CLASS 9 - CHAPTER 2 - Polynomials

                                                   EXERCISE 2.1

1. Which of the following expressions are polynomials in one variable and which are

not? State reasons for your answer.

(i)   4x² - 3x  + 7   => In one variable. Only unknown in this expression is x.

(ii)  y² + √2           => In one variable.Only unknown is y

(iii) 3√t + √2         => In one variable.Only unknown is t

(iv) y + 2/y            => In one variable.Only unknown is t

(v) x¹⁰+y³ + t⁵⁰     => Not in one variable.There are three variables x, y and t.

2. Write the coefficients of x² in each of the following:

(i) 2 + x² + x          => coefficient of  x² is 1.

(ii) 2 – x² + x³        => coefficient of  x² is -1.

(iii) (π/2)x² + x       => coefficient of  x² is π/2.

(iv)√2x - 1            => coefficient of  x² is 0.

3. Give one example each of a binomial of degree 35, and of a monomial of degree 100.

  Answer:

1.one example each of a binomial of degree 35 is x³⁵ + 1

2.one example each of a monomial of degree 100 is x^100.

4. Write the degree of each of the following polynomials:

(i) 5x³+ 4x²+7x        => degree is 3

(ii) 4-y²                    => degree is 2

(iii) 5t – √7               => degree is 1

 (iv) 3                       => degree is 0

5. Classify the following as linear, quadratic and cubic polynomials:

(i)  x² + x                      => Quadratic as degree is 2.

(ii) x-x³                         => Cubic as degree is 3.

(iii) y + y² + 4               =>  Quadratic as degree is 2.

(iv) 1 + x                      => Linear as degree is 1.

(v) 3t                            => Linear as degree is 1.

 (vi) r²                           =>  Quadratic as degree is 2.

(vii) 7x³                          => Cubic as degree is 3.

                                 EXERCISE 2.2

1. Find the value of the polynomial 5x - 4x² + 3 at

Let P(X)=  5x - 4x² + 3

(i) x = 0 

Then P(0) = 5(0) - 4(0) + 3 = 3
 

(ii) x = –1

P(-1) = 5(-1) - 4(-1)²+ 3 = -5 - 4 +3 = -6

 (iii) x = 2

P(2) = 5(2)  - 4 (2²) + 3 = 10 - 4x4 +3 = 10 - 16 +3 = -3

2. Find p(0), p(1) and p(2) for each of the following polynomials:

 
(i) p(y) = y²-y+1

Put y = 0

Then 

P(0) = 0 -0 +1 = 1

P(1) = 1 - 1 + 1 = 1

P(2) = 2² -2 + 1 = 4 - 2 + 1 = 3

 

(ii) p(t) = 2+ t +2t² - t³

P(0) = 2 + 0 + 2(0) - (0) = 2

p(1) = 2 + 1 + 2.1² - 1³ = 2 + 1 +2 -1= 5 - 1 = 4      

p(2) = 2 + 2 + 2.2² - 2³ = 4 + 8 - 8 = 4

(iii) p(x) = x³

p(0 ) = 0

p(1) = 1³ = 1

p(2) = 2³ = 8

(iv) p(x) = (x – 1) (x + 1)

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

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

p(2) = (2-1)(2+1) = 2

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

We have P(-1/3) = 3(-1/3) +1 = -1 + 1 = 0. Therefore verified to be true.

(ii) p(x) = 5x – π, x =4/5

We have p(4/5) = 5(4/5) -  π = 4 - π ≠ 0 . Therefore verified to be false.

(iii) p(x) = x² - 1 , x = 1, –1 

P(1) =  1 - 1 = 0.                                Therefore verified to be true. 

and p(-1) =  (-1)² - 1 = 1 - 1 = 0     Therefore verified to be true. 

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2

P(-1) = (-1+1)(-1-2) = (0)(-3) = 0.        Therefore, verified to be true.

P(2) = ( 2 +1)(2-2) = (3)(0) = 0.            Therefore verified to be true.

(v) p(x) = x2, x = 0 

p(0) = 0         Therefore verified to be true.

(vi) p(x) = lx + m, x = –m/l

P( -m/l) = l(-m/l) + m = -m+m = 0 . Therefore verified to be true.

(vii) p(x) = 3x² - 1, x =-1/√3,2/√3

p(-1/√3) = 3(-1/√3)² - 1  = 3/3 - 1 = 1-1 = 0  Therefore verified to be true.

 p(2/√3) = 3(2/√3)² - 1 = 4 - 1 = 3   Therefore verified to be true.

(viii) p(x) = 2x + 1, x =1/2

P(1/2) = 2(1/2) + 1 = 1 + 1 =0      Therefore verified to be true.

4. Find the zero of the polynomial in each of the following cases:
(i) p(x) = x + 5

Put x+ 5 = 0 

= > x = -5

Therefore -5 is the zero of p(x).

(ii) p(x) = x – 5 

put x - 5 = 0.

=> x = 5

Thus, 5 is the zero of p(x). 

(iii) p(x) = 2x + 5

put 2x + 5 = 0

=> x = -5/2

Thus -5/2 is the zero of p(x)

(iv) p(x) = 3x – 2 

Put 3x - 2 = 0

=>  x = 2/3

Thus -2/3 is the zero of p(x) 

(v) p(x) = 3x

Put 3x = 0

=>  x =0

Thus 0 is the zero of p(x)  

 

 (vi) p(x) = ax, a ≠ 0

Put ax = 0

=> x =0    ( given that  a ≠ 0)

Thus, 0 is the zero of p(x) 

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.

Put cx + d = 0

=>  x = -d/c

Thus,   -d/c is the zero of p(x)

                                   EXERCISE 2.3

1. Find the remainder when x³ + 3x² + 3x + 1 is divided by

 Let P(x) = x³ + 3x² + 3x + 1

(i) x + 1

Put x + 1 = 0

=> x = -1

p(-1) = (-1)³ + 3(-1)² + 3(-1) + 1 = -1 + 3 -3 +1 = 0 

Thus remainder is  0 when p(x) is divided by x + 1

(ii) x –1/2

put x - 1/2 = 0

Thus  x = 1/2

and p(-1/2) =  (-1/2)³ + 3(-1/2)² + 3(-1/2) + 1 = -1/8 + 3/4  -3/2 + 1

       = (-1 + 6 -12 +8)/8

       = 1/8

(iii) x 

 Put x = 0

=> p(0) = 1 

Thus, remainder is 1 when p(x) is divided by x.

(iv) x + π

Put x + π =  0 => x = -π

Thus, p(π) =  (-π )³ + 3(-π )² + 3(-π ) + 1 = -π ³ + 3π ²  - 3π  + 1 . This is the remainder.

(v) 5 + 2x

Put 5 + 2x = 0

thus. x= -5/2

Now p(-5/2 ) =  (-5/2 )³ + 3(-5/2 )² + 3(-5/2 ) + 1

                        =  -125/8 +  75/4 - 15/2 + 1 

                        = (-125 + 150 - 60 + 8)/8

                        =  -27/8     

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

Let p(x) = x³ - ax² + 6x  -a

Put x -a = 0 => x =a

Then find  p(a)

p(a) = a³ - a.a² + 6a  -a = a³ - a³ + 6a  - a= 5a

Therefore, 5a will be the remainder when p(x) is divided  by (x+a)

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

 Let p(x) = 3x³ + 7x.

now put 7 + 3x = 0

Thus  x = -7/3

Now find p(-7/3)

p(-7/3)  = 3(-7/3)³  + 7(-7/3) =  -343/9  -  49/3 =  (-343 - 147)/9 = -490/9

Thus, -490/9 is the remainder.

And therefore 7+3x is not a factor of p(x).

                                    EXERCISE 2.4

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

To find whether (x+1) is a factor of the following polynomials, first we put x+1 = 0.

Thus x = -1

Now, we will put x = -1 in each of the polynomials and for those polynomials which become x zero have (x+1) as the factor.

(i) x3 + x2 + x + 1

   put x=-1

   x³+x²+x+1 =  -1 +1 -1 + 1 = 0. Thus (x+1) is a factor. 

(ii) x⁴ + x³+x²+x+1

  put x = -1

  x⁴ + x³+x²+x+1 = (-1⁾⁴ + (-1)³ + (-1)²+(-1) + 1 = 1 - 1 + 1 - 1 + 1 = 1

Thus, (x+1) is not a factor.

(iii) x⁴ +3 x³+3x²+x+1

 Put x = -1

 x⁴ +3 x³+3x²+x+1 = (-1)⁴ +3(-1)³+3(-1)²+(-1)+1 = 1 - 3 + 3 -1 + 1 = 1

Thus. (x+1) is not a factor.

(iv) x³  - x² -( 2 + √2)x +√2.

  

 put x = -1

x³  - x² -( 2 + √2)x +√2 = -1 - 1 - (2+√2)(-1)  + √2 = -1 - 1 + 2 + √2 + √2 = 2√2

Thus, (x+1 ) is not a factor.

2. 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

Using Factor Theorem,

First put g(x) = 0

=>  x + 1 = 0

=>  x = -1

then,

find p(-1) =  2(-1)³ + (-1)²  - 2(-1) - 1 = - 2 + 1 +2 - 1

                  = -3 + 3

                  = 0

 Hence, g(x) is a factor of p(x).

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

Using factor theorem,

First put  g(x) = 0

=> x = -2

Then find 

p(-2) = (-2)³ + 3(-2)²  + 3(-2) + 1

          = -8 + 12 - 6 + 1

          = -14+13

          = -1      ≠ 0

Hence, g(x)  is not a factor of p(x).

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

 Put g(x) = 0

=> x = 3

Then find p(3)

p(3) = (3)³ - 4(3)² + 3 +6

        = 27 - 36    + 3 + 6

        = 36 -36 = 0

Thus g(x) is a factor of p(x).

3. 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 

 Given that (x - 1)  is a factor

=>  p(1) = 0

=>  x² + x + k = 0

=>   1 + 1 + k = 0

=> k = -2

(ii) p(x) = 2x² + kx + √2

 Given that (x - 1)  is a factor

=>  p(1) = 0

=>  2.1² + k + √2 = 0

=>  2 + k +  √2 =0

=> k = -(2 +  √2)

(iii) p(x) = kx² - √2 x + 1

 Given that (x - 1)  is a factor

=>  p(1) = 0

=>  k.1² - √2. 1 + 1 = 0

=>  k - √2 + 1 = 0

=> k =   √2 - 1

(iv) p(x) = kx² - 3x  + k

 Given that (x - 1)  is a factor

=>  p(1) = 0

=>  kx² - 3x  + k = 0

=>  k - 3.1 + k = 0

=> 2k - 3 = 0

=> k = 3/2

 
4. Factorize :

(i) 12x² - 7x + 1

Solution:

Here,

Product of Coefficient of  x² and the constant term is 12.

Now we have to break the co-efficient of x (-7) into two numbers such that the sum of those numbers is -7 and product is 12.We could do this by hit and trial method.By HIT and TRIAL  those numbers come out to be  -3 and -4. 

Thus ,

12x² - 7x + 1 can be written as 12x² - ( 4 + 3) x + 1

    12x² - ( 4 + 3) x + 1

 = 12x² - 4x - 3 x + 1

 =  4x(3x  - 1) - (3x -1)

 = (4x-1)(3x-1)

 
(ii) 2x² +7x + 3

Solution:

    2x² + 7x + 3 can be written as 

=  2x² + 6x + x +3

=  2x( x + 3) + ( x + 3)

=  (2x +1)( x + 3) 

(iii) 6x² +5x -6

 Solution:

 6x² +5x -6 can be written as 

= 6x² +9x - 4x -6

= 3x(2x + 3) - 2(2x+3)

= (3x-2)(2x + 3)

(iv) 3x² -x -4

Solution:
   3x² -x -4

= 3x² -4x+3x -4

= x(3x-4)  + (3x-4)

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

5. Factorize :
(i) x³-2x² - x +2

 Solution:
The constant term here is 2.Factors of the constant term 2 are ±1, ±2
Let p(x) =  x³-2x² - x + 2

Then, p(x=1) = 1³-2.1² -1 + 2
                   =  1 - 2 - 1 + 2
                   = 0
Thus, (x-1) is a factor of p(x)
And p(x=-1) =    (-1)³-2.(-1)² -(-1)1 + 2
                       =     -1   - 2 +1 + 2
                       =    0
Thus, (x+1) is a factor of p(x)
Now,
 p(x=2) = 2³-2.2² -2 + 2   
              = 8 - 8 - 2 + 2
              = 0
Thus, (x-2) is a factor of p(x).
A cubic equation has 3 roots. And we have found them all.
Thus,  p(x) = (x+1)(x-1)(x-2) 


(ii)  x³ -3x² - 9x - 5

Solution:
Let p(x) = x³ -3x² - 9x - 5

Factors of -5(the constant term) are ±1,±5
Now,
p(x= -1)   = (-1)³ -3(-1)² - 9(-1) - 5 
                  = -1 -3 + 9 -5 
                  = -9 +9  
                  = 0
Thus, (x+1) is a factor of p(x)

Now, dividing p(x) by (x+1) to get the third factor.

(iii) x³ + 13x² + 32x + 20
Solution:
 The constant term is 20 and its factors are ±1,±2,±4±5,±10,±20
By hit and trial, we can find that the one of the factors of x³ + 13x² + 32x + 20 is (x+1) as
when we find
  P(-1) = (-1)³ + 13(-1)² + 32(-1) + 20 = -1 +13 -32 + 20
                         =  -33+33 =0

Now divide the polynomial x³ + 13x² + 32x + 20 by (x+1) to get the rest of the factors.

We will get x²+12x+20 as the quotient. Now, we can factorize it to get
(x+10)(x+2).
Thus,  x³ + 13x² + 32x + 20 = (x+1)(x+2)(x+10)

 
(iv) 2y³+y² -2y -1

By Trial and error method the   2y³+y² -2y -1 is equal to zero for y = -1
P(-1) = 2(1)³+(1)² -2(1) -1
           = 2 +1 - 2 - 1
           = 0
Thus,  (y+1) is a factor of 2y³+y² -2y -1. Now, dividing 2y³+y² -2y -1 by (y-1) to get the other two factors.

Thus, 2y³+y² -2y -1 = (y-1)(2y²+ 3y +1) 

Again, 2y²+ 3y +1 = 2y²+ 2y + y +1 = 2y(y+1) + (y +1) = (2y+1)(y+1)

Thus,
 2y³+y² -2y -1 = (y-1)(2y+1)(y+1)

                                   EXERCISE 2.5

1. Use suitable identities to find the following products:

(i) (x + 4) (x + 10) 
Solution:
  (x + 4) (x + 10)  = x² + (4+10)x +40         ( using (x+a)(x+b)= x² + (a+b)x +ab)
                                    =  x² + 14x + 40

(ii) (x + 8) (x – 10) 
 Solution:
(x + 8) (x – 10)  = x² + (8-10)x -80             ( using (x+a)(x+b)= x² + (a+b)x +ab)
                                 = x² - 2 x - 80

(iii) (3x + 4) (3x – 5)
Solution:
Let 3x = y
Now,
(3x + 4) (3x – 5)=  (y + 4) (y – 5)
                                  =  y² + (4-5)y -20
                                  =  y²  - y -20
Now, put y =3x
Thus,
(3x + 4) (3x – 5) = (3x)²  - 3x -20                  
                                   = 9x²  - 3x -20


(iv)   (y2 +3/2) (y2 –3/2)
Solution:
    
 = (y²  + 3/2)(y² - 3/2)
       = y⁴ - (3/2)²                            (Using    (x-y)(x+y) = x²-y²)
       = y⁴   - 9/4


(v) (3 – 2x) (3 + 2x)
Solution:
      = 3² - (2x)²                                (Using    (x-y)(x+y) = x²-y²)
      = 9 - 4x²
 
  2. Evaluate the following products without multiplying directly:
(i) 103 × 107
     Solution:
     103 × 107 = (100+3)(100+7)
                           = 100² + (3+7)100 + 21              ( using (x+a)(x+b)= x² + (a+b)x +ab)
                           = 10000 + 1000 +21
                           = 11021 
(ii) 95 × 96
      Solution:
      95 × 96 = (100-5)(100-4)
                       = 100² + (-5-4)100 + 20              ( using (x+a)(x+b)= x² + (a+b)x +ab)
                       = 10000 - 900 + 20
                       = 9120

 (iii) 104 × 96
        Solution:
        104 x 96 = (100+4)(100-4)
                           = 100² + (4-4)100 -16              ( using (x+a)(x+b)= x² + (a+b)x +ab)
                           = 10000 - 16
                           = 9984
        

3. Factorise the following using appropriate identities:
(i) 9x² + 6xy +y²
  Solution:
    (3x)² + 2(3x)y + y²                               Using (a+b)² = a² + 2ab + b²
   = ( 3x + y) ²

(ii) 4y² – 4y + 1
    Solution:
     4y² – 4y + 1 = (2y)² - 2 (2y)(1)  + 1²
                                = (2y-1)²                         Using (a-b)² = a² - 2ab + b²


 (iii) x² –y² /100
Solution:
        x² –y² /100
    = x² –(y /10)²
    = (x - y/10)(x+y/10)                              Using (a-b)(a+b) = a² - b²
 

4. Expand each of the following, using suitable identities:

(i)  (x+2y+4z)²
Solution:
     (x+2y+4z)²
=  x² + (2y)² + (4z)² + 2(x)(2y) + 2(2y)(4z) + 2 (x)(4z)
=  x² + 4y² + 16z² + 4xy + 16yz + 8xz

(ii) (2x – y + z)²
Solution:
     (2x - y+z)²
=  (2x)² + (-y)² + (z)² + 2(2x)(-y) + 2(-y)(z) + 2 (2x)(z)
=  4x² + y² + z² - 4xy - 2yz + 4xz

(iii)  (-2x+3y+2z)²
Solution:
=  (-2x)² + (3y)² + (2z)² + 2(-2x)(3y) + 2(3y)(2z) + 2(-2x)(-2z)
=  4x² + 9y² + 4z² -12xy + 12yz + 8xz

(iv) (3a – 7b – c) ²
 Solution:
=  (3a)² + (-7b)² + (-c)² + 2(3a)(-7b) + 2(-7b)(-c) + 2(3a)(-c)
=  9a² + 49b² + c² -42xy + 14yz -6ac

(v) (–2x + 5y – 3z)²
Solution:
 =  (-2x)² + (5y)² + (-3z)² + 2(-2x)(5y) + 2(5y)(-3z) + 2(-2x)(-3z)
=  4x² + 25y² + 9z² -20xy -30yz +12xz

(vi) [(1/4)a -(1/2)b + 1) ]²
Solution:
=  ((1/4)a)² + (-(1/2)b)² + (1)² + 2((1/4)a)(-(1/2)b) + 2(-(1/2)b)(1) + 2((1/4)a)(1)
=  a²/16 + b²/16 + 1 -  (1/4)ab -b +(1/2)a


5. Factorise:

(i)4x²+9y²+16z²+12xy -24yz -16xz
Solution:
     4x²+9y²+16z²+12xy -24yz -16xz
=  (2x)²+(3y)²+(4z)²+2(2)(3)xy -2(3)(4)yz -2(2x)(4z)
= (2x + 3y - 4z)(2x + 3y - 4z)

 (ii) 2x²+y²+ 8z² -2√2xy +4√2yz -8xz
Solution:
(-√2x)²+(y)²+ (2√2z)² -2√2xy +4√2yz -8xz 
= (√2x +y + 2√2z)(√2x +y + 2√2z)

6. Write the following cubes in expanded form:
(i)(2x+1)³
Solution:
(2x+1)³ = (2x)³ +  1³ + 3(2x)(1)(2x +1)   ( using (a+b)³ = a³ + b³ + 3ab(a+b))
                 =  8x³     + 1 +  6x(2x+1)
                 =  8x³     + 1 + 12x² + 6x
                 =  8x³+ 12x² + 6x + 1 

 (ii) (2a – 3b)³
Solution:
(2a – 3b)³ = (2a)³ + (-3b)³ + 3(2a)(-3b)(2a-3b)
                              = 8a³  - 27b³  - 36a²b + 54ab²
 (iii)[ (3/2)x + 1]³
Solution:
[ (3/2)x + 1]³ = (3x/2)³ + 1³ + 3(3x/2)(1){(3/2)x + 1}
                            = 27x³/8 + 1 + 9x/2{3x/2 + 1 }
                            = 27x³/8 + 1 + 27x²/4 + 9x/2
                            =27x³/8  + 27x²/4 + 9x/2  + 1
 
(iv)[x- 2y/3]³
Solution:
[x- 2y/3]³ = (x)³ +(-2/3y)³ + 3(x)(-2y/3){x -2y/3}
                      = x³ -  (8/27 )y³- (6/3)x²y + (12/9)xy²
                      = x³ -  (8/27)y³- 2x²y + (4/3)xy²

  
7. Evaluate the following using suitable identities:

(i)       (99) ³
Solution:
      (99) ³ =    (100-1) ³ = (100)³ - 1³ - 3(100)(-1) (100-1)
                   = 1000000 - 1 -300(99) = 1000000 - 1 - 29700
                   = 970299
(ii)     (102)³  
 Solution:
           (102)³  = (100+ 2) ³ = 1000000 + 2 ³ + 3(100)(2)(100+2)
                           = 1000000 + 8 + 60000 + 1200
                          = 1061208

 (iii)   (998)³
Solution:
            (998)³  =  (1000 - 2)³ = (1000)³  - 2³  -3(1000)(-2)(1000-2)
                            =  1000000000 - 8 - 6000(998)
                            =  1000000000 - 8 - 5988000
                            =  994011992


 8. Factorise each of the following:

(i) 8a³ + b³ + 12aa²b + 6aba²
  Solution:
 8a³ + b³ + 12aa²b + 6aba² = (2a)³ + (b)³ + 2(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 =  (2a-b)(2a-b)(2a-b)


(iii) 27 – 125a³ – 135a + 225a² = (3)³ - (5a)³ - 3(3)²(-5a) +3(3)(5a)²
                                                                 = (3-5a)³
                                                                 = (3-5a)(3-5a) (3-5a)
(iv) 64a³ – 27b³ – 144a²b + 108ab² = (4a)³ - (3b)³ -3(4a)(-3b)(4a-3b)
                                                                            = (4a-3b)³ = (4a-3b) (4a-3b) (4a-3b)

 (v) 27p³ –1/216–  (9/2)p² +(1/4)p = (3p)³  - (1/6)³ -3(3p)(-1/6)(3p-1/6)
                                                                           = (3p - 1/6)³ = (3p - 1/6)(3p - 1/6)(3p - 1/6)





9. Verify :
 (i) x³ + y³ = (x + y) (x² – xy + y²)
Solution:
We know,
(x+y)³ = x³ + y³ + 3xy(x+y) 
Thus,
                 x³ + y³ = (x+y)³ - 3xy(x+y) 
                                 = (x+y)[(x+y)² - 3xy]
                                 = (x+y)[x²+y² + 2xy - 3xy]      
                                 = (x+y)[x²+y² - xy ]


 (ii)  x³ – y³ = (x – y) (x² + xy + y²)
Solution:
We know,
(x-y)³ = x³ - y³ - 3xy(x-y) 
Thus,
                 x³ - y³ = (x-y)³ + 3xy(x-y) 
                                 = (x-y)[(x-y)²+ 3xy]
                                 = (x-y)[x²+y² - 2xy + 3xy]      
                                 = (x-y)[x²+y² + xy ]       

10. Factorise each of the following:

(i) 27y³ + 125z³
Solution:
 27y³ + 125z³ = (3y)³ + (5z)³
                             = ( 3y+5z)(9y²+25z² - 15yz)          Using x³ + y³ = (x+y)[x²+y² - xy ]

 (ii) 64m³ – 343n³ 
Solution:
    = (4m)³ - (7n)³
                                       = (4m-7n) (16m²+49n²+28mn)    Using x³ - y³ = (x-y)[x²+y² + xy ]

[Hint : See Question 9.]
11. Factorise : 27x³ + y³ + z³ – 9xyz

Solution: 
27x³ + y³ + z³ – 9xyz
= {(3x)³ + y³ + z³ – 3(3x)yz}
Using  x³+ y³+z³ -3xyz= (x+y+z)[x²+y² + z²- xy-yz- zx ], we have

 = (3x+y+z){(3x)² +y²+z²-(3x)y -yz-(3x)z} 
=  (3x+y+z)(9x² + y² + z² -3xy - yz-3xz)


 12. Verify that x³ + y³ + z³ – 3xyz = (1/2)(x+y+z ) [(x-y )²+ (y-z )²+ (z-x)²]
Solution: 
We know , x³+ y³+z³ -3xyz= (x+y+z)[x²+y² + z²- xy-yz- zx ] 
Now, Mutliply and divide the RHS by 2, we get
x³+ y³+z³ -3xyz= (1/2)(x+y+z)[2x²+2y² + 2z²- 2xy-2yz- 2zx ]
Rearranging to get 
                                 = (1/2)(x+y+z)[x²+y²- 2xy+y² + z²-2yz+ z²+ x²- 2zx ] 
                                 = (1/2)(x+y+z)[(x-y)²+(y - z)²+( z²- x²) ] 

Hence, verified.


13. If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.
Solution:
Using   x³+ y³+z³ -3xyz= (x+y+z)[x²+y² + z²- xy-yz- zx ]
Given that (x+y+z)= 0
Thus,

 x³+ y³+z³ -3xyz=0
 => x³+ y³+z³ = 3xyz


14. Without actually calculating the cubes, find the value of each of the following:
(i) (–12)³ + (7)³ + (5)³
Solution:
We know if (x+y+z) = 0 , Then x³ + y³ + z³ = 3xyz.
Here, 
-12+7+5  = 0
Therefore,
(–12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260


 (ii) (28)³ + (–15)³ + (–13)³

Solution:
We know if (x+y+z) = 0 , Then x³ + y³ + z³ = 3xyz.
Here, 
28-15-13  = 0
Therefore,
(28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13) =16380


15. 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  
Solution:Area of a rectangle = length x breadth.
      Area = 25a² - 15a - 20a + 12
                 = 5a(5a - 3) -4(5a - 3)
                 = (5a-3)(5a-4)
Therefore, possible length = (5a-3)  and possible breadth = (5a-4)
          (ii)     Area : 35y2 + 13y –12
Solution:
    Area = 35y² + 13y –12
               = 35y² + 28y - 15y -12
               = (5y+4) (7y-3)
Therefore, possible length = (5y+4)  and possible breadth = (7y-3)

16. What are the possible expressions for the dimensions of the cuboids whose volumes
are given below?(i) Volume : 3x² – 12x 
Solution: Volume of cuboid = length x breadth x height
     Thus,  3x² – 12x  = 3x(3x-4)
Comparing with the formula for Volume we can conclude that one of the possible solution is 
length = 3, breadth = x and height = (3x-4)
           
 (ii)Volume : 12ky² + 8ky – 20k

Solution: Volume of cuboid = length x breadth x height
 Now, 12ky² + 8ky – 20k = 4k(3y² + 2y -5) = 4k(3y² - 3y+5y - 5) = 4k[3y(y-1)+5  (y-1)]
                                                     = 4k[(3y+5)(y-1)]
Comparing with the formula for the cuboid we can get one possible solution as
breadth = 4k
Height = (3y+5)
Length= (y-1)
                        
Comparing with the formula for Volume we can conclude that one of the possible solution is 
length = 3, breadth = x and height = (3x-4)