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

- Polynomials Class 10 Ex 2.1
- Polynomials Class 10 Ex 2.2
- Polynomials Class 10 Ex 2.3
- Polynomials Class 10 Ex 2.4

Board | CBSE |

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 2 |

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

**Ex 2.4 Class 10 Maths 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)**Â 2x

^{3}Â + x

^{2}Â â€“ 5x + 2; Â , 1, -2

**(ii)**Â x

^{3}Â â€“ 4x

^{2}Â + 5x â€“ 2; 2, 1, 1

**Solution:**

(i)Â Comparing the given polynomial with ax

(i)

^{3}Â + bx

^{2}Â + cx + d, we get:

a = 2, b â€“ 1, c = -5 and d = 2.

âˆ´Â Â p(x) = 2x

^{3}Â + x

^{2}Â â€“ 5x + 2

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

^{3}Â + bx

^{2}Â + cx + d, we get:

a = 1, b = -4, c = 5 and d = â€“ 2.

âˆ´ p (x) = x

^{3}Â â€“ 4x

^{2}Â + 5x â€“ 2

â‡’Â p(2) = (2)

^{3}Â â€“ 4(2)

^{2}Â + 5 x 2 â€“ 2

= 8 â€“ 16+ 10 â€“ 2 = 0

p(1) = (1)

^{3}Â â€“ 4(1)

^{2}Â + 5 x 1-2

= 1 â€“ 4 + 1 â€“ 2

= 6-6 = 0

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

^{3}Â â€“ 4x

^{2Â }+ 5x â€“ 2.

Hence verified.

Now we take Î± = 2, Î² = 1 and Î³ = 1.

Î± + Î² + Î³ = 2 + 1 + 1 =Â Â =Â

Î±Î² + Î²Î³ + Î³Î± = 2 + 1 + 2 =Â Â =Â

Î±Î²Î³ = 2 x 1 x 1Â =Â Â =Â .

Hence verified.

**Ex 2.4 Class 10 MathsÂ 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 Î± , Î² andÂ Î³ be the zeroes of the required polynomial.

Then Î± + Î² + Î³ = 2, Î±Î² + Î²Î³ + Î³Î± = -7 and Î±Î²Î³ = -14.

âˆ´ Cubic polynomial

= x

^{3}Â â€“ (Î± + Î² + Î³)x

^{2}Â + (Î±Î² + Î²Î³ + Î³Î±)x â€“ Î±Î²Î³

= x

^{3}Â â€“ 2x

^{2}Â â€“ 1x + 14

Hence, the required cubic polynomial is x

^{3}Â â€“ 2x

^{2}Â â€“ 7x + 14.

**Ex 2.4 Class 10 MathsÂ Question 3.**If the zeroes of the polynomial x

^{3}Â â€“ 3x

^{2}Â + x + 1 are a-b, a, a + b, find a and b.

**Solution:**

Let Î± , Î² andÂ Î³ be the zeroes of polynomial x

^{3}Â â€“ 3x

^{2}Â + x + 1.

Then Î± =Â a-b, Î² = a and Î³ = a + b.

âˆ´ Sum of zeroes = Î± + Î² + Î³

â‡’Â Â 3 = (a â€“ b) + a + (a + b)

â‡’Â (a â€“ b) + a + (a + b) = 3

â‡’Â a-b + a + a + b = 3

â‡’ Â Â Â 3a = 3

â‡’ a =Â Â Â = 1 â€¦(i)

Product of zeroes = Î±Î²Î³

â‡’ -1 = (a â€“ b) a (a + b)

â‡’ (a â€“ b) a (a + b) = -1

â‡’Â Â (a

^{2}Â â€“ b

^{2})a = -1

â‡’ Â a

^{3}Â â€“ ab

^{2}Â = -1Â Â â€¦ (ii)

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

(1)

^{3}-(1)b

^{2}Â = -1

â‡’ 1 â€“ b

^{2}Â = -1

â‡’ â€“ b

^{2}Â = -1 â€“ 1

â‡’Â Â b

^{2}Â = 2

â‡’ b = Â±âˆš2

Hence, a = 1 and b = Â±âˆš2.

**Ex 2.4 Class 10 MathsÂ Question 4.**If two zeroes of the polynomial x

^{4}Â â€“ 6x

^{3}Â â€“ 26x

^{2Â }+ 138x â€“ 35 are 2 Â± âˆš3, finnd other zeroes.

**Solution:**

Since two zeroes are 2 + âˆš3 and 2 â€“ âˆš3,

âˆ´Â [x-(2 + âˆš3)] [x- (2 â€“ âˆš3)]

= (x-2- âˆš3)(x-2 + âˆš3)

= (x-2)

^{2}â€“ (âˆš3)

^{2}x

^{2}Â â€“ 4x + 1 is a factor of the given polynomial.

Now, we divide the given polynomial by x

^{2}Â â€“ 4x + 1.

So, x

^{4}Â â€“ 6x

^{3}Â â€“ 26x

^{2}Â + 138x â€“ 35

= (x

^{2}Â â€“ 4x + 1) (x

^{2}Â â€“ 2x â€“ 35)

= (x

^{2}Â â€“ 4x + 1) (x

^{2}Â â€“ 7x + 5x â€“ 35)

= (x

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

= (x

^{2}Â â€“ 4x + 1) (x â€“ 7) (x + 5)

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

**Ex 2.4 Class 10 Maths Question 5.**If the polynomial x

^{4}â€“ 6x

^{3}+ 16x

^{2}â€“ 25x + 10 is divided by another polynomial x

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

**Solution:**

We have

p(x) = x

^{4}â€“ 6x

^{3}+ 16x

^{2}â€“ 25x + 10

Remainder = x + a â€¦ (i)

Now, we divide the given polynomial 6x

^{3}+ 16x

^{2}â€“ 25x + 10 by x

^{2}â€“ 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

â‡’ 2k = 10

â‡’ k = = 5 â€¦.(ii)

and 10 -8k + k

^{2}â€“ a â€¦.(iii)

Substituting the value of k = 5, we get:

10 â€“ 8(5) + (5)

^{2}= a

â‡’ 10 â€“ 40 + 25 = a

â‡’ 35 â€“ 40 = a

â‡’ a = -5

Hence, k = 5 and a = -5.

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