- FLashback of Class IX Material
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- Geometric Meaning of the Zero's of the polynomial
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- Relation between coefficient and zero's of the Polynomial
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- Formation of polynomial when the zeros are given
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- Division algorithm for Polynomial

- Ncert Solutions Exercise 2.1
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- Ncert Solutions Exercise 2.2
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- Ncert Solutions Exercise 2.3
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- Ncert Solutions Exercise 2.4

- Polynomials Problem and Solutions-1
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- Polynomials Problem and Solutions-2
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- Polynomial worksheets
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- Polynomial questions
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In this page we have *NCERT Solutions for Class 10 Maths Polynomials* for
EXERCISE 2.4 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

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) 2

(ii)

(i)

Now for verification of zeroes, putting the given value in x.

= (2×1/8) + 1/4 - 5/2 + 2

= 1/4 + 1/4 - 5/2 + 2

= 1/2 - 5/2 + 2 = 0

= (2×1) + 1 - 5 + 2

= 2 + 1 - 5 + 2 = 0

= (2 × -8) + 4 + 10 + 2

= -16 + 16 = 0

Thus, 1/2, 1 and -2 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also, k

Now,

-b/a = k

⇒ 1/2 = 1/2 + 1 - 2

⇒ 1/2 = 1/2

c/a = k

⇒ -5/2 = (1/2 × 1) + (1 × -2) + (-2 × 1/2)

⇒ -5/2 = 1/2 - 2 - 1

⇒ -5/2 = -5/2

-d/a = k

⇒ -2/2 = (1/2 × 1 × -2)

⇒ -1 = 1

Thus, the relationship between zeroes and the coefficients are verified.

(ii)

Now for verification of zeroes, putting the given value in x.

= 8 - 16 + 10 - 2

= 0

= 1 - 4 + 5 - 2

= 0

= 1 - 4 + 5 - 2

= 0

Thus, 2, 1 and 1 are the zeroes of the given polynomial.

Comparing the given polynomial with a

Also, k

Now,

-b/a = k

⇒ 4/1 = 2 + 1 + 1

⇒ 4 = 4

c/a = k

⇒ 5/1 = (2 × 1) + (1 × 1) + (1 × 2)

⇒ 5 = 2 + 1 + 2

⇒ 5 = 5

-d/a = k

⇒ 2/1 = (2 × 1 × 1)

⇒ 2 = 2

Thus, the relationship between zeroes and the coefficients are verified.

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.

Let the polynomial be

Then, k

k

k

∴ a = 1, b = -2, c = -7 and d = 14

So, one cubic polynomial which satisfy the given conditions will be

If the zeroes of the polynomial

Since, (a - b), a, (a + b) are the zeroes of the polynomial

Therefore, sum of the zeroes = (a - b) + a + (a + b) = -(-3)/1 = 3

⇒ 3a = 3 ⇒ a =1

∴ Sum of the products of is zeroes taken two at a time = a(a - b) + a(a + b) + (a + b) (a - b) =1/1 = 1

a

⇒ 3a

Putting the value of a,

⇒ 3(1)

⇒ 3 - b

⇒ b

⇒ b = ±√2

Hence, a = 1 and b = ±√2

If two zeroes of the polynomial x

2+√3 and 2-√3 are two zeroes of the polynomial p(x) = x

Let x = 2±√3

So, x-2 = ±√3

On squaring, we get x

⇒ x

Now, dividing p(x) by x

p(x) = x

= (x

= (x

= (x

= (x

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

Therefore

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

If the polynomial x

On dividing x

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

But the remainder is given as x+ a.

On comparing their coefficients,

2k - 9 = 1

⇒ k = 10

⇒ k = 5 and,

-(8-k)k + 10 = a

⇒ a = -(8 - 5)5 + 10 =- 15 + 10 = -5

Hence, k = 5 and a = -5

Download Polynomails EXERCISE 2.4 as pdf

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