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Sum of the zeroes = – 8/3

Product of the zeroes = 4/3

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x² – 8x/3 + 4/3

P(x)= 3x² – 8x + 4

Using splitting the middle term method,

3x² – 8x + 4 = 0

3x² – (6x + 2x) + 4 = 0

3x² – 6x – 2x + 4 = 0

3x(x – 2) – 2(x – 2) = 0

(x – 2)(3x – 2) = 0

⇒ x = 2, 2/3

Sum of the zeroes = -3/2√5x

Product of the zeroes = – ½

P(x) = x² – (sum of the zeroes) + (product of the zeroes)

Then, P(x)= x²  -3/2√5x – ½

P(x)= 2√5x² – 3x – √5

Using splitting the middle term method,

2√5x² – 3x – √5 = 0

2√5x² – (5x – 2x) – √5 = 0

2√5x² – 5x + 2x – √5 = 0

√5x (2x – √5) – (2x – √5) = 0

(2x – √5)(√5 – 1) = 0

⇒ x = – 1/√5, √5/2

Let the cubic polynomial be 

ax³ + bx³ + cx + d, and its zeroes be α, β and γ. 

Then, 

α + β + γ = 2 = \(\frac{-(-2)}{1}=\frac{-b}{a}\) 

αβ + βγ + γα = -7 = \(\frac{-7}{1}=\frac{c}{a}\)

αβγ = – 14 = \(\frac{-14}{1}=\frac{c}{a}\) 

a = 1, then b = -2, c = -7 and d = 14. 

So, one cubic polynomial which satisfies the given conditions will be x³ – 2x² – 7x + 14.

Let P(x) = x³ - 6x² + 3x + 10

And (a), (a + b) and (a + 2b) are the zeroes of P(x).

We know, 

Sum of the zeroes = - (coefficient of x²) ÷ coefficient of x³

α + β + γ = - b/a

a + (a + b) + (a + 2b) = - (- 6)

3a + 3b = 6

a + b = 2

⇒ a = 2 - b      (1)

Product of all the zeroes = - (constant term) ÷ coefficient of x³

αβγ = - 10

a(a + b)(a + 2b) = - 10

(2 - b) (2) (2 + b) = - 10

(2 - b) (2 + b) = - 5

4 - b² = - 5
⇒ b² = 9

⇒ b = +-3

When b = 3, a = 2 - 3 = - 1 (from (1))

⇒ a = - 1

When b = - 3,a = 2 - (- 3) = 5 (from(1))

⇒ a = 5

Case 1: when a = - 1 and b = 3

The zeroes of the polynomial are:

a = - 1 a + b = - 1 + 3 = 2 a + 2b = - 1 + 2(3) = 5

⇒ - 1, 2 and 5 are the zeroes

Case 2: when a = 5,b = - 3

The zeroes of the polynomial are:

a = 5 a + b = 5 - 3 = 2 a + 2b = 5 - 2(3) = - 1

⇒ - 1, 2 and 5 are the zeroes

By both the cases the zeroes of the polynomial are - 1, 2, 5

Let p(x) =a⁵ -4a²x³ +2x + 2a +3

Since, x + 2a is a factor of p(x), then put p(-2a) = 0

(-2a)⁵ – 4a² (-2a)³ + 2(-2a) + 2a + 3 = 0 

=> -32a⁵ + 32a⁵ -4a + 2a+ 3 = 0

=> -2a + 3=0

2a =3

a = 3/2.

Hence, the value of a is 3/2.

9y² − 66yz + 121z²

Using (a − b)² = a² + b² − 2ab

= (3y)² + (11z)² − 2 × 3y × 11

= (3y − 11z)²

(d) Both (a) and (c)

Let f(x) = ax⁴ + bx⁴ + cx² + dx + e be the given polynomial. 

Then, (x² – 1) is a factor of f(x). 

⇒ (x – 1) (x + 1) is a factor of f(x) 

⇒ (x – 1) and (x + 1) are factors of f(x) 

⇒ f(1) = 0 and f(–1) = 0 

⇒ a + b + c + d + e = 0 and a – b + c – d + e = 0. 

Adding and subtracting the two equations, we get 

2(a + c + e) = 0 and 2(b + d) = 0 

⇒ a + c + e = 0 and b + d = 0.

Let f(x) = ax⁴ + bx³ + cx² + dx + e 

As (x – 1) is a factor of f(x) we have 

x² – 1 = (x + 1) (x – 1) hence f(1) = 0 and f(-1) = 0 

f(1) = a + b + c + d + e = 0 ……………. (1) 

and f(-1) = a- b + c- d + e = 0 

⇒ a + c + e = b + d 

Substitute this value in equation (1) 

a + c + e + b + d=0 

b + d + b + d=0 

2 (b + d) = 0 

⇒ b + d = 0 

∴ a + c + e = b + d = 0

Correct option is (A) 2(3p – 5q)

Let \((x-\alpha)\) be common factor of f(x) and g(x).

\(\therefore\) \(x=\alpha\) be common zero of f(x) and g(x).

\(\therefore\) \(\alpha^2+5\alpha+p=0\) and \(\alpha^2+3\alpha+q=0\)

\(\Rightarrow\) \((\alpha^2+5\alpha+p)\) \(-(\alpha^2+3\alpha+q)=0\)

\(\Rightarrow\) \(2\alpha+p-q=0\)

\(\Rightarrow\) \(\alpha=\frac{q-p}2\) is a common root of \(\alpha^2+5\alpha+p=0\)

\(\therefore\) \((\frac{q-p}2)^2+5(\frac{q-p}2)+p=0\)

\(\Rightarrow\) \((q-p)^2+10(q-p)+4p=0\)

\(\Rightarrow\) \((q-p)^2+10q-6p=0\)

\(\Rightarrow\) \((q-p)^2=6p-10q\) = 2 (3p – 5q)

Correct option is A) 2(3p – 5q)

Let f(x) = ax² + bx + c and g(x) = bx² + ax + c 

given that (x + 1) is a common factor for both f(x) and g(x). 

∴ f(-1) = g(- 1) 

⇒a(- 1)² + b(- 1) + c 

= b(- 1)² + a (- 1) + c 

⇒ a – b + c = b – a + c 

⇒ a + a = b + b 

⇒ 2a = 2b 

⇒ a = b 

Also f(- 1) = a – b + c = 0 

⇒ b – b + c = 0 

⇒ c = 0

Let f(x) = x² – x – 6 and

g(x) = x² + 3x – 18

Given that (x – a) is a factor of both f(x) and g(x). 

f(a) = g(a) = 0 

⇒ a² – a – 6 = a² + 3a – 18 

⇒ – 4a = – 18 + 6 

⇒ – 4a = – 12 

∴ a = 3

(i) 103³

= (100 + 3)³

= (100)³ + (3)³ + 3 x 100 x 3 x (100 + 3)

= 1000000 + 27 + 900 (103)

= 1000027 + 92700

= 1092727

(ii) 101 x 102

= (100 + 1) (100 + 2)

= (100) + 100(1 + 2) + 1 x 2

= 10000 + 300 + 2

= 10302

(iii) (999)²

= (1000 - 1)²

= (1000)² + (1)² - 2 x 1000 x 1

= 1000000 + 1 - 2000

= 998001 

Correct option is C) LCM and HCF are equal

Let the zeroes of the quadratic polynomial be

α = 3 + √7, β = 3 – √7

Then, α + β = 3 + √7 + 3 – √7 = 6

αβ = (3 + √(7)) × (3 – √7) = 9 – 7 = 2

Sum of zeroes = α + β = 6

Product of zeroes = αβ = 2

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (6)x + 2

= x² – 6x + 2

Let larger part be x

Smaller part = 16 – x

According to question,

2x² = (16 – x)² + 164

⇒ 2x² – (16 – x)² – 164 = 0

⇒ 2x² – [256 – 32x + x²] – 164 = 0

⇒ 2x² – 256 + 32x – x² – 164 = 0

⇒ x² + 32x – 420 = 0

⇒ x² + 42x – 10x – 420 = 0

⇒ x(x + 42) – 10(x + 42) = 0

⇒ (x + 42) (x – 10) = 0

⇒ x = -42 or x = 10

⇒ x = 10 [x = -42 not possible]

Larger number (x) = 10

Smaller number (16- x) = 16 – 10 = 6

Hence two numbers are 10, 6