NCERT Solutions Maths Class 10 Chapter 2

# Go To:  Exercise 2.1    Exercise 2.2    Exercise 2.3

## NCERT Solutions Maths Class 10 Exercise 2.4 (Optional)

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

Solution: (i) 2x3 + x2 − 5x + 2; 1/2, 1, −2

Comparing the given polynomial with the standard form of a cubic polynomial ax3 + bx2 + cx + d, we get

a = 2, b = 1, c = −5 and d = 2

= 0

P(1) = 2(1)3 + (1)2 – 5(1) + 2

= 2 + 1 – 5 + 2 = 0

p(−2) = 2(−2)3 + (−2)2 – 5(−2) + 2

= 2(−8) + 4 + 10 + 2 = −16 + 16 = 0

Thus, 1/2, 1 and −2 are the zeroes of 2x3 + x2 – 5x + 2.

Now, Î± + Î² + Î³

=

And Î±Î² + Î²Î³ + Î³Î±

= (1/2)(1) + (1)( −2) + (−2)(1/2)

= ½ 2 1 = 5/2 = c/a

And Î±Î²Î³ = ½ × 1 × (−2) = −1 = −2/2 = −d/a

(ii) x3 − 4x2 + 5x − 2; 2, 1, 1

Comparing the given polynomial with the standard form of a cubic polynomial ax3 + bx2 + cx + d, we get

a = 1, b = −4, c = 5 and d = −2

p(2) = (2)3 − 4(2)2 + 5(2) −2

= 8 – 16 + 10 − 2 = 0

p(1) = (1)3 −4(1)2 + 5(1) −2

= 1 – 4 + 5 − 2 = 0

Thus, 2, 1 and 1 are the zeroes of x3 − 4x2 + 5x – 2.

Now, Î± + Î² + Î³ = 2 + 1 + 1 = 4 = –(–4)/1 = –b/a

And Î±Î² + Î²Î³ + Î³Î± = (2)(1)+(1)(1)+(1)(2)

= 2 + 1 + 2 = 5/1 = c/a

And Î±Î²Î³ = 2 × 1 × 1 = 2 = –(–2)/1 = –d/a

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 are 2, −7, −14 respectively.

Solution:  Let the cubic polynomial be ax3 + bx2 + cx + d and its zeroes be Î±, Î² and Î³.

Then Î± + Î² + Î³ = 2 = –(–2)/1 = –b/a and Î±Î² + Î²Î³ + Î³Î± = –7 = –7/1 = c/a

And Î±Î²Î³ = –14 = –14/1 = –d/a

Here, a = 1, b = −2, c = −7 and d = 14

Hence, the cubic polynomial will be x3 − 2x2 − 7x + 14.

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

Solution:  Since (a − b), a, (a + b) are the zeroes of the polynomial x3 − 3x2 + x + 1,

therefore, Î± + Î² + Î³ = a – b + a + a + b = –(–3)/1 = 3

3a = 3

a = 1

And Î±Î² + Î²Î³ + Î³Î± = (a – b)a + a(a + b) + (a + b)(a – b) = 1/1 = 1

a2 – ab + a2 + ab + a2 − b2 = 1

3a2 − b2 = 1

3(1)2 – b2 = 1 (Since, a = 1)

3 – b2 = 1

b2 = 2

b = ±2

Hence, a = 1 and b = ±2.

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

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

Since (2 ± √3) are two zeroes of the polynomial p(x), x = 2 ± √3 will satisfy the given polynomial.

Thus, x – 2 = ±√3

Squaring both sides, we get

x2 – 4x + 4 = 3

x2 – 4x + 1 = 0

Now, we divide the polynomial p(x) by x2 – 4x + 1 to obtain the other zeroes.

P(x) = x4 – 6x3 – 26x2 + 138x – 35

= (x2 – 4x + 1)(x2 – 2x – 35)

= (x2 – 4x + 1)(x2 – 7x + 5x – 35)

= (x2 – 4x + 1)[x(x –  7) + 5(x – 7)]

= (x2 – 4x + 1)(x + 5)(x – 7)

(x + 5) and (x 7) are the other factors of p(x).

Thus, (x + 5)(x – 7) = 0

⇒ x = –5, 7

Hence, –5 and 7 are the other zeroes of the given polynomial.

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.

Solution: Let us divide the given polynomial x4 – 6x3 + 16x2 – 25x + 10 by  x2 – 2x  + k, we get

Remainder = (2k − 9)x – (8 – k)k + 10

On comparing this remainder with the given remainder, i.e. x + a, we get

2k – 9 = 1

2k = 10

k = 5

And –(8 − k)k + 10 = a

a = −(8 − 5)5 + 10 = −5

Hence, k = 5 and a = −5.