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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

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