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(i) Let p(x) = 3x² + 7x + 2

Here we have to split middle term of p(x)

i.e. 7 into two parts such that their sum is 7 and product is 2 x 3 = 6.

6 and 1 are such numbers whose sum is 7 and product is 6

∴ 3x² + 7x + 2 = 3x² + 6x + x + 2

= 3x(x + 2) + 1(x + 2)

= (x + 2)(3x + 1)

(ii) Let p(x) = 4x² – x – 3

Here we have to split middle term – 1 into two parts such that their sum is – 1 product is 

3 × – 4 = – 12

– 4 and 3 are such numbers whose sum is – 1 and product is – 12

4x² – x – 3 = 4x² – 4x + 3x – 3

= 4x(x – 1) + 3(x – 1)

= (x – 1 )(4x + 3)

(iii) Let p(x) = 12x² – 7x + 1

Here we have to split middle term – 7 into two parts such that their sum is – 7 and product is 

12 × 1 = 12

  – 4 and – 3 are such numbers whose sum is – 7 and product is 12

12x² – 7x + 1 = 12x² – 4x – 3x + 1

= 4x(3x – 1) – 1(3x – 1)

= (3x – 1)(4x – 1)

(iv) Let p(x) = 6x² + 5x – 6

Here we have to split middle term 5 into two parts such that their sum is 5 and product is 

6 × – 6 = – 36

9 and – 4 are such numbers whose sum is 5 and product is – 36. .

6x² + 5x – 6 

= 6x² + 9x – 5x – 6

= 3x(2x + 3) – 2(2x + 3) 

= (2x + 3)(3x – 2)

(i) Let p(x) = x⁴ – 2x³ – 3x² + 2x + 2 and
g(x) = x – 1

If x – 1 i.e. g(x) be a factor of p(x) then

P(1) = 0

Now p(1) = (1)⁴ – 2(1)³ – 3(1)² + 2(1) + 2

= 1 – 2 – 3 + 2 + 2 = 0

∵ p(1) = 0

=> x – 1 is a factor of

p(x) = x⁴ – 2x³ – 3x² + 2x + 2.

(ii) Let p(x) = x⁴ + x³ + x² + x + 1

If x – 1 is a factor of p(x) then p(1) = 0

p(1) = (1)⁴ + (1)³ + (1)² + (1) + 1

= 1 + 1 + 1 + 1 + 1 = 5

p(1) ≠ 0

∴ x – 1 is a factor of x⁴ + x³ + x² + x + 1.

(iii) Let p(x) = x⁴ + 3x³ – 3x² + x – 2

If x – 1 is  a factor of p(x) then p(1) = 0

∴ p(1) = (1)⁴ + 3(1)³ – 3(1)² + (1) – 2

= 1 + 3 – 3 + 1 – 2 = 0

p(1) = 0

∴ x – 1 is a factor of x⁴ + 3x³ – 3x² + x – 2.

(iv) Let p(x) = x³ – x² – (2 + √3)x + √3

If x – 1 is a factor of p(x) then p(1) = 0

∴ p(1)= (1)³ – (1)² – (2 + √3)(1) + √3

= 1 – 1 – 2 – √3 + √3 = -2

∴ p( 1) ≠ 0

Hence, x – 1 is not a factor of x³ – x² – (2 + √3)x + √3

Let p(x) = x³ – 3x² + kx – 10

When x – 5 is a factor of p(x) then P( 5) = 0

⇒ (5)³ – 3(5)² + k x 5 – 10 = 0

⇒ 125 – 75 + 5k – 10 = 0

⇒ 5k – 40 = 0

k = \(\frac { 40 }{ 5 }\) = 8

Hence, k = 8

Let p(x) = 4x⁴ – 3x³ + 2x² – 5x + 1

Now

P(1) = 4(1)⁴ – 3(1)³ + 2(1)² – 5(1) + 1 

= 4 – 3 + 2 – 5 + 1 = -1

p(1) ≠ 0

1 is not a zero of p(x).

Answer is (A) - √3

We know that

(a – b)³ = a³ – b³ – 3ab(a – b)

Using this identity,

(3a – 2b)³ = (3a)³ – (2b)³ -3 x 3a x 2b (3a – 2b)

⇒ (3a – 2b)³ = 27a³ – 8b³ – 18ab(3a – 2b)

Now substituting 3a – 2b = 11 and ab = 12,

we get

(11)² = 27a³ – 8b³ – 18 x 12 x 11

⇒ 1331 = 27a³ – 8b³ – 2376

⇒ 27a³ – 8b² = 1331 + 2376

⇒ 27a³ – 8b³ = 3707

Let p(x) = ax³ + bx² + x – 6

(x + 2) is a factor of p(x)

⇒ p(-2) = 0 [∴ x + 2 = 0 ⇒ x = -2]

⇒ a(-2)³ + b(-2)² + (-2) – 6 = 0

⇒ -8a + 4b – 2 – 6 = 0

⇒ -8a + 4b = 8

⇒ -2a + b = 2 …(i)

It is given that p(x) leaves the remainder 4 when it divided by (x – 2) i.e. p(2) = 4.

⇒ a(2)³ + b(2)² + (2) – 6 = 4

⇒ 8a + 4b – 4 = 4

⇒ 8a + 46 = 8

⇒ 2a + b = 2 …(ii)

Solving (i) and (ii), we get a = 0 and b = 2.

Let p(x) = (2√2x² + 5x + √2)

If x + √2 is a factor of p(x), then according to factor theorem p(√2) = 0.

p(-√2) = 2√2(-√2)² + 5(-√2) + √2 

= 2√2 x 2 – 5√2 + √2 = 5√2 – 5√2 = 0

p(-√2) = 0

(x + √2) is a factor p(x) 

i.e. (2√2 x² + 5x + √2).

Let p(x) = x³ + 10x² + ax + b

p(x) i.e. x³ + 10x² + ax + 6 is exactly divisible by (x – 1) and (x + 2)

Therefore, p(1) and p(-2) must equal to zero.

p(1) = (1)³ + 10(1)² + a x 1 + b = 0

⇒ 1 + 10 + a + b = 0

⇒ a + b = – 11 …(i)

Also, p(-2) = 0

(-2)³ + 10(-2)² + a x (- 2) + b = 0

-8 + 40 – 2a + b = 0

⇒ – 2a + b = – 32 …(ii)

Solving (i) and (ii), we get

a = 7 and b = -18.

Let the given polynomials are 

p(x) = 3x³ + ax² + 3x + 5 

and f(x) = 4x³ + x² – 2x + a

According to question p(2) = f(2)

3(2)³ + a(2)² + 3(2) + 5 

= 4(2)³ + (2)² – 2(2) + a

⇒ 3 x 8 + 4a + 6 + 5 

= 32 + 4 – 4 + a

⇒ 24 + 4a + 11 = 32 + a

⇒ 4a – a = 32 – 35

⇒ 3a = -3

⇒ a = -1

As the remainder is same, 

so p(2) or f(2) would be same when a = – 1

p(2) = 3(2)³ + (-1)(2)² + 3 x 2 + 5 [∴ a = -1]

= 24 – 4 + 6 + 5

= 35 – 4 = 31

Thus, a = – 1 and p(2) or f(2) = 31

We have,

L.H.S.

= (a + b + c)³ – a³ -b³ – c³

= {(a + b + c)³ – (a)³} – (b³ + c³)

= (a + b + c – a){(a + b + c)² + a(a + b + c) + a²} – (b + c){b² – bc + c²)

[∵ x³ – y³ = (x – y)(x² + xy + y²) and x³ + y³ = (x + y)(x² – xy + y²)]

= (b + c){a² + b² + c² + 2ab + 2bc + 2ca + a² + ab + ac + a²} – (b + c)(b² – bc + c²)

= (b + c)[3a² + b² + c² + 3ab + 2bc + 3ca – b² + bc – c²]

= (b + c)[3a² + 3ab + 3bc + 3ca]

= 3(b + c)[a² + ab + bc + ca]

= 3(b + c)[a(a + b) + c(a + b)]

= 3(b + c)(a + b)(a + c)

= 3(a + b)(b + c)(c + a)

= R.H.S.

Answer is (A) 3√2 - x³ + 2x

Answer is (D) 1