Class 10 - Maths - Polynomials

Exercise 2.1

Question 1:

The graphs of y = p(x) are given in Fig. 2.10 below, for some polynomials p(x). Find the number of zeroes of p(x), in each case.

      Class_10_Maths_Polynomials_Division_Of_Graphs         

Answer:

(i) The graph does not intersect the x axis, so there is no zero in this graph.

(ii) The graph intersects the x axis at one place, so there is only one zero in this graph.

(iii) The graph intersects the x axis at three places, so there are three zeros in this graph.

(iv) The graph intersects the x axis at two places, so there are two zeros in this graph.

(v) The graph intersects the x axis at four places, so there are four zeros in this graph.

(vi) The graph intersects the x axis at three places, so there are three zeros in this graph.

 

 

 

                                                                        Exercise 2.1

Question 1:

Find the zeroes of the following quadratic polynomials and verify the relationship between the zeroes and the coefficients.

(i) x2 – 2x – 8                                (ii) 4s2 – 4s + 1                               (iii) 6x2 – 3 – 7x

(iv) 4u2 + 8u                                 (v) t2 – 15                                        (vi) 3x2 – x – 4

Answer:

(i) x2 – 2x – 8                       

= x2 – 4x + 2x - 8

= x(x - 4) + 2(x - 4)

= (x - 4)(x + 2)

The value of x2 – 2x – 8 is zero if x + 2 = 0 or x – 4 = 0

So, x = -2, 4

Therefore zeroes of x2 – 2x – 8 are -2 and 4

Now, Sum of zeroes = -2 + 4 = 2 = -(-2)/1 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = (-2) * 4 = -8 = -8/1 = 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

So, s = 1/2, 1/2

Therefore zeroes of 4s2 – 4s + 1 are 1/2 and 1/2

Now, Sum of zeroes = 1/2 + 1/2 = 1 = -(4)/4 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = 1/2 * 1/2 = 1/4 = Constant term/ (Coefficient of x2)

(iii) 6x2 – 3 – 7x

= 6x2 – 7x - 3

= 6x2 – 9x + 2x - 3

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

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

The value of 6x2 – 7x – 2 is zero if 2x - 3 = 0 or 3x + 1 = 0

So, x = -1/3, 3/2

Therefore zeroes of 6x2 – 7x – 3 are -1/3 and 3/2

Sum of zeroes = -1/3 + 3/2 = (-2 + 9)/6 = 7/6 = -(-7)/6 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = (-1/3) * 3/2 = -1/2 = -3/6 = Constant term/ (Coefficient of x2)

(iv) 4u2 + 8u

= 4u(u + 2)

The value of 4u2 + 8u is zero if 4u = 0 or u + 2 = 0

=> u = 0, -2

Therefore, the zeroes of 4u2 + 8u are 0 and -2.

Now, Sum of zeroes = 0 + (-2) = -2 = -(8)/4 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = 0 * (-2) = 0 = 0/4 = Constant term/ (Coefficient of x2)

(v) t2 – 15 

= t2 – (√15)2                      

= (t - √15)(t + √15)

The value of t2 – 15 is zero if t - √15 or t + √15 = 0

=> t = √15, -√15

Therefore, the zeroes of t2 – 15 are √15 and -√15.

Now, Sum of zeroes = -√15 + √15 = 0 = -(0)/1 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = (-√15) * √15 = -15 = -15/1 = Constant term/ (Coefficient of x2)

(vi) 3x2 – x – 4

= 3x2 – 4x + 3x - 4

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

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

The value of 6x2 – 7x – 2 is zero if 3x - 4 = 0 or x + 1 = 0

So, x = 4/3, -1

Therefore zeroes of 3x2 – x – 4 are 4/3 and -1

Sum of zeroes = 4/3 + (-1) = (4 - 3)/3 = 1/3 = -(-1)/3 = -(Coefficient of x)/ (Coefficient of x2

Product of zeroes = 4/3 * 1 = -4/3 = Constant term/ (Coefficient of x2)

Question 2:

Find a quadratic polynomial each with the given numbers as the sum and product of its zeroes respectively. 

 (i) 1/4, -1              (ii) √2, 1/3             (iii) 0, √5            (iv) 1, 1         (v) -1/4, 1/4            (vi) 4, 1

Answer:

(i) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 1/4 = -b/a

and α * β = -1 = -4/4 = c/a

On comparing, we get

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

Hence, the required quadratic polynomial is 4x2 - x - 4.

(ii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = √2= 3√2/3 = -b/a

and α * β = 1/3 = c/a

On comparing, we get

a = 3, b = -3√2, c = 1

Hence, the required quadratic polynomial is 3x2 - 3√2x + 1.

(iii) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 0 = 0/1 = -b/a

and α * β = √5 = √5/1 = c/a

On comparing, we get

a = 1, b = 0, c = √5

Hence, the required quadratic polynomial is x2 – 0*x + √5 = x2 + √5.

(iv) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 1 = 1/1 = -b/a

and α * β = 1 = 1/1 = c/a

On comparing, we get

a = 1, b = -1, c = 1

Hence, the required quadratic polynomial is x2 - x + 1.

(v) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = -1/4 = -b/a

and α * β = 1/4 = c/a

On comparing, we get

a = 4, b = 1, c = 1

Hence, the required quadratic polynomial is 4x2 + x + 1.

(vi) Let α and β are the zeroes of the quadratic polynomial ax2 + bx + c.

Given, α + β = 4 = 4/1 = -b/a

and α * β = 1 = 1/1 = c/a

On comparing, we get

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

Hence, the required quadratic polynomial is x2 - 4x + 1.

                                             Exercise 2.3

Question 1:

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:

(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2              (ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

Answer:

(i) p(x) = x3 – 3x2 + 5x – 3, g(x) = x2 – 2

                Class_10_Maths_Polynomials_Division_Of_Polynomial9                               

So, quotient = x - 3 and remainder = 7x - 9

(ii) p(x) = x4 – 3x2 + 4x + 5, g(x) = x2 + 1 – x

             Class_10_Maths_Polynomials_Division_Of_Polynomial8                      

So, quotient = x2 + x - 3 and remainder = 8

(iii) p(x) = x4 – 5x + 6, g(x) = 2 – x2

 

 Class_10_Maths_Polynomials_Division_Of_Polynomial7

 

So, quotient = -x2 - 2 and remainder = -5x + 10

Question 2:

Check whether the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:

(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12                             (ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

Answer:

(i) t2 – 3, 2t4 + 3t3 – 2t2 – 9t – 12                            

Since remainder is 0

Hence t2 – 3 is a factor of 2t4 + 3t3 – 2t2 – 9t – 12

 

 Class_10_Maths_Polynomials_Division_Of_Polynomial6

 

 

(ii) x2 + 3x + 1, 3x4 + 5x3 – 7x2 + 2x + 2

            Class_10_Maths_Polynomials_Division_Of_Polynomial5          

Since remainder is 0, hence x2 + 3x + 1 is a factor of 3x4 + 5x3 – 7x2 + 2x + 2

(iii) x3 – 3x + 1, x5 – 4x3 + x2 + 3x + 1

    Class_10_Maths_Polynomials_Division_Of_Polynomial4                 

Since remainder is not equal to 0, hence x3 – 3x + 1 is not a factor of x5 – 4x3 + x2 + 3x + 1

Question 3:

Obtain all other zeroes of 3x4 + 6x3 – 2x2 – 10x – 5, if two of its zeroes are √(5/3) and - √(5/3)                             

Answer:

Class_10_Maths_Polynomials_Division_Of_Polynomial3

Given, two zeroes are √(5/3) and - √(5/3)

Let x = √(5/3) and x =  -√(5/3)=> x - √(5/3) = 0 and x + √(5/3) = 0

=> {x - √(5/3)}{x + √(5/3)} = 0

=> x2 – 5/3 = 0

=> 3x2 – 5 = 0

Apply division lemma, we get

Quotient = x2 + 2x + 1 and remainder = 0

So, 3x2 – 5 is a factor of 3x4 + 6x3 – 2x2 – 10x – 5

Now, 3x4 + 6x3 – 2x2 – 10x – 5 = (3x2 – 5)(x2 + 2x + 1)

                                                      = 3(x2 – 5/3)(x2 + 2x + 1)

                                                      = 3{x - √(5/3)}{x + √(5/3)} (x + 1)2

                                                      = 3{x - √(5/3)}{x + √(5/3)} (x + 1)(x + 1)

Hence, the zeroes are: -1, -1, √(5/3) and  -√(5/3)

Question 4:

On dividing x3 – 3x2 + x + 2 by a polynomial g(x), the quotient and remainder were x - 2 and -2x + 4, respectively. Find g(x).

Answer:

According to division algorithm,

Dividend = Divisor * Quotient+ Remainder

=> p(x) = g(x) * q(x) + r(x),

Put the value in formula we get

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

Add 2x and subtract 4 both side we get

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

=> g(x) = (x3 – 3x2 + 3x– 2)/(x - 2)

        Class_10_Maths_Polynomials_Division_Of_Polynomial2                                               

 Hence, we get g(x) = x2 – x +1

Question 5:

Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and

(i) deg p(x) = deg q(x)                 (ii) deg q(x) = deg r(x)                   (iii) deg r(x) = 0

Answer:

According to the division algorithm,

if p(x) and g(x) are two polynomials with g(x) ≠ 0,

then we can find polynomials q(x) and r(x) such that p(x) = g(x) * q(x) + r(x),

where r(x) = 0 or degree of r(x) < degree of g(x)

Degree of a polynomial is the highest power of the variable in the polynomial.

(i) deg p(x) = deg q(x)

Degree of quotient will be equal to degree of dividend when divisor is constant i.e. when any

polynomial is divided by a constant.

Let us assume the division of 6x2 + 2x + 2 by 2

Here, p(x) = 6x2 + 2x + 2, g(x) = 2

q(x) = 3x2 + x + 1 and r(x) = 0

Degree of p(x) and q(x) is the same i.e. 2

Checking for division algorithm, p(x) = g(x) * q(x) + r(x)

6x2 + 2x + 2 = 2(3x2 + x + 1) + 0

Thus, the division algorithm is satisfied.

(ii) deg q(x) = deg r(x)

Let us assume the division of x3 + x by x2

Here, p(x) = x3 + x,  g(x) = x2, q(x) = x and r(x) = x

Clearly, the degree of q(x) and r(x) is the same i.e. 1

Checking for division algorithm, p(x) = g(x) * q(x) + r(x)

x3 + x = x2 * x + x = x3 + x

Thus, the division algorithm is satisfied.

(iii) deg r(x) = 0

Degree of remainder will be 0 when remainder comes to a constant.

Let us assume the division of x3 + 1 by x2

Here, p(x) = x3 + 1, g(x) = x2, q(x) = x and r(x) = 1

Clearly, the degree of r(x) is 0

Checking for division algorithm,

p(x) = g(x) * q(x) + r(x)

x3 + 1 = x2 * x + 1 = x3 + 1

Thus, the division algorithm is satisfied.

 

 

 

                                                              Exercise 2.4 (Optional)*

Question 1:

Verify that the numbers given alongside of the cubic polynomials below are their zeroes.

Also verify the relationship between the zeroes and the coefficients in each case:

(i) 2x3 + x2 – 5x + 2; 1/2, 1, -2                      (ii) x3 – 4x2 + 5x – 2; 2, 1, 1

Answer:

(i) Let p(x) = 2x3 + x2 – 5x + 2                

Now, p(1/2) = 2(1/2)3 + (1/2)2 – 5(1/2) + 2

                       = 2 * 1/8 + 1/4 – 5/2 + 2

                       = 1/4 + 1/4 – 5/2 + 2

                       = 1/2 + 2 – 5/2

                       = 5/2 – 5/2

                       = 0

So, 1/2 is one of the zero of p(x).

p(1) = 2(1)3 + (1)2 – 5(1) + 2

         = 2 + 1 – 5 + 2

         = 5 – 5

         = 0

So, 1 is one of the zero of p(x).      

p(-2) = 2(-2)3 + (-2)2 – 5(-2) + 2

          = -2 * 8 + 4 + 10 + 2

          = -16 + 16

          = 0             

So, -2 is one of the zero of p(x).  

Therefore, 1/2, 1 and -2 are the zeroes of p(x).

Let a = 1/2, b = 1, c = -2

Now,

a + b + c = 1/2 + 1 + (-2) = 1/2 + 1 -2 = -1/2 = -(1)/2 = -(coefficient of x2)/( coefficient of x3)  

ab + bc + ca = 1/2 * 1 + 1 * (-2) + (-2) * 1/2

                     = 1/2 - 2 – 1

                     = 1/2 - 3

                     = -5/2

                     = -(coefficient of x)/( coefficient of x3

abc = 1/2 * 1 * (-2) = -1 = -2/2 = -(constant term)/( coefficient of x3)  

Hence, the relation between zeroes and coefficients is verified.

(ii) Let p(x) = x3 – 4x2 + 5x – 2

Now, p(2) = 23 – 4 * 22 + 5 * 2 – 2

                   = 8 - 16 + 10 - 2

                   = 18 - 18

                    = 0

So, 2 is one of the zero of p(x).

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

         = 1 - 4 + 5 + 2

         = 6 – 6

         = 0

So, 1 is one of the zero of p(x).      

Therefore, 2, 1 and 1 are the zeroes of p(x).

Let a = 2, b = 1, c = 1

Now,

a + b + c = 2 + 1 + 1 = 4 = -(-4)/1 = -(coefficient of x2)/( coefficient of x3)  

ab + bc + ca = 2 * 1 + 1 * 1 + 1 * 2

                     = 2 + 1 + 2

                     = 5

                     = 5/1  

                     = coefficient of x/coefficient of x3 

abc = 2 * 1 * 1 = 2 = -(-2)/1 = -(constant term)/( coefficient of x3)  

Hence, the relation between zeroes and coefficients is verified.

Question 2:

Find a cubic polynomial with the sum, sum of the product of its zeroes taken two at a time, and the product of its zeroes as 2, -7, -14 respectively.

Answer:

Let p(x) = ax3 + bx2 + cx + d be a cubic polynomial whose zeroes are α, β and γ.

Given that, α + β + γ = 2

αβ + βγ + αγ = -7

αβγ = -14

We know that

α + β + γ = -(coefficient of x2)/( coefficient of x3)  

αβ + βγ + αγ = coefficient of x/coefficient of x3 

αβγ = -(constant term)/( coefficient of x3)  

Therefore,

α + β + γ = -(coefficient of x2)/( coefficient of x3) = -b/a = 2/1 

αβ + βγ + αγ = coefficient of x/coefficient of x3 = c/a = -7/1

αβγ = -(constant term)/( coefficient of x3) = -d/a = -14/1

On comparing, we get

a = 1, b = -2, c = -7 and d = 14  

Hence, the required cubic polynomial is p(x) = x3 - 2x2 - 7x + 14

Question 3:

If the zeroes of the polynomial x3 – 3x2 + x + 1 are a – b, a, a + b, find a and b.

Answer:

We know that

      Sum of zeroes = -(coefficient of x2)/( coefficient of x3)

=> a – b + a + a + b = -(-3)/1

=> 3a = 3

=> a = 3/3

=> a = 1

Product of zeroes = -(constant term)/( coefficient of x3)

=> (a - b)(a)(a + b) = -1/1

=> (1 - b)(1)(1 + b) = -1                 [By putting value of a]

=> 1 – b2 = -1

=> b = 1 + 1

=> b2 = 2

=> b = ±√2

Hence, a = 1 and b = ±√2

 

Question 4:

If two zeroes of the polynomial x4 – 6x3 – 26x2 + 138x – 35 are 2 ± √3, find other zeroes.

Answer:

Let p(x) = x4 – 6x3 – 26x2 + 138x – 35

Given that 2 + √3 and 2 - √3 are two of the zeroes of p(x)

So, (x - 2 - √3) and (x - 2 + √3) are the factors of p(x)

Therefore, (x - 2)2 - (√3)2 is a factor of p(x)

=> x2 – 2x + 4 – 3 is a factor of p(x)

=> x2 – 2x + 1 is a factor of p(x)

Now, divide x4 – 6x3 – 26x2 + 138x – 35 by x2 – 2x + 1, we get

       Class_10_Maths_Polynomials_Division_Of_Polynomial1                               

Quotient = x2 – 2x – 35 and remainder = 0

Therefore,

p(x) = x4 – 6x3 – 26x2 + 138x – 35

        = (x2 – 2x + 1)( x2 – 2x – 35)

       = (x2 – 2x + 1)( x2 – 7x + 5x – 35)

       = (x2 – 2x + 1){x(x – 7) + 5(x – 7)}

       = (x2 – 2x + 1)(x - 7)(x + 5)

To get the other zeroes, put x + 5 = 0 and x – 7 = 0

=> x = -5, 7

Hence, the other two zeroes are -5 and 7     

Question 5:

If the polynomial x4 – 6x3 + 16x2 – 25x + 10 is divided by another polynomial x2 – 2x + k, the remainder comes out to be x + a, find k and a.

Answer:

Given that

Divisor = x2 – 2x + k

Dividend = x4 – 6x3 + 16x2 – 25x + 10

Remainder = x + a

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 = (x2 – 2x + k) * Quotient

=> Quotient = (x4 – 6x3 + 16x2 – 26x + 10 - a)/ (x2 – 2x + k)

So, if the polynomial x4 – 6x3 + 16x2 – 26x + 10 – a is divisible by x2 – 2x + k, the remainder will

be zero.

 Class_10_Maths_Polynomials_Division_Of_Polynomial

 

                                

On comparing,

     -10 + 2k = 0

=> 2k = 10

=> k = 10/2

=> k = 5

and 10 – a – 8k + k2 = 0

=> 10 – a – 8 * 5 + 52 = 0

=> 10 – a – 40 + 25 = 0

=> 35 – a – 40 = 0

=> -5 – a = 0

=> a = -5

Hence, k = 5 and a = -5

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