Exercise 2.1

Q.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+2–√
              (iii) 3t√+t2–√   
           (iv) y+2y
              (v) x10+y3+t50
Sol.

(i) In 4x2−3x+7, all the indices of x are whole numbers so it is a polynomial in one variable x.
(ii) In y2+2–√, the index of y is a whole number so it is a polynomial in one variable y.
(iii) 3t√+t2–√=3t12+2–√t, here the exponent of the first term is 12, which is not a whole number.  Therefore it is not a polynomial.
(iv) y+2y=y+2y−1, here the exponent of the second term is –1, which is not a whole number and so it is not a polynomial.
(v) x10+y3+t50 is not a polynomial in one variable as three variables x, y, t occur in it.

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Q.2       Write the coefficients of x2 in each of the following :
             (i) 2+x2+x                   
             (ii) 2−x2+x3
             (iii) π2x2+x   
             (iv) 2x−−√−1
Sol.

Coefficient of x2:
(i)  in2+x2+xis1
(ii) in2−x2+x3is−1
(iii) inπ2x2+xisπ2
(iv) in2x−−√−1is0.

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

Example :
Binomial of degree 35 may be taken as x35+4x
Monomial of degree 100 may be taken as 5x100

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

(i) The highest power term is 5x3 and the exponent is 3. So, the degree is 3.
(ii) The highest power term is −y2 and the exponent is 2. So, the degree is 2.
(iii) The highest power term is 5t and the exponent is 1. So, the degree is 1.
(iv) The only term here is 3 which can be written as 3x0 and so the exponent is 0. Therefore the degree is 0.

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Q.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
Sol.

(i) The highest degree of x2+x is 2, so it is a quadratic polynomial .
(ii) The highest degree of x−x3 is 3, so it is a cubic polynomial .
(iii) The highest degree of y+y2+4 is 2, so it is a quadratic polynomial .
(iv) The highest degree of x in (1 + x) is 1.  So it is a linear polynomial.
(v) The highest degree of t in 3t is 1. So it is a linear polynomial.
(vi) The highest degree of r in r2 is 2. So, it is a quadratic polynomial.
(vii) The highest degree of x in 7x3 is 3. So, it is cubic polynomial.

Exercise 2.2

Q.1      Find the value of the polynomial 5x−4x2+3 at
               (i) x = 0          (ii) x = – 1        (iii) x = 2
Sol.

Let p(x)=5x−4x2+3
(i)    At x = 0 : p(0)=5(0)−4(0)2+3
         =0−0+3=3
(ii)   At x = –1 : p(−1)=5(−1)−4(−1)2+3
         =−5−4+3=−6
(iii)  At x = 2 : p(2)=5(2)−4(2)2+3=10−16+3
         =13−16=−3

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Q.2       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)
Sol.

(i)    We have , p(y)=y2−y+1
         p(0)=(0)2−0+1=0−0+1=1
         p(1)=(1)2−1+1=1−1+1=1
         and p(2)=(2)2−2+1=4−2+1=3
(ii)   We have p(t)=2+t+2t2−t3
         Therefore p(0)=2+0+2(0)2−(0)3
          =2+0+0−0=2
          p(1)=2+1+2(1)2−(1)3
          =2+1+2−1=5−1=4
           and p(2)=2+2+2(2)2−(2)3
           =2+2+8−8=4
(iii)   We have, p(x)=x3
           Therefore p(0)=(0)3=0
           p(1)=(1)3=1
           and p(2)=(2)3=8
(iv)    We have p(x)=(x−1)(x+1)
           Therefore  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

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Q.3        Verify whether the following are zeroes of the polynomial, indicated against them.
               (i) p(x)=3x+1,x=−13
                   (ii) p(x)=5x−π,x=45
                   (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=−mℓ
                   (vii) p(x)=3x2−1,x=−13√,23√
                   (viii) p(x)=2x+1,x=12
Sol.

(i)     We have , p(x) = 3x + 1
          At x=−13, p(−13)=3(−13)+1=−1+1=0
         Therefore −13 is a zero of polynomial = 3x + 1.
(ii)    We have p(x)=5x−π
          At  x=45
          P(45)=5(45)−π=4−π
          Therefore 45 is not a zero of polynomial 5x−π
(iii)   We have, p(x)=x2−1
           At x = 1, p(1)=(1)2−1=1−1=0
           Therefore 1 is a zero of p (x).
           Also , at x = -1 p(−1)=(−1)2−1
           =1−1=0
           Therefore -1 is a zero of polynomial x2−1
(iv)     We have , p(x)=(x+1)(x−2)
            At x = –1, p(−1)=(−1+1)(−1−2)
            =(0)(−3)=0
            Therefore -1 is  a zero of p (x) .
            Also , at x = 2 , p(2)=(2+1)(2−2)=(3)(0)=0
            Therefore 2 is zero of polynomial (x + 1)(x + 2).
(v)       We have, p(x)=x2
            At x = 0 , p(0)=(0)2=0
            Therefore 0 is a zero of polynomial x2.
(vi)      We have, p(x)=ℓx+m
             At x=−mℓ p(−mℓ)=ℓ(−mℓ)+m
             =−m+m=0
             Therefore −mℓ is a zero of polynomial lx + m.
(vii)     We have p(x)=3x2−1
             At x=−13√ p(−13√)=3(13√)2−1
             =3×−13−1=1−1=0
             Therefore 13√ is a zero of polynomial 3x2−1
             At x=23√ , p(23√)=3(23√)2−1=3×43−1
             =4−1=3
             Therefore 23√ is not a zero of polynomial 3x2−1
(viii)    We have p(x)=2x+1
              At x=12 p(12)=2(12)+1=1+1=2
              Therefore 12 is not a zero of polynomial 2x + 1.

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Q.4        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
               (vi) p(x)=ax,a≠0
               (vii) p(x)=cx+d,c≠0,c,d are real numbers.
Sol.

(i)     We have to solve p(x) = 0
          ⇒ x + 5 = 0 ⇒ x = – 5
          Therefore – 5 is a zero of the polynomial x + 5 .
(ii)     We have to solve p(x) = 0
           ⇒ x – 5 = 0 ⇒ x = 5
           Therefore 5 is a zero of the polynomial x – 5.
(iii)    We have to solve p(x) = 0
            ⇒ 2x + 5 = 0 ⇒ x=−52
            Therefore −52 is a zero of the polynomial 2x + 5.
(iv)     We have to solve p(x) = 0
            ⇒ 3x – 2 = 0 ⇒ x=23
            Therefore 23 is a zero of the polynomial 3x – 2.
(v)       We have to solve p(x) = 0
            ⇒ 3x = 0 ⇒ x = 0
            Therefore 0 is a zero of the polynomial 3x.
(vi)      We have to solve p(x) = ax, a≠0
             ⇒ ax = 0 ⇒ x = 0
             Therefore 0 is a zero of the polynomial ax.
(vii)     We have to solve p(x) = 0 , c≠0
             ⇒ cx + d = 0 ⇒ x=−dc
             Therefore −dc is a zero of the polynomial cx + d.

Exercise 2.3

Q.1      Find the remainder when x3+3x2+3x+1 is divided by
                (i) x + 1                          (ii) x−12
                (iii) x                              (iv) x+π            (v) 5 + 2x
Sol.

(i) By remainder theorem, the required remainder is equal to p (–1).
Now, p(x)=x3+3x2+3x+1
Therefore  p(−1)=(−1)3+3(−1)2+3(−1)+1
=−1+3−3+1=0
Hence, required remainder = p(–1) = 0

(ii) By remainder theorem, the required remainder is equal to p(12).
Now, p(12)=x3+3x2+3x+1
=(12)3+3(12)2+3(12)+1
=18+34+32+1=1+6+12+88
=278
Hence, required remainder =p(12)=278

(iii) By remainder  theorem, the required remainder is equal to p (0).
Now, p(x)=x3+3x2+3x+1
Therefore  p(0)=0+0+0+1=1
Hence, the required remainder = p(0) = 1

(iv) By remainder theorem the required remainder is p(−π)
Now , p(x)=x3+3x2+3x+1
Therefore p(−π)=(−π)3+3(−π)2+3(−π)+1
=−π3+3π2−3π+1
Hence, the required remainder = p(x) =−π3+3π2−3π+1

(v) By remainder theorem, the required remainder is p(−52)
Now, p(x)=x3+3x2+3x+1
Therefore p(−52)
=(−52)3+3(−52)2+3(−52)+1
=−1258+754−152+1
=−125+150−60+88=−278
Hence, required remainder =p(−52)=−278

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

Let p(x)=x3−ax2+6x−a
By remainder theorem, when p(x) is divided by x – a.
Then remainder = p(a)
Therefore p(a)=a3−a.a2+6a−a
=a3−a3+6a−a=5a
Hence, required remainder = p(x) = 5a

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

7 + 3x will be a factor of  p(x)=3x3+7xifp(−73)=0
Now, p(−73)=3(−73)3+7(−73)
=3×(−34327)−493=−3439−493≠0
Therefore 7 + 3 x is not a factor  of  3x3+7x

 

Exercise 2.4

Q.1     Determine which of the following polynomials has (x + 1) a factor :
              (i) x3+x2+x+1
              (ii) x4+x3+x2+x+1
              (iii) x4+3x3+3x2+x+1
              (iv) x3+x2−(2+2–√)x+2–√ 
Sol.

(i) In order to prove that x + 1 is a factor of p(x)=x3+x2+x+1, it is sufficient to show that p(−1)=0
Now, p(−1)=(−1)3+(−1)2+(−1)+1
=−1+1−1+1=0
Hence, (x + 1) is a factor of  p(x)=x3+x2+x+1

(ii) In order to prove that (x + 1) is a factor of  p(x)=x4+x3+x2+x+1, it is sufficient to show that p (–1) = 0 .
Now,  p(−1)=(−1)4+(−1)3+(−1)2+(−1)+1
=1−1+1−1+1=1≠0
Therefore (x + 1) is not a factor of   x4+x3+x2+x+1

(iii) In order to prove that (x + 1) is a factor of p(x)=x4+3x3+3x2+x+1, it is sufficient to show that p(–1) = 0.
Now, p(−1)=(−1)4+3(−1)3+3(−1)2+(−1)+1
=1−3+3−1+1=1≠0
Therefore (x + 1) is not a factor of   x4+3x3+3x2+x+1.

(iv) In order to prove that (x + 1) is a factor of  p(x)=x3−x2−(2+2–√)x+2–√,  it is sufficient to show that p(-1)= 0
p(−1)=(−1)3−(−1)2−(2+2–√)(−1)+2–√
=−1−1+2+2–√+2–√=22–√≠0
Therefore   (x + 1) is not a factor of   x3−x2−(2+2–√)x+2–√

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Q.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
            (ii) p(x)=x3+3x2+3x+1,g(x)=x+2
            (iii) p(x)=x3−4x2+x+6,g(x)=x−3
Sol.

(i) In order to prove that g(x) = x + 1 is a factor of p(x)=2x3+x2−2x−1, it is sufficient to show that p(–1) = 0.
Now, p(−1)=2(−1)3+(−1)2−2(−1)−1
=−2+1+2−1=0
Therefore g(x) is a factor of p(x).

(ii) In order to prove that g(x) = x + 2 is a factor of  p(x)=x3+3x2+3x+1, it is sufficient to show that p(–2) = 0
Now, p(−2)=(−2)3+3(−2)2+3(−2)+1
=−8+12−6+1
=−1≠0
Therefore g(x) is not a factor of p(x) .

(iii) In order to prove that g(x) = x – 3 is a factor of p(x)=x3−4x2+x+6. It is sufficient to show that p(+3) = 0
Now,  p(3)=(3)3−4(3)2+3+6
=27−36+3+6
36 – 36 = 0
Therefore g (x) is  a factor of p(x).

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Q.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
            (ii) p(x)=2x2+kx+2–√
            (iii) p(x)=kx2−2–√x+1
            (iv) p(x)=kx2−3x+k
Sol.

(i) If (x -1) is a factor of p(x)=x2+x+k, then
p(1)=0
⇒ (1)2+1+k=0
⇒ 1+1+k=0
⇒ k=−2
Hence, k = -2

(ii) If (x – 1) is a factor of   p(x)=2x2+kx+2–√ , then
p(1)=0
⇒ 2(1)2+k(1)+2–√=0
⇒ 2+k+2–√=0
⇒ k=−(2+2–√)

(iii) If (x -1) is a factor of  p(x)=kx2−2–√x+1,  then
p(1)=0
⇒ k(1)2−2–√(1)+1=0
⇒ k−2–√+1=0
⇒ k=2–√−1

(iv) If (x -1) is a factor of p(x)=kx2−3x+k, then
p(1)=0
⇒ k(1)2−3(1)+k=0
⇒ k−3+k=0
⇒ 2k=3
⇒ k=32

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Q.4      Factorise :
               (i) 12x2−7x+1                            
            (ii) 2x2+7x+3
               (iii) 6x2+5x−6                          
            (iv) 3x2−x−4
Sol.

(i) Here p + q = coeff. of x = – 7
pq = coeff. of x2× constant term
= 12 × 1 = 12
Therefore p + q = – 7 = – 4 – 3
and pq = 12 = (– 4)(–3)
Therefore 12x2−7x+1=12x2−4x−3x+1
=4x(3x−1)−1(3x−1)
=(3x−1)(4x−1)

(ii) Here p + q = coeff. of x = 7
pq = coeff. of x2× constant term
= 2 × 3 = 6
Therefore p + q = 7 = 1 + 6
and pq = 6 = 1 × 6
Therefore 2x2+7x+3=2x2+x+6x+3
=x(2x+1)+3(2x+1)
=(2x+1)(x+3)

(iii) Here p + q = coeff. of x = 5
pq = coeff. of x2× constant term
= 6 × (– 6)  = – 36
Therefore p + q = 5 = 9 + (– 4)
and pq = – 36 = 9 × (– 4)
Therefore 6x2+5x−6=6x2+9x−4x−6
=3x(2x+3)−2(2x+3)
=(2x+3)(3x−2)

(iv) Here p + q = coeff. of x = –1
pq = coeff. of x2× constant term
= 3 × (– 4) = – 12
Therefore p + q = –1 = 3 + (– 4)
and pq = – 12 = 3 × (–4)
Therefore 3x2−x−4=3x2+3x−4x−4
=3x(x+1)−4(x+1)
=(x+1)(3x−4)

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Q.5      Factorise :
               (i) x3−2x2−x+2                                
            (ii) x3−3x2−9x−5
            (iii) x3+13x2+32x+20                
            (iv) 2y3+y2−2y−1
Sol. 

(i) Let f(x)=x3−2x2−x+2
The constant term in f(x) is + 2 and factors of + 2 are ±1,±2
Putting x = 1 in f(x),we have
f(1)=(1)3−2(1)2−1+2
=1−2−1+2=0
Therefore (x–1) is a factor of f(x) .
putting x = – 1 in f(x) , we have
f(−1)=(−1)3−2(−1)2−(−1)+2
=−1−2+1+2=0
Therefore x + 1 is a factor of f (x).
Putting x = 2 in f(x), we have
f(2)=(2)3−2(2)2−(2)+2
=8−8−2+2=0
Therefore (x + 2) is a factor of f(x)
Putting x = – 2 in f(x), we have
f(−2)=(−2)3−2(−2)2−(−2)+2
=−8−8+2+2
=−12≠0
Therefore x – 2 is not a factor of f (x).
Therefore The factors of f(x) are (x – 1), (x + 1) and (x – 2).
Let f(x) = k (x – 1) (x + 1) (x – 2)
⇒ x3−2x2−x+2=k(x−1)(x+1)(x−2)
Putting x = 0 on both sides, we have
2 = k (–1) (1) (–2) ⇒ k = 1
Therefore x3−2x2−x+2=(x−1)(x+1)(x−2)

(ii) Let p(x)=x3−3x2−9x−5
We shall look for all factors of – 5. Therefore are ±1,±5.
By trial, we find p (–1) = – 1 – 3 + 9 – 5 = 0 . So (x + 1) is a factor of p(x) .
Now, divide p (x) by (x + 1)

Therefore p(x)=(x+1)(x2−4x−5)
=(x+1)(x2+x−5x−5)
=(x+1)[x(x+1)−5(x+1)]
=(x+1)(x+1)(x−5)
Therefore x3−2x2−x+2=(x+1)(x+1)(x−5)

(iii) Let p(x)=x3+13x2+32x+20
We shall look for all factors of + 20, these are ±1,±2,±4,±5,±10and±20
By trial , we find
p(−2)=(−2)3+13(2)2+32(−2)+20
p(– 2) = – 8 + 52 – 64 + 20 = 0
Therefore (x + 2) is a factor of p(x)
Now, divide p (x) by x + 2

Therefore  p(x)=(x+2)(x2+11x+10)
=(x+2)(x2+x+10x+10)
=(x+2)[x(x+1)+10(x+1)]
=(x+2)(x+1)(x+10)

(iv) Let p(y)=2y3+y2−2y−1
By trial we find p(1) = 2 + 1 – 2 – 1 = 0
So, (y – 1) is a factor of p(y)
Now, divide p(y) by y – 1

Therefore p(y)=(y−1)(2y2+3y+1)
=(y−1)(2y2+2y+y+1)
=(y−1)[2y(y+1)+1(y+1)]
=(y−1)(y+1)(2y+1)

Exercise 2.5

Q.1     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+32)(y2−32)
           (v) (3 – 2x) (3 + 2x)
Sol.

(i) (x + 4) (x + 10) = x2+(4+10)x+4×10
                                 =x2+14x+40
(ii) (x + 8) (x –10)  =x2+(8−10)x+8×(−10)
                                  =x2−2x−80
(iii) (3x + 4) (3x – 5) = 3x(3x−5)+4(3x−5)
                                      =3x×3x−3x×5+4×3x−4×5
                                      =9x2−15x+12x−20
                                      =9x2−3x−20
(iv) (y2+32)(y2−32)  =(y2)2−(32)2
                                                 =y4−94
(v) (3 – 2x) (3 + 2x) = (3)2−(2x)2
                                    =9−4x2

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Q.2      Evaluate the following products without multiplying directly :
            (i) 103 × 107      (ii) 95 × 96       (iii) 104 × 96
Sol.

(i) 103 × 107 = (100+3)(100+7)
                         = (100)2+(3+7)(100)+3×7
                         = 100×100+(10)(100)+21
                         = 10000+1000+21=11021
(ii) 95 × 96 = (100−5)(100−4)
                      = (100)2+(−5−4)(100)+(−5)(−4)
                      = 100×100+(−9)(100)+20
                       = 10000−900+20=9120
(iii) 104 × 96 = (100+4)(100−4)
                          = (100)2−(4)2
                          = 10000−16=9984

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Q.3      Factorise the following using appropriate identifies :
             (i) 9x2+6xy+y2                        
             (ii) 4y2−4y+1
             (iii) x2−y2100
Sol.

(a) 9x2+6xy+y2 = (3x)2+2(3x)(y)+(y)2 [(a+b)2=a2+2ab+b2]
                                            = (3x+y)2=(3x+y)(3x+y)

(b) 4y2−4y+1 = (2y)2−2(2y)(1)+(1)2 [(a−b)2=a2−2ab+b2]
                                    = (2y−1)2=(2y−1)(2y−1)
(c) x2−y2100 = (x)2−(y10)2 [(x+y)(x−y)=x2−y2]
                        =(x−y10)(x+y10)

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Q.4      Expand each of the following using suitable identifies :
            (i) (x+2y+4z)2
            (ii) (2x−y+z)2
            (iii) (−2x+3y+2z)2
            (iv) (3a−7b−c)2
            (v) (−2x+5y−3z)2
            (vi) [14a−12b+1]2
Sol.

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

(ii) (2x−y+z)2
=[2x+(−y)+z]2
=(2x)2+(−y)2+z2+2(2x)(−y)+2(−y)(z)+2(z)(2x)
=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)
=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)2+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) [14a−12b+1]2
=[14a+(−12b)+1]2
=(14a)2+(−12b)2+(1)2+2(14a)(−12b)+2(−12b)(1)+2(1)(14a)
=116a2+14b2+1−14ab−b+12a

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Q.5      Factorise :
               (i) 4x2+9y2+16z2+12xy−24yz−16xz
               (ii) 2x2+y2+8z2−22–√xy−42–√yz−8xz
Sol.

(i) 4x2+9y2+16z2+12xy−24yz−16xz
=(2x)2+(3y)2+(−4z)2+2(2x)(3y)+2(3y)(−4z)+2(2x)(−4z)
= [2x+3y+(−4z)]2=(2x+3y−4z)2

(ii) 2x2+y2+8z2−22–√xy+42–√yz−8xz
=(2–√x)2+(−y)2+(−22–√z)2+2(2x−−√)(−y)+2(−y)(−22–√z)+2(2x−−√)(−22z−−√)
=[2–√x+(−y)+(−22–√z)]2
=(2–√x−y−22–√z)2

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Q.6      Write the following cubes in expanded form :
               (i) (2x+1)3
               (ii) (2a−3b)3
               (iii) [32x+1]3
               (iv) [x−23y]3
Sol.

(i) (2x+1)3 = (2x)3+3(2x)2(1)+3(2x)(1)2+(1)3 =8x3+12x2+6x+1

(ii) (2a−3b)3 = (2a)3−3(2a)2(3b)+3(2a)(3b)2−(3b)3 = 8a3−36a2b+54ab2−27b3

(iii)[32x+1]3=(32x)3+3(32x)2(1)+3(32x)(1)2+13 =278x3+274x2+92x+1

(iv) [x−23y]3=x3−3(x)2(23y)+3(x)(23y)2−(23y)3 =x3−2x2y+43xy2−827y3

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Q.7      Evaluate the following using suitable identities
                (i) (99)3
                (ii) (102)3
                (iii) (998)3
Sol.

(i) (99)3=(100−1)3
           =(100)3−13−3(100)(1)(100−1)
           =1000000−1−29700=970299

(ii) (102)3=(100+2)3
(100)3+(2)3+3(100)(2)(100+2)
=1000000+8+61200=1061208

(iii) (998)3=(1000−2)3
               =(1000)3−(2)3−3(1000)(2)(1000−2)
               =1000000000−8−5988000=994011992

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Q.8      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) 27p3−1216−92p2+14p
Sol.

(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−b3−3(2a)(b)(2a−b)
=(2a−b)3
=(2a−b)(2a−b)(2a−b)

(iii) 27−125a3−135a+225a2
=(3)3−(5a)3−3(3)(5a)(3−5a)
=(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) 27p3−1216−92p2+14p
=(3p)3−(16)3−3(3p)(16)(3p−16)
=(3p−16)3=(3p−16)(3p−16)(3p−16)

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Q.9      Verify :
             (i) x3+y3=(x+y)(x2−xy+y2)
                (ii) x3−y3=(x−y)(x2+xy+y2)
Sol.

(i) R.H.S =(x+y)(x2−xy+y2)
                 =x(x2−xy+y2)+y(x2−xy+y2)
                 =x3−x2y+xy2+x2y−xy2+y3
                 =x3+y3=L.H.S.
Thus, verified.
(ii) R.H.S. =(x−y)(x2+xy+y2)
                   =x(x2+xy+y2)−y(x2+xy+y2)
                   =x3+x2y+xy2−x2y−xy2−y3
x3−y3=L.H.S.
Thus, verified.

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Q.10     Factorise each of the following :
                 (i) 27y3+125z3
                 (ii) 64m3−343n3
Sol.

(i) 27y3+125z3
=(3y)3+(5z)3
=(3y+5z)[(3y)2−(3y)5z+(5z)2]
=(3y+5z)(9y2−15yz+25z2)

(ii) 64m3−343n3 = (4m)3−(7n)3
                          =(4m−7n)[(4m)2+(4m)(7n)+(7n)2]
                          =(4m−7n)(16m2+28mn+49n2)

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Q.11     Factorise : 27x3+y3+z3−9xyz
Sol.

27x3+y3+z3−9xyz
=(3x)3+y3+z3−3(3x)(y)(z)
=(3x+y+z)[(3x)2+y2+z2−(3x)y−yz−z(3x)]
=(3x+y+z)(9x2+y2+z2−3xy−yz−3zx)

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Q.12    Verify that x3+y3+z3−3xyz=12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
Sol.

R.H.S =12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]
           =12(x+y+z)(x2−2xy+y2+y2−2yz+z2+z2−2zx+x2)
           =12(x+y+z)(2x2+2y2+2z2−2xy−2yz−2zx)
           =22(x+y+z)(x2+y2+z2−xy−yz−zx)
           =(x+y+z)(x2+y2+z2−yz−zx−xy−xy−yz−zx)
           =x3+y3+z2−3xyz
           = L.H.S.
Hence verified.

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Q.13     If x + y + z = 0 , show that x3+y3+z3=3xyz.
Sol.

We have, x + y + z = 0
⇒ x + y = – z
Cubing both sides, we have
(x+y)3=(−z)3
⇒ x3+y3+3xy(x+y)=−z3
⇒ x3+y3−3xyz=−z3 [since x + y = – z]
⇒ x3+y3+z3=3xyz, which stands proved.

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Q.14     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
Sol.

(i) Let x = – 12, y = 7 and z = 5
Here x+y+z=−12+7+5=0
⇒ x3+y3+z3=3xyz
⇒ (−12)3+(7)3+(5)3=3×(−12)×7×5=−1260

(ii) Let x = 28, y = – 15 and z = – 13
Here, x+y+z=28−15−13=0
⇒ x3+y3+z3=3xyz
⇒ (28)3+(−15)3+(−13)3
=3(28)(−15)(−13)=16380

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Q.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   (ii) Area:35y2+13y−12
Sol.

Possible length and breadth of the rectangle are the factors of its given area.
Area = length × breadth
(i) Area:25a2+35a+12=25a2−15a−20a+12
                                                           =5a(5a−3)−4(5a−3)=(5a−3)(5a−4)
Therefore Possible length and breadth are (5a – 3) and (5a – 4) units.
(ii) Area=35y2+13y−12
                    =35y2+28y−15y−12
                    =7y(5y+4)−3(5y+4)
                    =(5y+4)(7y−3)
Therefore possible length and breadth are (5y + 4) and (7y – 3) units.

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Q.16     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
Sol.

Possible expressions for the dimensions of the cuboids are the factors of their volumes.
(i) Volume =3x2−12x=3x(x−4)
Therefore Possible dimensions of cuboid are 3, x and (x – 4) units

(ii) Volume =12ky2+8ky−20k
                     =4k(3y2+2y−5)
                     =4k[3y(y−1)+5(y−1)]
                     =4k(y−1)(3y+5)
Therefore Possible dimensions of cuboid are
4k, (y – 1) and (3y + 5) units.