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

Last Updated : 03 Apr, 2024

This exercise has been deleted as per new NCERT Syllabus

### (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

Î±, Î², Î³ are the zeroes of the cubic polynomial ax3+bx2+cx+d

Î± +Î²+Î³ = â€“b/a

Î±Î²+Î²Î³+Î³Î± = c/a

Î± Î²Î³ = â€“ d/a.

Î±+Î²+Î³ = Â½+1+(-2)

= -1/2 = â€“b/a

Î±Î²+Î²Î³+Î³Î± = (1/2Ã—1)+(1 Ã—-2)+(-2Ã—1/2)

= -5/2 = c/a

Î± Î² Î³ = Â½Ã—1Ã—(-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,

Î± + Î² + Î³ = â€“b/a

= 2+1+1

= 4

â€“b/a = -(-4)/1

Î±Î² + Î²Î³ + Î³Î± = c/a

= 2Ã—1+1Ã—1+1Ã—2

= 5

c/a = 5/1

Î± Î² Î³ = â€“ d/a.

= 2Ã—1Ã—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, â€“7, â€“14 respectively.

Solution:

Let us consider the cubic polynomial as ax3+bx2+cx+d and zeroes of the polynomials be Î±, Î², Î³.

Î±+Î²+Î³ = -b/a = 2/1

Î±Î² +Î²Î³+Î³Î± = c/a = -7/1

Î± Î²Î³ = -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 â€“ b, a, a + b, find a and b.

Solution:

p(x) = x3-3x2+x+1

Zeroes are given as a â€“ 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 â€“ 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 = âˆš2

Therefore,1-âˆš2, 1,1+âˆš2 are the zeroes of x3-3x2+x+1.

### Question 4. If two zeroes of the polynomial x4-6x3-26x2+138x-35are 2 Â±âˆš3,find other zeroes.

Solution:

Degree of polynomial is 4

Therefore, it has four roots

f(x) = x4-6x3-26x2+138x-35

As 2 +âˆš3 and 2-âˆš3 are zeroes of given polynomial f(x).

Therefore, [xâˆ’(2+âˆš3)] [xâˆ’(2-âˆš3)] = 0

(xâˆ’2âˆ’âˆš3)(xâˆ’2+âˆš3) = 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 â€“2xâˆ’35)

On factorizing (x2â€“2xâˆ’35) by splitting the middle term

x2â€“(7âˆ’5)x âˆ’35 = x2â€“ 7x+5x-35

=x(x âˆ’7)+5(xâˆ’7)

(x+5)(xâˆ’7) = 0

x= âˆ’5 and x = 7.

Therefore, all four zeroes of given polynomial equation are: 2+âˆš3, 2-âˆš3, âˆ’5 and 7.

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