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(i) According to the question,

Let p(x) =3x² + 6x−24 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Therefore, zero of g(x) = 2

So, substituting the value of x in p(x), we get,

p(2) = 3(2)² + 6 (2) – 24

= 12 + 12 – 24

= 0

Since, the remainder = zero,

We can say that,

g(x) = x – 2 is factor of p(x) = 3x² + 6x−24

(ii) According to the question,

Let p(x) = 4x² + x − 2 and g(x) = x – 2

g(x) = x – 2

zero of g(x) ⇒ g(x) = 0

x – 2 = 0

x = 2

Therefore, zero of g(x) = 2

So, substituting the value of x in p(x), we get,

p(2) = 4(2)² + 2−2

= 16 ≠ 0

Since, the remainder = zero,

We can say that,

g(x) = x – 2 is factor of p(x) = 4x² + x −2

6x² + 19x + 15 = 6x² + 10x + 9x + 15 

= 2x (3x + 5) + 3 (3x + 5) 

= (3x + 5) (2x + 3)

10m² – 31m – 132 

= 10m² – 55m + 24m – 132 

= 5m (2m- 11) + 12 (2m- 11) 

= (2m – 11) (5m + 12)

2x² − 7x − 15

= 2x² − 10x + 3x − 15

= 2x(x − 5) +3(x − 5)

= (x − 5)(2x + 3)

12x² + 11x + 2

= 12x² + 8x + 3x + 2 

= 4x (3x + 2) + 1 (3x + 2) 

= (3x + 2) (4x + 1)

Correct option is (C) p(x) = 2x + 3

If a polynomial equation has n degree then number of zeros of that polynomial equation is n.

In among all options p(x) = 2x+3 is a linear polynomial which is a polynomial of degree 1.

Thus, it has only 1 zero.

Correct option is C) p(x) = 2x + 3

Correct option is B) A(x) = – x² + 7x + 30

Correct option is (C) a = -9, b = 20, c = -12

\(\because\) f(1) = 0

\(\Rightarrow\) \(1^3+a.1^2+b.1+c=0\)

\(\Rightarrow\) a+b+c = -1  ________(1)

\(\because\) f(2) = 0

\(\Rightarrow\) \(2^3+a.2^2+b.2+c=0\)

\(\Rightarrow\) 4a+2b+c = -8  ________(2)

\(\because\) f(4) = f(0)

\(\Rightarrow\) \(4^3+a.4^2+b.4+c=c\)

\(\Rightarrow\) 64+16a+4b = 0

\(\Rightarrow\) 16a+4b = -64

\(\Rightarrow\) 4a+b = -16

\(\Rightarrow\) b = -16 - 4a   ________(3)

By putting b = -16 - 4a from (3) into (1) and (2), we get

a - 16 - 4a + c = -1

\(\Rightarrow\) 3a - c = -15   ________(4)

and 4a - 32 - 8a + c = -8

\(\Rightarrow\) 4a - c = -24   ________(5)

Subtract (4) from (5), we get

(4a - c) - (3a - c) = -24 - (-15)

\(\Rightarrow\) a = -24+15 = -9

Put a = -9 in equation (4), we get

3 \(\times\) -9 - c = -15

\(\Rightarrow\) c = -27+15 = -12

Put a = -9 in equation (3), we get

b = -16 - 4 \(\times\) -9 = -16+36 = 20

\(\therefore\) a = -9, b = 20, c = -12

Correct option is C) a = -9, b = 20, c = -12

(c) –9, 20, –12 

Given, f(x) = x³ + lx² + mx + n. 

f(1) = f(2) = 0 ⇒ (x – 1) and (x – 2) are factors of f(x). 

Since, f(x) is polynomial of degree 3, it shall have three linear factors. So, let the third factor be (x – k). 

Then, f(x) = (x – 1) (x – 2) (x – k) 

⇒ f(x) = x³ + lx² + mx + n = (x – 1) (x – 2) (x – k) 

Given, f(4) = f(0) 

⇒ (4 – 1) (4 – 2) (4 – k) = (–1) (–2) (–k) 

⇒ 24 – 6k = – 2k ⇒ 4k = 24 ⇒ k = 6

∴ f(x) = (x – 1) (x – 2) (x – 6) = (x² – 3x + 2) (x – 6) 

= x³ – 9x² + 20x – 12

∴ x³ + lx² + mx + n = x³ – 9x² + 20x – 12 

⇒ l = – 9, m = 20, n = – 12.

(b) 3

Since a + b + c = 0 ⇒ a³ + b³ + c³ = 3abc. 

So, as (a – b) + (b – c) + (c – a) = 0 

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

∴ Given expression = \(\frac{3(a-b)(b-c)(c-a)}{(a-b)(b-c)(c-a)}\) = 3.

(a) (2²⁰⁰ – 1) x + (–2²⁰⁰ + 2) 

Let x²⁰⁰ = (x² – 3x + 2). Q(x) + lx + m                 ...(i) 

where, Q(x) = quotient and (lx + m) is the remainder 

Now (x² – 3x + 2) = 0 ⇒ (x – 1) (x – 2) = 0 ⇒ x = 1, 2. 

Substituting x = 1 in (i), we have, 

1²⁰⁰ = 0. Q.(x) + l + m                       ...(ii) 

Similarly, for x = 2, 

2²⁰⁰ = 0. Q(x) + 2l + m                     ...(iii)

∴ l + m = 1, 2l + m = 2²⁰⁰ 

Solving we get, l = 2²⁰⁰ – 1 and m = 2 – 2²⁰⁰ 

Hence remainder = lx + m = (2²⁰⁰ – 1) x + (–2²⁰⁰ + 2).

(d) None of these

Hint. 

ak² – ak + b = 0

k² – ck + d = 0 

Solving by the rule of cross multiplication,

\(\frac{k^2}{-ad+bc}\) = \(\frac{k}{b-ad}\) = \(\frac{1}{-ac+a}\)

⇒ k = \(\frac{b-ad}{a(1-c)},\frac{bc-ad}{b-ad}.\)

(b) -1

Since x + k is the HCF of the given expressions, therefore, x = – k will make each expression zero. 

k² – ak + b = 0                         ...(i) 

k² – bk + a = 0                       ...(ii) 

Solving (i) and (ii) by the rule of cross multiplication,

\(\frac{k^2}{-a^2+b^2}=\frac{k}{b-a}=\frac{1}{-b+a}\)

From last two relations, k = \(\frac{b-a}{-(b-a)}=-1\)

∴ \(\frac{k^2}{-a^2+b^2}=\frac{1}{-b+a}⇒\frac{(-1)^2}{-(a^2+b^)}=\frac{1}{-b+a}\)

⇒ \(\frac{1}{(b-a)(b+a)}=\frac{1}{-b+a}\) ⇒ a+b = -1.

(i) f(x) = x² - 2x - 8

factorize the given polynomial by splitting the middle term:

⇒ x² - 4x + 2x – 8

⇒ x (x - 4) + 2 (x - 4)

For zeros of f(x), f(x) = 0

⇒(x + 2) (x - 4) = 0x+2 = 0x = - 2x - 4 = 0x = 4

⇒x = - 2, 4

Therefore zeros of the polynomial are - 2 & 4

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = - 2 + 4 = 2

And - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) is - (- 2) = 2

Hence the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and sum of zeroes are same.

Product of zeros = - 2 × 4 = - 8

\(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) is - 8.

Hence the value of \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and product of zeroes are same.

(ii) g(s) = 4s² - 4s + 1 

factorize the given polynomial by splitting the middle term:

⇒ 4s² - 2s - 2s + 1

⇒ 2s (2s - 1) - 1 (2s - 1)

For zeros of g(s), g(s) = 0 (2s - 1) (2s - 1) = 0

2s - 1 = 0 s = \(\frac{1}{2}\)

Therefore zeros of the polynomial are \(\frac{1}{2}\), \(\frac{1}{2}\)

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1  \(\frac{-Coefficient\,of\,s}{Coeficient\,of\,s^2}\) = \(\frac{4}{4}\) = 1

Hence the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) and sum of zeroes are same.

Product of zeros = - \(\frac{1}{2}\) × - \(\frac{1}{2}\) = \(\frac{1}{4}\) \(\frac{Constant\,term}{coefficient\,of\,s^2}\) = \(\frac{1}{4}\)

Hence the value of \(\frac{Constant\,term}{coefficient\,of\,x^2}\) and product of zeroes are same. 

(iii) h(t) = t² - 15

use the formula a² - b² = (a+ b)(a - b) to solve the above equation,

Here a is t and b is √15.

Solve the given expression as:

t² - (√15)² = (t + √15)(t - √15)

For zeros of h(t), h(t) = 0

t + √15 = 0

t = - √15

t - √15 = 0

t = √15

Therefore zeros of the given polynomial are t = √15 & -√15

In a polynomial the relations hold are as follows:

sum of zeroes is equal to - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)

For the given polynomial,

Sum of zeros = √15 + (- √15) = 0

The value of - \(\frac{Coefficient\,of\,t}{Coeficient\,of\,t^2}\) is 0.

Hence, the value of - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)and sum of zeroes are same.

Product of zeros = - √15 x √15

The value of \(\frac{Constant\,term}{coefficient\,of\,t^2}\) is - √15

Hence the value of \(\frac{Constant\,term}{coefficient\,of\,x^2}\) and product of zeroes are same.

(iv) f(x) =  6x² - 3 - 7x

Write the equation in the form of ax² +bx+c as:

6x² - 7x -3

factorize the given polynomial by splitting the middle term:

⇒ 6x² - 9x + 2x - 3

⇒ 3x(2x - 3) +1(2x - 3)

⇒ (3x + 1) (2x - 3)

For zeros of f(x),f(x) = 0

⇒ (3x + 1) (2x - 3) = 0

x = \(\frac{-1}{3}\), \(\frac{3}{2}\)

Therefore zeros of the polynomial are   \(\frac{-1}{3}\), \(\frac{3}{2}\)

In a polynomial the relations hold are as follows:

sum of zeroes is equal to -  \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\) product of zeroes is equal to \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

Sum of zeros = \(\frac{-1}{3}\) + \(\frac{3}{2}\) = \(\frac{-2+9}{2}\) = \(\frac{7}{6}\) = \(\frac{7}{6}\) = - \(\frac{Coefficient\,of\,x}{Coeficient\,of\,x^2}\)

Product of zeros = \(\frac{-1}{3}\)× \(\frac{3}{2}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\) = \(\frac{Constant\,term}{coefficient\,of\,x^2}\)

(c) y² + y – 6 = 0

x⁴ + x³ – 4x² + x + 1 = 0

⇒ x² + x - 4 + \(\frac{1}{x}\) + \(\frac{1}{x^2}\) = 0

⇒ \(x^2+\frac{1}{x^2}\) + \(x+\frac{1}{x}\) - 4 = 0 

⇒ \(\big(x+\frac{1}{x}\big)^2\) - 2 + \(\big(x+\frac{1}{x}\big)\) - 4 = 0

⇒ y² + y – 6 = 0                      \(\big(∵x+\frac{1}{x}=y\big)\)

(a) 1

x² = y + z ⇒ x² + x = x + y + z 

⇒ x (x + 1) = x + y + z ⇒ \(\frac{x}{x+y+z}\) = \(\frac{1}{x+1}\)

Similarly, \(\frac{1}{y+1}\) = \(\frac{y}{x+y+z}\) and \(\frac{1}{z+1}\) = \(\frac{z}{x+y+z}\)

∴ \(\frac{1}{x+1}\) + \(\frac{1}{y+1}\) + \(\frac{1}{z+1}\) = \(\frac{x}{x+y+z}\) + \(\frac{y}{x+y+z}\) +  \(\frac{z}{x+y+z}\) 

= \(\frac{x+y+z}{x+y+z}\) = 1.

α and β are the zeros of the quadratic polynomial p(x) = 4x² - 5x - 1

Sum of the roots = α + β = \(\frac{constant\,term}{coefficient\,of\,x^2}\) = - \(\frac{(-5)}{4}\) = \(\frac{5}{4}\)

Product of the roots = α x β = \(\frac{constant\,term}{coefficient\,of\,x^2}\) = \(\frac{(-1)}{4}\)

Now,

α²β + αβ² =αβ(α + β)

On substituting values from above, we get

= \(\frac{-1}{4}\) × \(\frac{5}{4}\) = - \(\frac{5}{16}\)

(c) (a + b) (b + c) (c + a) 

Use the identity, if a + b + c = 0, then a³ + b³ + c³ = 3abc.

Let f(x) = 5x² + 8x - 4 and α and β be its zeroes

Here a = 5, b = 8 and c = -4

5x² + 8x - 4 = 0

5x² + 10x - 2x - 4 = 0

5x(x + 2) - 2(x + 2) = 0

(x + 2)(5x - 2) = 0

x + 2 = 0 or 5x - 2 = 0

x = -2 or x= 2/5

So, the zeroes are α = -2 and β = 2/5

Sum of zeroes = α + β = -3

Product of zeroes = αβ = 2

Then, the quadratic polynomial = x² – (sum of zeroes)x + product of zeroes

= x² - (-3)x + 2

= x² + 3x + 2

Given: α and β are zeroes of f(x) = x² + x – 2

To find: (1/α – 1/β)

α + β = Sum of zeros = -(coefficient of x)/(coefficient of x²) = -1

α β = Product of zeros = (constant term)/(coefficient of x²) = -2

Now,

(1/α – 1/β) = (β – α)²) / αβ

= (β + α)² – 4αβ) / (αβ)²

= 9/ 4

Given: Zeros of the polynomial x³ – 3x² + x + 1 are (a – b), a and (a + b).

Now by using the relationship between the zeros of the quadratic polynomial we have:

Sum of zeros = -(coefficient of x)/(coefficient of x²) = -1

a – b + a + a + b = -1

3a = 3

a = 1

Product of zeros = (constant term)/(coefficient of x²) = -1

(a – b) (a) (a + b) = -1

(1 – b) (1) (1 + b) = – 1

1 – b² = – 1

b = ±√2

Let f(x) = 3x² + x ˗ 4

= 3x² + 4x - 3x ˗ 4

= x (3x + 4) - 1(3x + 4)

= (3x + 4) (x ˗ 1)

To find the zeroes, set f(x) = 0

(3x + 4) (x ˗ 1) = 0

3x + 4 = 0 or x ˗ 1 = 0

x = -4/3 or x = 1

Answer: (D) √2x² - 3√3x + √6

An expression of the form p(x) = a₀ + a₁x + a₂x² + ….. + a_nx_n, where a_n ≠ 0, is called a polynomial in x of degree n. 
Here, a₀, a₁, a₂, ……, an are real numbers and each power of x is a non-negative integer.
Hence, √2x² - 3√3x + √6 is a polynomial. 

(d) \(\frac{q-m}{p-l}\)

Let f(x) = x² + p\(x\) + q 

g(x) = x² + l\(x\) + m. 

Since (x + k) is a common factor of 

f(x) and g(x), f(–k) = k² – pk + q = 0 

g(–k) = k² – lk + m = 0 

⇒ k² – px + q = k² – lk + m 

⇒ q – m = (p – l)k ⇒ k = \(\frac{q-m}{p-l}\)

(c) 3

Put x⁴ = – 1 in f(x) = x⁴⁰ + 2 

Remainder = (x⁴)10 + 2 = (–1)10 + 2 = 3.

(b) 1008.

Required remainder = f(1) 

= 1 + 1 + 1 + 1 + 1 ......... + 1 (1008 times) 

= 1008 × 1 = 1008.

(d) 1 – a₀ – a₂ – ...... = a₁ + a₃ + ......

Let f(x) = a₀ + a₁x + a₂x² + ......... + a_nxⁿ 

Given, f(1) = 1 

⇒ a₀ + a₁ + a₂ + ........ + a_n = 1 

⇒ 1 – a₀ – a₂ – ........ = a₁ + a₃ + ........ .

Given p(x) = x² + 5x + 6 is a quadratic polynomial. 

It has atmost two zeroes. 

To find zeroes, let p(x) = 0 

⇒ x² + 5x + 6 = 0 

⇒ x² + 3x + 2x + 6 = 0 

⇒ x(x + 3) + 2 (x + 3) = 0 

⇒ (x + 3) (x + 2) = 0 

⇒ x + 3 = 0 or x + 2 = 0 

⇒ x = -3 or x = -2 

Therefore the zeroes of the polynomial are -3 and -2.

Given p(x) = (x + 2) (x + 3) 

It is a quadratic polynomial. 

It has atmost two zeroes. 

Let p(x) = 0 

⇒ (x + 2) (x + 3) = 0 

⇒ (x + 2) = 0 or (x + 3) = 0 

⇒ x = -2 or x = -3

Therefore the zeroes of the polynomial are -2 and – 3.

Given p(x) = x⁴ – 16 is a biquadratic polynomial. It has almost two zeroes. 

Let p(x) = 0 

⇒ x⁴ – 16 = 0 

⇒ (x²)² – 4² = 0 

⇒ (x² – 4) (x² + 4) = 0 

⇒ (x + 2) (x – 2) (x² + 4) = 0 

⇒ (x + 2) = 0 or (x – 2) = 0 or (x² + 4) = 0 

⇒ x = -2 (or) x = 2 (or) x² = -4 

Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Given that a + b + c = 9 

Squaring on both sides, (a + b + c)² = 9² 

⇒ a² + b² + c² + 2 (ab + be + ca) = 81 

⇒ a² + b² + c² = 81 – 2 (ab + be + ca) (by problem) 

= 81 – 2 x 26 

= 81 – 52 = 29

Given polynomial f(x) = x² – 8x + k

Let α and β are zeros of polynomial f(x) then

Sum of zero (α + β) = - b/a

= -(-8)/1 = 8  ....(i)

and product of zeros

(α x β) = c/a = k/1 = k  ....(ii)

Now, from equation (i)

(α + β) = 8

Squaring both sides,

(α + β)² = 8²

⇒ a² + β² + 2αβ = 64 …(iii)

It is given that sum of square of zeros is 40.

i.e., α² + β² = 40

Putting values from equation (i) and (ii) in (iii)

40 + 2k = 64

⇒ 2k = 64 – 40

⇒ 2k = 24

⇒ k = 12

Thus k = 12

Answer is (A) 100 – 25p²

First expression = (2 + p)

and other expression = 100 – 25p²

= (10 – 5p) (10 + 5p)

= 5(2 – p) × 5(2 + p)

= 25(2 – p) (2 + p)

Thus L.C.M. of two expression = 100 – p²

Answer is (A) (x² – 1)(x – 2)

x² – 1 = (x – 1)(x + 1) …(i)

x² – x – 2 = x² – 2x + x – 2 

= x (x – 2) + 1 (x – 2) 

= (x + 1)(x – 2) …(ii)

from eqn (i) and (ii)

(x + 1)(x – 1 )(x – 2)

L.C.M. = (x² – 1)(x – 2)

Answer is (B) x + 1

x² – 5x + 6 = 0 

⇒ x² – 3x – 2x + 6 = 0 

⇒ x (x – 3) – 2 (x – 3) = 0 

⇒ (x – 2) (x – 3) = 0 

⇒ x – 2 = 0 or x – 3 = 0 

⇒ x = 2 or x = 3 

∴ x = 2 or 3 are the zeroes of x² – 5x + 6.

x + 5 = 0

x = – 5 

∴ x = – 5

2x – 3 = 0 

2x = 3 

x = 3/2

∴ x = 3/2 is the zero of 2x – 3