NCERT Solutions Class 9 Mathematics Polynomials Exercise 2.2

Exercise 2.2

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Question :

Find the value of the polynomial 5x – 4x² + 3 at

(i) x = 0                             (ii) x = -1                         (iii) x = 2

Answer :

Let p(x) = 5x – 4x² + 3

(i) Put x = 0, we get

p(0) = 5 * 0 – 4 * 0 + 3 = 0 – 0 + 3 = 3                             

(ii) x = -1    

p(-1) = 5 * (-1) – 4 * (-1)² + 3 = -5 – 4 + 3 = -9 + 3 = -6                    

(iii) x = 2

p(2) = 5 * 2 – 4 * 2² + 3 = 10 – 4 * 4 + 3 = 10 – 16 + 3 = 13 – 16 = -3

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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)

Answer.

(i)   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

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

(ii)   p(t) = 2 + t + 2t² – t³

        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 = 4

         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

         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

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

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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) = 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

Answer :

(i) Put x = -1/3, we get

p(-1/3) = 3 * (-1/3) + 1 = -1 + 1 = 0

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

(ii) put x = 4/5, we get

p(x) = 5 * 4/5 – π = 4 – π

Hence, x = 4/5 is a zero of the polynomial p(x) = 5x – π              

(iii) Put x = 1, we get

p(1) = 1² – 1 = 1 – 1 = 0

Hence, x = 1 is a zero of the polynomial p(x) = x² – 1

Put x = -1, we get

p(1) = (-1)² – 1 = 1 – 1 = 0

Hence, x = -1 is a zero of the polynomial p(x) = x² – 1                                   

(iv) Put x = -1, we get

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

Hence, x = -1 is a zero of the polynomial p(x) = (x + 1)(x – 2)

Put x = 2, we get

p(2) = (2 + 1) (2 – 2) = 3 * 0 = 0

Hence, x = 2 is a zero of the polynomial p(x) = (x + 1)(x – 2)

(v) p(x) = x², x = 0

Put x = 0, we get

p(0) = 0² = 0

Hence, x = 0 is a zero of the polynomial p(x) = x²                         

(vi) Put x = -m/I, we get

p(-m/l) = l * (-m/l) + m = -m + m = 0

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

(vii) Put x = -1/√3, we get

p(-1/√3) = 3 * (-1/√3)² – 1 = 3 * 1/3 – 1 = 1 – 1 = 0

Hence, x = -1/√3 is a zero of the polynomial 3x² – 1

Put x = 2/√3, we get

p(2/√3) = 3 * (2/√3)² – 1 = 3 * 4/3 – 1 = 4 – 1 = 3

It in not equal to zero. Hence, x = 2/√3 is not a zero of the polynomial 3x² – 1      

(viii) Put x = 1/2, we get

p(1/2) = 2 * (1/2) + 1 = 1 + 1 = 2

It in not equal to zero. Hence, x = 1/2 is not a zero of the polynomial 2x + 1.