Class 10 NCERT Solutions - Chapter 2 Polynomials - Exercise 2.4

Exercise 2.4 of Chapter 2 (Polynomials) in Class 10 NCERT Mathematics focuses on the division algorithm for polynomials. This exercise helps students understand how to divide polynomials and find quotients and remainders, which are essential skills in algebra.

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

Solution:

p(x) = 2x³+x²-5x+2

p(1/2) = 2(1/2)³+(1/2)²-5(1/2)+2 

           = (1/4)+(1/4)-(5/2)+2 

           = 0

p(1) = 2(1)³+(1)²-5(1)+2 = 0

p(-2) = 2(-2)³+(-2)²-5(-2)+2 = 0

Therefore, 1/2, 1, -2 are the zeroes of 2x³+x²-5x+2.

Now, comparing the given polynomial with general expression

ax³+bx²+cx+d = 2x³+x²-5x+2

a=2, b=1, c= -5 and d = 2

α, β, γ are the zeroes of the cubic polynomial ax³+bx²+cx+d

α +β+γ = –b/a

αβ+βγ+γα = c/a

α βγ = – d/a.

α+β+γ = ½+1+(-2) 

            = -1/2 = –b/a

αβ+βγ+γα = (1/2×1)+(1 ×-2)+(-2×1/2) 

                  = -5/2 = c/a

α β γ = ½×1×(-2) 

         = -2/2 = -d/a

Hence, the relationship between the zeroes and the coefficients are satisfied.

(ii) x³ - 4x² + 5x - 2 ;2, 1, 1

Solution:

p(x) = x³-4x²+5x-2

Zeroes are 2,1,1.

p(2)= 2³-4(2)²+5(2)-2 

      = 0

p(1) = 1³-(4)(1² )+(5)(1)-2 = 0

Therefore, proved, 2, 1, 1 are the zeroes of x³-4x²+5x-2

On comparing the given polynomial with general expression

ax³+bx²+cx+d = x³-4x²+5x-2

a = 1, b = -4, c = 5 and d = -2

Therefore,

α + β + γ = –b/a

                = 2+1+1 

                = 4 

         –b/a = -(-4)/1

αβ + βγ + γα = c/a

                      = 2×1+1×1+1×2 

                      = 5 

                c/a = 5/1 

α β γ = – d/a.

          = 2×1×1 

          = 2 

 -d/a = -(-2)/1

Hence, the relationship between the zeroes and the coefficients is satisfied.

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.

Solution:

Let us consider the cubic polynomial as ax³+bx²+cx+d and zeroes of the polynomials be α, β, γ.

α+β+γ = -b/a = 2/1

αβ +βγ+γα = c/a = -7/1

α βγ = -d/a = -14/1

On comparing

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

Therefore, the cubic polynomial is x³-2x²-7x+14

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

Solution:

p(x) = x³-3x²+x+1

Zeroes are given as a – b, a, a + b

px³+qx²+rx+s = x³-3x²+x+1

On comparing

p = 1, q = -3, r = 1 and s = 1

Sum of zeroes = a – b + a + a + b

-q/p = 3a

Putting the values q and p.

-(-3)/1 = 3a

a=1

Therefore, zeroes are 1-b, 1, 1+b.

Product of zeroes = 1(1-b)(1+b)

-s/p = 1-b²

-1/1 = 1-b²

b² = 1+1 = 2

b = √2

Therefore,1-√2, 1,1+√2 are the zeroes of x³-3x²+x+1.

Question 4. If two zeroes of the polynomial x⁴-6x³-26x²+138x-35are 2 ±√3,find other zeroes.

Solution:

Degree of polynomial is 4

Therefore, it has four roots

f(x) = x⁴-6x³-26x²+138x-35

As 2 +√3 and 2-√3 are zeroes of given polynomial f(x).

Therefore, [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

Therefore, x²-4x+1 is a factor of polynomial f(x).

Let it be g(x) = x²-4x+1

By dividing f(x) by g(x) we get another factor of f(x)

x⁴-6x³-26x²+138x-35 = (x²-4x+1)(x² –2x−35)

On factorizing (x²–2x−35) by splitting the middle term

x²–(7−5)x −35 = x²– 7x+5x-35

                         =x(x −7)+5(x−7)

(x+5)(x−7) = 0

x= −5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+√3, 2-√3, −5 and 7.

Summary

Exercise 2.4 in Chapter 2 of Class 10 NCERT Mathematics focuses on the division algorithm for polynomials. This crucial concept teaches students how to divide polynomials of various degrees, find quotients and remainders, and apply the division algorithm effectively. The exercise covers a range of problems, from simple divisions to more complex ones, helping students build a strong foundation in polynomial manipulation. Understanding this topic is essential for advanced algebraic concepts and problem-solving in mathematics.