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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 $ax^3 + bx^2 + cx + d$, 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) 2x

(ii)x

(i)p(x) = 2x

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

P (1/2) = 2(1/2)

=(2×1/8)+ 1/4 - 5/2+ 2

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

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

P (1) = 2(1)

=(2×1)+ 1 - 5 + 2

= 2+ 1 - 5 + 2 = 0

P (-2) = 2(-2)

=(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 ax

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) p(x) =x

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

p(2) = 2

= 8 - 16 +10 - 2

= 0

p(1) = 1

= 1 - 4+5 - 2

= 0

p(1) = 1

= 1 - 4+5 - 2

= 0

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

Comparing the given polynomial with ax

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 beax

Then, k

k

k

We can take any value of a and then find the values of b,c and d

Taking a = 1

b = -2, c = -7 and d = 14

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

If the zeroes of the polynomialx

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

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

So, $(x- 2 + \sqrt 3)(x -2 - \sqrt 3)$ is a factor of the polynomial

Multiplying them

$x^2- 4x + 1$ is a factor of polynomial p(x)

Another Method to get this would be

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

- NCERT Solutions for Class 10 Maths Polynomials Exercise 2.4 has been prepared by Expert with utmost care. If you find any mistake.Please do provide feedback on mail. You can download the solutions as PDF in the below Link also

Download Polynomials Class 10 EXERCISE 2.4 as pdf - This chapter 2 has total 4 Exercise 2.1 ,2.2,2.3 and 2.4. This is the Fourth exercise in the chapter.You can explore previous exercise of this chapter by clicking the link below

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