Class 10 Mathematics Chapter 2: Polynomials - NCERT Solutions

Total no of zeroes of a polynomial equation = the number of times the curve intersect x-axis

1. In this graph the number of zeroes of p(x) is 0, because the graph is parallel to x- axis and does not intersect at any point on x-axis.
2. In this graph the number of zeroes of p(x) is 1, because the curve intersects x-axis only at one point.
3. In this graph the number of zeroes of p(x) is 3, because the curve intersects x-axis at three points. 
4. In this graph the number of zeroes of p(x) is 2, because the curve intersects x-axis at two points.
5. In this graph the number of zeroes of p(x) is 4, because the curve intersects x-axis at four points.
6. In this graph the number of zeroes of p(x) is 3, because the curve intersects x-axis at three points.

(i)   x² – 2x – 8

                 = x – 4x + 2x – 8                  

                = x(x – 4) + 2(x – 4)

                 = (x + 2) (x – 4)

                The value of x² – 2x – 8 is zero if (x + 2) = 0 and (x – 4) = 0

                      x = -2 or   x = 4

                       Sum of zeroes = (-2 + 4) = 2 = - coefficient of x

                                                                        coefficient of x²      

                       Product of zeroes = (-2) × 4 = -8 = Constant term

                                                                             coefficient of x²

(ii)          4s² – 4s + 1

                          = 4s² – 2s – 2s + 1

                        = 2s (2s – 1) – 1 (2s – 1)

                         = ( 2s – 1 )  ( 2s – 1 )      

                     The value of 4s² – 4s + 1 is zero , if (2s-1) = 0 and (2s-1 ) = 0

                       s = 1/2 , 1/2

                       Sum of zeroes = (1/2 + 1/2) = 1   - coefficient of x

                                                                            coefficient of x²

                      Product of zeroes =1/2 × 1/2 = 1/4 =  constant term

                                                                                coefficient of x²             

(iii)    6x² –7x – 3  

                        = 6x – 9x + 2x – 3                     

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

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

                     The value of  6x² –7x – 3 is zero, if (3x + 1) = 0 and (2x – 3) = 0

                            X = -1 /3 , 3/2

                            Sum of zeroes = ( -1/3 + 3/2) = 7/6 =  - coefficient of x

                                                                                        coefficient of x²

                           Product of zeroes = -1/3 × 3/2 = -3/2 =  constant term

                                                                                        coefficient of x²            

(iv)        4u²+8u

                       4u(u+2)

                       The value of 4u²+8u is zero, if 4u = 0 and (u+2) =0

                        u   = 0,  - 2

                       Sum of zeroes = ( 0+ (-2)) = -2 =  - coefficient of x

                                                                               coefficient of x²

                         Product of zeroes = (-2) × 0 = 0 =   constant term

                                                                                 coefficient of x²            

(v)

(vi) 

3x²–x–4

                            3x – 4x + 3x – 4 

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

                       The value of 3x – x + 4 is zero, if (3x – 4) = 0 and (x + 1) = 0

                             Sum of zeroes = [4/3 + ( -1)] = 1/3 = - coefficient of x

                                                           coefficient of x²

                          Product of zeroes = (-1) × 4/3 = -4/3 = constant term

                                                                                          coefficient of x²           

(i) Let α, β are the zeroes of the polynomial ax²+ bx +c, therefore

                               Sum of zeroes (α + β) = 1/4 =  -b/a

                            Product of zeroes (αβ) = -1 = c/a

                            On comparing,

                             a = 4, b = -1 and c = - 4

                         Hence, the required polynomial is 4x² –x – 4

(ii)

 

(iii) 

Let α, β are the zeroes of the polynomial ax²+ bx +c, therefore

                                Sum of zeroes (α + β) = 0 = -b/a

                         Product of zeroes (αβ) = √5 = √5/1 = c/a

                      On comparing,

                        a = 1, b = 0 and c = √5

                       Hence, the required polynomial is x² + √5

(i) Given ,

                     Dividend = p(x) = x² – 3x² + 5x – 3

                         Divisor = g(x) = x² – 2

                

                      Quotient =  x – 3

                             Reminder = 7x – 9

(ii) Given,

                           Dividend  = p(x) = x⁴ – 3x²  + 4x + 5

                          Divisor = g(x) = x² + 1 – x  

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Note: If on dividing the second polynomial by first we get zero remainder then we say that  first Is factor of second polynomial.

(i) Given ,

                      First polynomial = t²-3

                     Second polynomial =  2t⁴ +3t³-2t² -9t-12

As we can see the remainder is 0. Thereofre we can say that  first  polynomial is a factor of second polynomial.

(ii) Given,

   First polynomial = x²+3x+1

   Second polynomial = 3x⁴ + 5x^{3 –} 7x² + 2x + 2

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                   Given  ,

                   Dividend  p(x) =  x³ - 3x² + x + 2

                   Quotient = x – 2

                   Remainder = -2x + 4

             Let divisor  = g(x)

             As we know ,

             Dividend = Divisor × Quotient + Remainder

                      x³- 3x² +x+ 2 = g(x) × (x – 2) + (-2x + 4)

                      x ³- 3x² + x + 2 – (-2x + 4) = g(x) × (x – 2 )

                     Therefore, g(x) × (x – 2 ) = x³ – x² + x + 2

       Now we will divide p(x) by quotient x – 2 to find divisor g(x)

                                

            Therefore ,  g(x) = ( x² – x + 1 )

we know that 

              

(i) Deg P(x) = deg g (x)

    The degree of dividend or quotient can be equal, only if the divisor is a constant (degree 0)

    Then, let p(x) = 3x² –  6x + 5

     Let g(x) = 3

     Therefore,  q(x) = x² – 2x +1  and r(x) = 2

(ii) Deg  q(x)  =  deg r(x)    

     Let p(x) = x² + 1

     Let g(x) = x

     Therefore, q(x) = x + 1 and r(x) = 0

     Here, we can see the degree of quotient is equal to the degree of remainder.

     Hence, division algorithm is satisfied here.

   (iii)    deg r(x) = 0

The degree of remainder is zero, only if the remainder left after division algorithm is Constant.

Let p(x) = x² + 1

Let g(x) = x

                Therefore, q(x)= x  and r(x) = 1 

Here we can see the degree of remainder is zero.

Hence division algorithm is satisfied here.

(i)                         Here f(x) = 2x³ + x² - 5x + 2

                            Given roots of f(x) are ½, 1, -2

F(1/2) = 2×(1/2)³ + (1/2)² - 5(1/2 ) + 2 = 0

F(1) = 2(1)³ + 1² - 5(1) + 2 = 0

F(-2) = 2(-2)³ + (-2)² - 5(-2) + 2 = 0

Hence, ½, 1 and -2 are the zeroes of f(x).

Therefore, sum of zeroes = -b/a -1/2

Sum of product of zeroes taken two at a time = c/a = -5/2

Product of zeroes = -d/a = 2

(ii)                    Let the f(x) = ax³ + bx² + c + d

Let α, β and γ be the zeroes of the polynomial f(x).

Then, sum of zeroes = -b/a = 2/1 ………………(i)

Sum of product of zeroes taken two at a time = c/a = -7. ………………..(ii)

Product of zeroes = -d/a = -14 ……………….(iii)

From equation (i), (ii) and (iii) we have

a = 1 , b = -2 , c = -7 and d = 14

Therefore the required polynomial on putting the values of a, b, c and d

F(x) = x³ - 2x² - 7x + 14

                            Let the p(x) = ax³+bx²+ cx + d

                            Sum of zeroes and α, β and γ be the zeroes.

                            Then, α, β and γ = -b/ a = 2/1    …………………..(i)

                            αβ + βγ + γα= c/a = -7      …………………….(ii)

‘                           αβγ = -d /a = -14               ……………………………..(iii)

                            From equation (i), (ii) and (iii), we get

                            a = 1, b = -2, c = -7 and d = 14

                            Therefore, the required polynomial on putting the value of a, b, c and d is P(x) = x³ - 2x² – 7x + 14

Given, p(x) = x³ - 3x² + x + 1

And zeroes are given as a – b, a, a + b

Now, comparing the given polynomial with general expression, we get;

∴ ax³+bx²+ cx + d = x³ – 3x²+ x + 1

a = 1, b = -3, c = 1 and d = 1

Sum of zeroes = a – b + a + a + b

-b/a = 3a

Putting the values b and a

- (-3)/1 = 3a

                a = 1

Thus, the zeroes are 1 - b, 1, 1 + b.

Now, product of zeroes = 1(1 – b) (1 + b)

d/a = 1 - b²

-1/1 = 1- b²

b² = 1 + 1 = 2

b = √2

Hence, 1, -√2, 1, 1 + √2 are the zeroes of x³ – 3x² + x + 1

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Given,

Divisor = x² – 2x + k

Dividend = x⁴ – 6x³ + 16x² – 25x + 10

Remainder = x + a

As we know that,

Dividend = divisor quotient + remainder

x⁴ – 6x³ + 16x² – 25x + 10 = x² – 2x + k quotient + (x + a)

x⁴ – 6x³ + 16x² – 25x + 10 –  (x + a)  = x² – 2x + k quotient

x⁴ – 6x³ + 16x² – 26x + 10 – a  = quotient

               x² – 2x + k

If the polynomial x⁴ – 6x³ + 16x² – 26x + 10 –  a is divided by x² – 2x + k remainder comes out to be zero.

Therefore, By equating the remainder with zero, we have

(-10 + 2k) = 0 => 2k = 10  => k = 5

Or, 10 – a – 8k + k² = 0

Putting the value of k, we get

10 – a – 8(5) + (5)² = 0

10 – a – 40 + 25 = 0

- a – 5 = 0 => a = -5

Hence, k = 5 and a = -5