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

Solution :-

(i) At x = 0 
   p(x) = 5x – 4x² + 3
   p(0) = 5 x 0 - 4 x 0 + 3
        = 0 - 0 + 3
        = 3  
        

(ii) At x = -1
   p(x) = 5x – 4x² + 3
   p(-1) = 5 x (-1) – 4 x (-1)² + 3
        = -5 - 4 x 1 + 3    
        = -5 - 4 + 3
        = -9 + 3
        = -6
    

(iii) At x = 2
   p(x) = 5x – 4x² + 3
   p(2) = 5 x 2 - 4 x 2² + 3   
        = 10 - 4 x 4 + 3
        = 13 - 16
        = -3
    

Question-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).

Solution :-

(i) p(y) = y² – y + 1 at

    p(0) = 0² – 0 + 1 
         = 0 - 0 + 1
         = 1

    p(1) = 1² – 1 + 1
         = 1 - 1 + 1
         = 2 - 1
         = 1

    p(2) = 2² – 2 + 1
         = 4 - 2 + 1
         = 5 - 2
         = 3
  

      
(ii) p(t) = 2 + t + 2t² – t³ 
     
    p(0) = 2 + 0 + 2 x 0² – 0³ 
         = 2 + 0
         = 2

    p(1) = 2 + 1 + 2 x 1² – 1³
         = 3 + 2 - 1
         = 5 - 1
         = 4
        
    p(2) = 2 + 2 + 2 x 2² – 2³
         = 4 + 2 x 4 - 8
         = 4 + 8 - 8
         = 12 - 8
         = 4  
   

     
(iii) p(x) = x³

    p(0) = 0³ = 0

    p(1) = 1³ = 1

    p(2) = 2³ = 8
        

(iv) p(x) = (x – 1) (x + 1)

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

Question-3 :-  Verify whether the following are zeroes of the polynomial, indicated against them.
(i) p(x) = 3x + 1, x = -1/3
(ii) p(x) = 5x - π, x = 4/5
(iii) p(x) = 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 = -m/l
(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3
(viii) p(x) = 2x + 1, x = 1/2

Solution :-

(i) p(x) = 3x + 1, x = -1/3
    p(-1/3) = 3 x (-1/3) + 1
            = -1 + 1
            = 0
    Hence, x = -1/3 is a zero of polynomial.

(ii) p(x) = 5x - π, x = 4/5
    p(4/5) = 5 x 4/5 - π
           = 4 - π 
    Hence, x = 4/5 is not a zero of polynomial.
        

(iii) p(x) = x² – 1, x = 1, –1
    p(1) = (1)² – 1
         = 1 - 1
         = 0
    p(-1) = (-1)² – 1
          = 1 - 1
          = 0
    Hence, x = 1, -1 are zeroes of polynomial.
        

(iv) p(x) = (x + 1) (x – 2), x = – 1, 2
    p(-1) = (-1 + 1)(-1 - 2)
          = 0 x (-3)
          = 0
    p(2) = (2 + 1)(2 - 2)
         = 3 x 0
         = 0
    Hence, x = -1, 2 are zeroes of polynomial. 
        

(v) p(x) = x², x = 0
    p(0) = 0² = 0
    Hence, x = 0 is a zero of polynomial.
        

(vi) p(x) = lx + m, x = -m/l
    p(-m/l) = l x (-m/l) + m
            = -m + m
            = 0
    Hence, x = -m/l is a zero of polynomial.
        

(vii) p(x) = 3x² – 1, x = -1/√3, 2/√3
    p(-1/√3) = 3 x (-1/√3)² – 1
             = 3 x 1/3 - 1
             = 1 - 1
             = 0   
    p(2/√3) = 3 x (2/√3)² – 1
            = 3 x 4/3 - 1
            = 4 - 1
            = 3
    Hence, x = -1/√3 is a zero of polynomial. 
    But x = 2/√3 is not a zero of polynomial. 
        

(viii) p(x) = 2x + 1, x = 1/2
    p(1/2) = 2 x 1/2 + 1
         = 1 + 1
         = 2
    Hence, x = 2 is not a zero of polynomial. 
    

Question-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.

Solution :-

    Zero of a polynomial is that value of the variable at which 
    the value of the polynomial is obtained as 0. So, p(x) = 0.   

(i) p(x) = x + 5 
    p(x) = 0
    x + 5 = 0
    x = -5
    Therefore, x = -5 is a zero of polynomial p(x) = x + 5.
        

(ii) p(x) = x – 5 
    p(x) = 0
    x - 5 = 0
    x = 5
   Therefore, x = 5 is a zero of polynomial p(x) = x - 5.
        

(iii) p(x) = 2x + 5 
    p(x) = 0
    2x + 5 = 0
    2x = -5
     x = -5/2
   Therefore, x = -5/2 is a zero of polynomial p(x) = 2x + 5.
        

(iv) p(x) = 3x – 2 
    p(x) = 0
    3x - 2 = 0
    3x = 2
     x = 2/3 
   Therefore, x = 2/3 is a zero of polynomial p(x) = 3x - 2.
        

(v) p(x) = 3x 
    p(x) = 0
    3x = 0
     x = 0
   Therefore, x = 0 is a zero of polynomial p(x) = 3x.
        

(vi) p(x) = ax, a ≠ 0 
    p(x) = 0
    ax = 0
     x = 0/a
     x = 0
   Therefore, x = 0 is a zero of polynomial p(x) = ax.
        

(vii) p(x) = cx + d, c ≠ 0, c, d are real numbers.
    p(x) = 0
    cx + d = 0
    cx = -d
     x = -d/c
    Therefore, x = -d/c is a zero of polynomial p(x) = cx + d.