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EXERCISE 2.4 on pages 36 and 37 . Hope you like them and do not forget to like , social_share
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If $k_1$, $k_2$, $k_3$ are the zeroes of the cubic polynomial $ax^3 + bx^2 + c^x + d$, then

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 Polynomials EXERCISE 2.4 as pdf

**Notes****NCERT Solutions****Assignments**

Given below are the links of some of the reference books for class 10 math.

- Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
- Mathematics for Class 10 by R D Sharma
- Pearson IIT Foundation Maths Class 10
- Secondary School Mathematics for Class 10
- Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.

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