2 = z

                          = (z)2 -  ∵ Using Identity III

                          = z2 - = (y2)2 -     ∵
Substituting the value of z

                          = y4 -

(v) (3 – 2x) (3 + 2x) = (3)2 - (2x)2    ∵ Using Identity III

                                   = 9 – 4x2

Question-19

Evaluate the following products without multiplying directly:
(i) 103 × 107 (ii) 95 × 96 (iii) 104 × 96.

Solution:
(i) 103 × 107
103 × 107 = (100 + 3) × (100 + 7)

              = (100)2 + (3 + 7) (100 + (3)(7)

              = 10000+1000 + 21 = 11021.

 

(ii) 95 × 96

95 x 96 = (90 + 5) × (90 + 6)

           = (90)2 + (5 + 6)(90) + (5)(6) | Using Identity IV

           = 8100 + 990 + 30 = 9120

Another Method:
 

95 × 96 = (100- 5) × (100 - 4) = {100 + (- 5)}{100 + (- 4)}

                                          = (100)2 + {(-5) + (-4)}(100) + (-5)(-4)

                                          = 10000-900 + 20

                                          = 9120.

(iii) 104 × 96

104 × 96 = (100 + 4) x (100 - 4) = (100)2 - (4)2 |Using Identity III

             = 10000 - 16 = 9984.

Question-20

Factorise the following using appropriate identities:
(i) 9x2 + 6xy + y2 (ii) 4y2 - 4y + 1 (iii) x2 - 

Solution:
(i) 9x2 + 6xy + y2

9x2 + 6xy + y2 = (3x)2 + 2(3x)(y) + (y)2 = (3x + y)2 = (3x + y)(3x + y)      (Using Identity I)


(ii) 4y2 - 4y + 1

4y2 - 4y + 1 = (2y)2 - 2(2y)(1) + (1)2 = (2y- 1)2 = (2y- 1)(2y - 1)              (Using Identity II)


(iii) x2 -

x2 - = (x)2 - =                                                (Using Identity II)

Question-21

Expand each of the following using suitable identities:
(i) (x + 2y + 4z)2 (ii) (2x - y + z)2 (iii) (-2x + 3y + 2x)2
(v) (3a - 7b - c)2 (v) (-2x + 5y-3z)2 (vi)

Solution:
(i) (x + 2y + 4z)2= (x + 2y + 4z) (x + 2y + 4z) 

           = (x)2 + (2y)2 + (4z)2 + 2(x)(2y) + 2(2y)(4z) + 2(4z)(x) ......... Using Identity V

           = x2 + 4y2 + 16z2 + 4xy + 16yz + 8zx


(ii) (2x – y + z)2= (2x + (-y) + z)2

           = (2x)2 + (- y)2 + (z)2 + 2(2x)(- y) + 2(- y)(z) + 2(z)(2x) ........ Using Identity V

           = 4x2 + y2 + z2 - 4xy -2yz + 4zx


 (iii) (- 2x + 3y + 2z)2 = {(- 2x) + 3y + 2z)}2

            = (- 2x)2 + (3y)2 + (2z)2 + 2(- 2x)(3y) + 2(3y)(2z) + 2(2z)(- 2x)......... Using Identity V

            = 4x2 + 9y2 + 4z2 – 12xy + 12yz - 8zx


(iv) (3a - 7b - c)2 = {3a + (-7b) + (-c)}2

            = (3a)2 + (-7b)2 + (- c)2 + 2(3a)(- 7b) + 2(- 7b)(- c) + 2(- c)(3a)

            = 9a2 + 49b2 + c2 - 42ab + 14bc – 6ca


(v) (-2x + 5y - 3z)2 = {(- 2x) + 5y + (- 3z)}2

            = (- 2x)2 + (5y)2 + (- 3z)2 + 2(- 2x)(5y) + 2(5y)(- 3z) + 2(- 3z)(- 2x)

            = 4x2 + 25y2 + 9z2 - 20xy - 30yz + 12zx


(vi)   

=

=

Question-22

Factorise: (i) 4x2 + 9y2 + 16z2 + 72xy - 24yz - 16xz
(ii) 2x2 + y2 + 8z2 – 2xy + 4
 yz - 8zx.

Solution:
(i) 4x2 + 9y2 + 16z2 + 12xy – 24 yz - 16xz

4x2 + 9y2 + 16z2 + 12xy - 24yz - 16xz

= (2x)2 + (3y)2 + (- 4z)2 + 2(2x)(3y) + 2(3y)(- 4z) + 2(- 4z)(- 2x)

= {2x + 3y + (- 4z)}2

= (2x + 3y - 4z)2

= (2x + 3y- 4z)(2x + 3y - 4z) 


(ii) 2x2 + y2 + 8z2 – 2 xy + 4 yz - 8zx

2x2 + y2 + 8z2 – 2
 xy + 4 yz - 8zx

= (-2
 x)2 + y2 + (2 z)2 + 2(- x)y + 2y(2 z)y + 2(2 z)(-  x)

= (-
 x + y + 2 z)2

= (-
 x + y + 2 z)( - x + y + 2 z)

Question-23

Write the following cubes in expanded form:
(i)(2x+1)3 (ii) (2a - 3b)3 (iii) (iv)

Solution:
(i)(2x +1)3 = (2x)3 + (1)3 + 3(2x)(1)(2x + 1)          |Using Identify VI

                = 8x3 +1 +6x(2x + 1) = 8x3 + 1+12x3 +6x

                = 8x3 + 12x3 + 6x +1

(ii) (2a – 3b)3 = (2a)3 – (3b)3 – 3(2a)(3b)(2a – 3b)      |Using Identify VII

                     = 8a3 – 27b3 – 18ab(2a – 3b)

                     = 8a3 – 27b3 – 36a2b + 54ab2

(iii) =

=

=


(iv) |Using Identity VII

= x3 -

= x3 -

= x3 – 2x2y+

Question-24

Evaluate the following using suitable:(i) (99)3 (ii) (102)3 (iii) (998)3

Solution:
(i) (99)3 = (100 -1)3 = (100)3+(1)3 - 3(100)(1)(100 -1) ...........Using Identity VII

             = 1000000 - 1 - 300(100 - 1) = 1000000 - 1 - 30000 + 300
             = 970299

(ii) (102)3 = (100 + 2)3 = (100)3 + (2)3 + 3(100)(2)(100 +2) ..............Using Identity VI

               = 1000000 + 8 +(600(100 +2) = 1000000 + 8 + 60000+ 1200
               = 1061208

(iii) (998)3 = (1000 - 2)3
                = (1000)3 - (2)2 - 3(1000)(2)(1000 - 2) ...................... Using Identity VII

                = 1000000000 - 8 - 6000(1000 - 2)


                = 1000000000 - 8 - 6000000 + 12000

                = 994011992.

Question-25

Factorise each of the following:
 (i) 8a3 + b3 + 12a2b + 6ab2 (ii) 8a3 -b3 - 12a2b + 6ab2


(iii) 27 - 125a3 - 135a + 225a2 (iv) 64a3 - 27b3 –144a2b + 108ab2

(v) 27p3
  -

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

                                               = (2a + b)3

                                               = (2a + b)(2a + b)(2a + b)

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

                                              = (2a - b)3        .....................  (Using Identity VII)

                                              = (2a - b)(2a - b)(2a - b)


(iii) 27 - 125a3 - 135a + 225a2 = 27 - 125a3 - 135a + 225a2

                                                     = (3)3 - (5a)3 - 3(3)(5a)(3 - 5a)  

                                                     = (3 - 5a)3


                                                     = (3 - 5a)(3 - 5a)(3 - 5a)

(iv)  64a3 - 27b3 - 144a2 + 108ab2 = (4a)3 - (3b)3 - 3(4a)(3b) (4a - 3b)

                                                           = (4a - 3b)3 ........................(Using Identity VII)

                                                           = (4a - 3b)(4a - 3b)(4a - 3b)

(v) 27p3 - = (3p)3 -

                                                      =  .........................(Using Identity VII)

                                                      =

 

Question-26

Verify: (i) x3 + y3 = (x + y)(x2 -xy + y2)  (ii) x3 - y3 = (x - y)(x2 + xy + y2).

Solution:
(i) We know that
 (x + y)3 = x3+ y3+ 3xy(x + y) .................... Using Identity VI
⇒ x3 + y3 = (x + y)3 - 3xy(x + y) ⇒ x3 + y3 = (x + y)[(x + y)2 - 3xy)
⇒ x3 +y3 = (x + y)(x2 + 2xy + y2 - 3xy) ..............  (Using Identity I)
⇒x3 + y3 = (x + y) (x2 - xy + y2)

(ii) We know that
(x - y)3 = x3 - y3 - 3xy(x - y)                                (Using Identity VII)
⇒ x3 - y3 = (x - y)3 + 3xy(x - y) ⇒ x3 - y3 = (x - y)[(x - y)2 + 3xy)
⇒ x3 - y3 = (x - y)(x2 - 2xy + y2 + 3xy) ..............  (Using Identity I)
⇒ x3 + y3 = (x - y) (x2 + xy + y2)

 

Question-27

Factorise each of the following:
(i) 27y3 + I25z3 (ii) 64m3 - 343n3.

 

Solution:
(i) 27y3 + 125z3
27y3 + I25z3  = (3y)3 + (5z)3 = (3y + 5z){(3y)2- (3y)(5z) + (5z)2}

                     = (3y + 5z)(9y2 - 15yz + 25z2)

(ii) 64m3 - 343n3
64m3 – 343n3 = (4m)3 - (7n)3 = (4m - 7n) {(4m)2 + (4m)(7n) + (7n)2}


                       = (4m - 7n)(l6m2 + 28mn + 49n2).
 

Question-28

Factorise : 27x3 + y3 + z3 - 9xyz.

Solution:
27x3 + y3 + z3 - 9xyz = (3x)3 + (y)3 + (z)3 - 3(3x)(y)(z)
                               = {(3x)2 + (y)2 + (z)2 - (3x)(y) - (y)(z) - (z)(3x)}         (Using Identity VIII) 

                               = (3x + y + z) (9x2 +y2 + z2 - 3xy - yz - 3zx).

Question-29

Verify that x3 + y3 + z3 - 3xyz = (x + y + z)[(x - y)2 + (y - z)2 + (z - x)2].

Solution:
L.H.S.
 x3 + y3 + z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)................(Using Identity VIII)
                            = (x+ y + z){2(x2 + y2 + z2 -xy - yz - zx)}
                            = (x + y + z)(2x2 + 2y2 + 2z2 - 2xy -2yz - 2zx)
                            = (x + y + z){(x2 - 2xy + y2) + (y2- 2yz + z2) + (z2 -2zx + x2)}
                            = (x + y + z)[(x - y)2 + (y - z)2 + (z - x)2 ].................(Using Identity II)                                                

Question-30

If x + y + z = O, show that x3 + y3 + z3 = 3xyz

Solution:
We know that
x3 +y3+z3 - 3xyz = (x + y + z) (x2 + y2 + z2 - xy - yz - zx)       (Using Identity VIII)
                          = (0) (x2 + y2 + z2 - xy - yz - zx)                    (∵ x + y + z = 0 )
                          = 0
⇒  x3 + y3 + z3 = 3xyz

Question-31

Without actually calculating the cubes, find the value of each of the following:
(i) (- 12)3 + (7)3 + (5)3     (ii) (28)3 + (-15)3 + (-13)3.

Solution:
(i) (-12)3 + (7)3 + (5)3
(-12)3+(7)3+(5)3 = 0 + 3(-12)(7)(5) ...........( (- 12) + (7) + (5) = 0) Using identity VIII 
                          = -1260

(ii) (28)3+(-15)3+(-13)3 = 0 + 3(28)(-15)(-13)...............  (  (28)+(-15)+(-13) = 0) - Using identity VIII

                                    = 16380

Question-32

Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:
(i) Area : 25a2 - 35a + 12 (ii) Area: 35y2 + 13y - 12.

Solution:
(i) 25a2 - 35a + 12 = 25a2 - 20a - 15a + 12
                   = 5a(5a - 4) - 3(5a - 4)
                   = (5a - 4)(5a - 3)

∴ The possible expressions for the length and breadth of the rectangle are
5a - 3 and 5a + 4



(ii) 35y2 + 13y - 12 = 35y2 + 28y – 15y - 12
         = 7y (5y + 4) - 3(5y + 4)
         = (5y + 4)(7y-3)

The possible expressions for the length and breadth of the rectangle are
7y - 3 and 5y + 4.

Question-33

What are the possible expressions for the dimensions of the cuboids whose volumes are given below ?
(i) Volume: 3x2 - 12x (ii) Volume :12ky2 + 8ky - 20k.

Solution:
(i)3x2 – 12x = 3x(x - 4)
∴ The possible expressions for the dimensions of the cuboid are 3, x and x-4.


(ii) 12ky2 + 8ky - 20k = 4k(3y2 + 2y - 5)
                                     = 4k(3y2 + 5y-3y-5)
                                     = 4k{y(3y + 5) - 1(3y + 5)}
                                     = 4k(3y + 5)(y- 1)
The possible expressions for the dimensions of the cuboid are 4k, 3y + 5 and y - 1




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