# Class 10 NCERT Solutions – Chapter 2 Polynomials – Exercise 2.4

**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) 2x**^{3 }+ x^{2 }– 5x + 2; 1/2, 1, -2

^{3 }+ x

^{2 }– 5x + 2; 1/2, 1, -2

**Solution:**

p(x) =

2x^{3}+x^{2}-5x+2p(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 = 0p(-2) = 2(-2)

^{3}+(-2)^{2}-5(-2)+2 = 0Therefore, 1/2, 1, -2 are the zeroes of 2x

^{3}+x^{2}-5x+2.Now, comparing the given polynomial with general expression

ax

^{3}+bx^{2}+cx+d = 2x^{3}+x^{2}-5x+2a=2, b=1, c= -5 and d = 2

α, β, γ are the zeroes of the cubic polynomial ax

^{3}+bx^{2}+cx+dα +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) x**^{3 }– 4x^{2 }+ 5x – 2 ;**2, 1, 1**

^{3 }– 4x

^{2 }+ 5x – 2

**Solution:**

p(x) = x

^{3}-4x^{2}+5x-2Zeroes are 2,1,1.

p(2)= 2

^{3}-4(2)^{2}+5(2)-2= 0

p(1) = 1

^{3}-(4)(1^{2 })+(5)(1)-2 = 0Therefore, proved, 2, 1, 1 are the zeroes of x

^{3}-4x^{2}+5x-2On comparing the given polynomial with general expression

ax

^{3}+bx^{2}+cx+d = x^{3}-4x^{2}+5x-2a = 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 ax

^{3}+bx^{2}+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 x

^{3}-2x^{2}-7x+14

**Question 3. If the zeroes of the polynomial x**^{3 }– 3x^{2 }+ x + 1 **are a – b, a, a + b, find a and b.**

^{3 }– 3x

^{2 }+ x + 1

**Solution:**

p(x) = x

^{3}-3x^{2}+x+1Zeroes are given as a – b, a, a + b

px

^{3}+qx^{2}+rx+s = x^{3}-3x^{2}+x+1On 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-b

^{2}-1/1 = 1-b

^{2}b

^{2}= 1+1 = 2b = √2

Therefore,1-√2, 1,1+√2 are the zeroes of x

^{3}-3x^{2}+x+1.

**Question 4. If two zeroes of the polynomial x**^{4}-6x^{3}-26x^{2}+138x-35 **are 2 ±**√**3,** **find other zeroes.**

^{4}-6x

^{3}-26x

^{2}+138x-35

**Solution:**

Degree of polynomial is 4

Therefore, it has four roots

f(x) = x

^{4}-6x^{3}-26x^{2}+138x-35As 2 +√3

and 2-√3are zeroes of given polynomial f(x).Therefore, [x−(2+√3)] [x−(2-√3)] = 0

(x−2−√3)(x−2+√3) = 0

Therefore, x

^{2}-4x+1 is a factor of polynomial f(x).Let it be g(x) = x

^{2}-4x+1By dividing f(x) by g(x) we get another factor of f(x)

x

^{4}-6x^{3}-26x^{2}+138x-35 = (x^{2}-4x+1)(x^{2}–2x−35)On factorizing (x

^{2}–2x−35) by splitting the middle termx

^{2}–(7−5)x −35 = x^{2}– 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.

## Please

Loginto comment...