NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 2.4.

Board |
CBSE |

Textbook |
NCERT |

Class |
Class 10 |

Subject |
Maths |

Chapter |
Chapter 2 |

Chapter Name |
Polynomials |

Exercise |
Ex 2.4 |

Number of Questions Solved |
5 |

Category |
NCERT Solutions |

### NCERT Solutions for Class 10 Maths Chapter 2 Polynomials Ex 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; \(\frac { 1 }{ 4 }\), 1, -2

**(ii)**x

^{3}– 4x

^{2}+ 5x – 2; 2, 1, 1

**Solution:**

**(i)**Comparing the given polynomial with ax

^{3}+ bx

^{2}+ cx + d, we get:

a = 2, b – 1, c = -5 and d = 2.

∴ p(x) = 2x

^{3}+ x

^{2}– 5x + 2

**(ii)**Compearing the given polynomial with ax

^{3}+ bx

^{2}+ cx + d, we get:

a = 1, b = -4, c = 5 and d = – 2.

∴ p (x) = x

^{3}– 4x

^{2}+ 5x – 2

⇒ p(2) = (2)

^{3}– 4(2)

^{2}+ 5 x 2 – 2

= 8 – 16+ 10 – 2 = 0

p(1) = (1)

^{3}– 4(1)

^{2}+ 5 x 1-2

= 1 – 4 + 1 – 2

= 6-6 = 0

Hence, 2, 1 and 1 are the zeroes of x

^{3}– 4x

^{2 }+ 5x – 2.

Hence verified.

Now we take α = 2, β = 1 and γ = 1.

α + β + γ = 2 + 1 + 1 = \(\frac { 4 }{ 1 }\) = \(\frac { -b }{ a }\)

αβ + βγ + γα = 2 + 1 + 2 = \(\frac { 5 }{ 1 }\) = \(\frac { c }{ a }\)

αβγ = 2 x 1 x 1 = \(\frac { 2 }{ 1 }\) = \(\frac { -d }{ a }\).

Hence 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.

**Solution:**

Let α , β and γ be the zeroes of the required polynomial.

Then α + β + γ = 2, αβ + βγ + γα = -7 and αβγ = -14.

∴ Cubic polynomial

= x

^{3}– (α + β + γ)x

^{2}+ (αβ + βγ + γα)x – αβγ

= x

^{3}– 2x

^{2}– 1x + 14

Hence, the required 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.

**Solution:**

Let α , β and γ be the zeroes of polynomial x

^{3}– 3x

^{2}+ x + 1.

Then α = a-b, β = a and γ = a + b.

∴ Sum of zeroes = α + β + γ

⇒ 3 = (a – b) + a + (a + b)

⇒ (a – b) + a + (a + b) = 3

⇒ a-b + a + a + b = 3

⇒ 3a = 3

⇒ a = \(\frac { 3 }{ 3 }\) = 1 …(i)

Product of zeroes = αβγ

⇒ -1 = (a – b) a (a + b)

⇒ (a – b) a (a + b) = -1

⇒ (a

^{2}– b

^{2})a = -1

⇒ a

^{3}– ab

^{2}= -1 … (ii)

Putting the value of a from equation (i) in equation (ii), we get:

(1)

^{3}-(1)b

^{2}= -1

⇒ 1 – b

^{2}= -1

⇒ – b

^{2}= -1 – 1

⇒ b

^{2}= 2

⇒ b = ±√2

Hence, a = 1 and b = ±√2.

**Question 4.
**If two zeroes of the polynomial x

^{4}– 6x

^{3}– 26x

^{2 }+ 138x – 35 are 2 ± √3, finnd other zeroes.

**Solution:**

Since two zeroes are 2 + √3 and 2 – √3,

∴ [x-(2 + √3)] [x- (2 – √3)]

= (x-2- √3)(x-2 + √3)

= (x-2)

^{2}– (√3)

^{2 }x

^{2}– 4x + 1 is a factor of the given polynomial.

Now, we divide the given polynomial by x

^{2}– 4x + 1.

So, x

^{4}– 6x

^{3}– 26x

^{2}+ 138x – 35

= (x

^{2}– 4x + 1) (x

^{2}– 2x – 35)

= (x

^{2}– 4x + 1) (x

^{2}– 7x + 5x – 35)

= (x

^{2}-4x + 1) [x(x- 7) + 5 (x-7)]

= (x

^{2}– 4x + 1) (x – 7) (x + 5)

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

**Question 5.
**If the polynomial x

^{4}– 6x

^{3}+ 16x

^{2}– 25x + 10 is divided by another polynomial x

^{2}– 2x + k, the remainder comes out to be x + a, find k and a.

**Solution:**

We have

p(x) = x

^{4}– 6x

^{3}+ 16x

^{2}– 25x + 10

Remainder = x + a … (i)

Now, we divide the given polynomial 6x

^{3}+ 16x

^{2}– 25x + 10 by x

^{2}– 2x + k.

Using equation (i), we get:

(-9 + 2k)x + 10-8 k + k

^{2}= x + a

On comparing the like coefficients, we have:

-9 + 2k = 1

⇒ 2k = 10

⇒ k = \(\frac { 10 }{ 2 }\) = 5 ….(ii)

and 10 -8k + k

^{2}– a ….(iii)

Substituting the value of k = 5, we get:

10 – 8(5) + (5)

^{2}= a

⇒ 10 – 40 + 25 = a

⇒ 35 – 40 = a

⇒ a = -5

Hence, k = 5 and a = -5.

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