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Given;

f(x) = ax² + bx + c has no real zeroes, and a + b + c 0.

Suppose

a = – 1, b = 1, c = – 1

Then a + b + c = – 1,

b ² – 4ac = – 3

Therefore it is possible that c is less tha zero.

Suppose c = 0

Then b² – 4ac = b² ≥ 0

So,

f(x) has at least one zero.

Therefore c cannot equal zero.

Suppose c > 0.

It must also be true that b² ≥0

Then,

b ² – 4ac < 0 only if a > 0.

Therefore,

a + b + c < 0.

– b > a + c > 0

b ² > (a + c)²

b ² > a² + 2ac + c²

b ² – 4ac > (a – c)² ≥ 0

As we know that the discriminant can’t be both greater than zero and less than zero,

So,

C can’t be greater than zero.

Given

f(x) = x² – p (x + 1) – c

α and β are the zeros

Then,

f(x) = x² – p (x + 1) – c

= x² – px – (p + c)

As

(α + 1)(β + 1) = αβ + α + β + 1

= – p – c + p + 1

= 1 – c

So,

The value of c,

c = 1

Given,

f(x) = x² – p (x + 1) – c

α and β are the zeros

So,

f(x) = x² – p (x + 1) – c

= x² – px – (p + c)

As,

(α + 1)(β + 1) = αβ + α + β + 1

= – p – c + p + 1

= 1 – c

4p² + 2p ÷ 2 = \(\frac {4p^2}{2p} + \frac {2p}{2p} + \frac {2}{2p}\)

= 2p + 1 + 1/P

Division fact :

(2p + 1 + 1/P). 2p = 4p² + 2p + 2

Correct option is (A) -q/p

px + q = 0

\(\Rightarrow\) x = \(\frac{-q}p\)

Thus, \(\frac{-q}p\) is a zero of linear equation px+q.

Correct option is A) -q/p

Let α, β and γ are the zeroes of the required polynomial. 

Then we have: 

α + β + γ = 3 + 5 + (-2) = 6 

αβ + βγ + γα = 3 × 5 + 5 × (-2) + (-2) × 3 = -1 

and αβγ = 3 × 5 × -2 = -30 

Now, p(x) = x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ 

= x³ – x² × 6 + x × (-1) – (-30) 

= x³ – 6x² – x + 30 

So, 

the required polynomial is p(x) = x³ – 6x² – x + 30.

It is given that the two roots of the polynomial are 2 and -5. 

Let α = 2 and β = -5 

Now, the sum of the zeroes, α + β = 2 + (-5) = -3 

Product of the zeroes, αβ = 2 × (-5) = -15 

∴ Required polynomial = x² – (α + β)x + αβ

= x² – (-3)x + 10 

= x² + 3x – 10

Let p(x) = x³ + 4x² – 3x – 18 

Now, p(2) = 2³ + 4 × 2² – 3 × 2 – 18 = 0

∴ 2 is a zero of p(x).

The given polynomial = x³ – 3x² + x + 1 and its roots are (a – b), a and (a + b). 

Comparing the given polynomial with Ax³ + Bx² + Cx + D, 

we have: 

A = 1, B = -3, C = 1 and D = 1 

Now, (a – b) + a + (a + b) = \(\frac{-B}A\)

⇒ 3 a = – \(\frac{-3}1\)

⇒ a = 1 

Also, (a – b) × a × (a + b) = \(\frac{-D}A\)

⇒ a (a² – b² ) = \(\frac{-1}1\)

⇒ 1 (1² – b² ) = -1 

⇒ 1– b² = -1 

⇒ b² = 2 

⇒ b = ±√2 

∴ a = 1 and b = ±√2

By using the relationship between the zeroes of he quadratic polynomial. 

We have, 

Sum of zeroes = \(\frac{-(coefficient\,of\,x^2)}{coefficient\,of\,x^3}\)

∴ a – b + a + a + b = \(\frac{-(-3)}1\)

⇒ 3a = 3

⇒ a = 1 

Now, 

Product of zeroes = \(\frac{-(constant\,term)}{coefficient\,of\,x^3}\)

∴ (a – b) (a) (a + b) = \(\frac{-1}1\) 

⇒ (1 – b) (1) (1 + b) = –1 [∵a =1] 

⇒ 1 – b² = –1 

⇒ b² = 2 

⇒ b = ±√2

(d) none of these 

A polynomial in x of degree n is an expression of the form p(x) = a₀ + a₁x + a₂x² + ……+ a_nxⁿ , where a_n ≠ 0.

Correct option is (B) 2

Sum of zeroes \(=\frac{\text{- coefficient of }x^2}{\text{coefficient of }x^3}\)

\(=\frac{-(-2)}1\) = 2

Correct option is B) 2

Correct option is (C) 1/3

3x – 1 = 0

\(\Rightarrow\) \(x=\frac13\)

Thus, zero of polynomial 3x-1 is \(\frac13.\)

Correct option is C) 1/3

Correct option is (A) 4/3

\(\frac{3x^3-2x^2+x+2}{3x+1}\) \(=x^2-x+\frac23+\frac{\frac43}{3x+1}\)

Thus, when we divides \((3x^3-2x^2+x+2)\) by (3x+1) then quotient \(=x^2-x+\frac23\) and remainder \(=\frac{4}{3}.\)

Correct option is A) 4/3

Correct option is C) -2, 2

True

Let a polynomial

f(x) = x² + 9

f(x) = (x + 3i)(x – 3i)

So,

The zeroes are always imaginary and not real.

General form of a cubic polynomial whose zeroes are a, b and c is:

x³ – (a + b + c) x² + (ab + bc + ca)x – abc …(1)

To Find: Cubic polynomial whose zeroes are 1/2, 1 and -3

Let us say, a = 1/2, b = 1 and c = -3

Putting the values of a, b and c in the equation (1) we get:

= x³ – (1/2 + 1 – 3) x² + (1/2 – 3 – 3/2)x – (- 3/2)

= x³ – (-3/2) x² – 4x + 3/2

= 2x³ + 3x² – 8x + 3

Which is required polynomial.

As we know, General form of a cubic polynomial whose zeroes are a, b and c is:

x³ – (a + b + c) x² + (ab + bc + ca)x – abc

Also written as :

x³ – (Sum of the zeros) x² + (Sum of the product of the zeros taking two at a time) x – (Product of Zeros) …(1)

Given:

Sum of the zeros = 5

Sum of the product of its zeros taken two at a time -2

Product of its zeros = –24

Putting these values in the equation (1), we get:

x³ – 5x² – 2x + 24

Which is required polynomial.

Let f(x) = x² – 3x – m (m + 3)

Above polynomial can be written as,

f(x) = x² – (m + 3)x + mx – m(m + 3)

= x(x – m – 3) + m(x – m – 3)

= (x – m – 3)(x + m)

To find the zeroes of f(x), put f(x) = 0

(x – m – 3)(x + m) = 0

Either x – m – 3 = 0 or x + m = 0

x = m + 3 or x = -m

Required Zeros are (m + 3), -m

Let f(x) = x² + x ‒ p (p + 1)

Above polynomial can be written as,

= x² + (p + 1) x – px – p (p + 1)

= x (x + (p + 1)) – p (x + (p + 1))

= (x – p)(x + (p + 1))

To find the zeroes of f(x), put f(x) = 0

(x – p)(x + (p + 1)) = 0

either (x – p) = 0 or (x + (p + 1)) = 0

x = p or x = – (p + 1)

Hence, the zeros of the given polynomial are p and – (p + 1)

Given: (2 + √3) is one of zero of polynomial x² ‒ 4x + 1.

To find: Other zero

Since given polynomial is a quadratic so it has only two zeros.

Let other zero be x.

Now,

Sum of zeros = -(coefficient of x)/(coefficient of x^2)

x + (2 + √3) = -(-4/1) = 4

x + 2 + √3 = 4

or x = 2 – √3

Hence, the other zero is (2 -√3).

Given: x = –4 is one zero of the polynomial x² – x – (2k + 2) 

Therefore, it will satisfy the above polynomial. 

Now, we have 

(–4)² – (–4) – (2k + 2) = 0 

⇒ 16 + 4 – 2k – 2 = 0 

⇒ 2k = – 18 

⇒ k = 9

f(x) = x² – 3x – m (m + 3) 

By adding and subtracting mx, we get 

f(x) = x² – mx – 3x + mx – m (m + 3) 

= x[x – (m + 3)] + m[x – (m + 3)] 

= [x – (m + 3)] (x + m) 

f(x) = 0 ⇒ [x – (m + 3)] (x + m) = 0 

⇒ [x – (m + 3)] = 0 or (x + m) = 0 

⇒ x = m + 3 or x = –m 

So, the zeroes of f(x) are –m and +3.

Given: x = 3 is one zero of the polynomial 2x² + x + k 

Therefore, it will satisfy the above polynomial. 

Now, we have 

2(3)² + 3 + k = 0 

⇒ 21 + k = 0 

⇒ k = – 21

i. x³ – 2 = x³ + 0x² + 0x – 2 

∴ Coefficient form of the given polynomial = (1, 0, 0, -2) 

ii. 5y = 5y + 0 

∴Coefficient form of the given polynomial = (5,0)

p(y) = y² – 3√2 y + 1 

Put p = 3√2 in the given polynomial. 

∴ p(3√2) = (3√2)² – 3√2(3√2 ) + 1 

= 9 x 2 – 9 x 2 + 1 

= 18 – 18 + 1 

∴ p(3√2) = 1

p(x) = x² – 5x + 5 

Put x = 0 in the given polynomial. 

∴ P(0) = (0)² – 5(0) + 5

= 0 – 0 + 5 

∴ p(0) = 5

p(y) = y² – 3√2 y + 1 

Put p= 3√2 in the given polynomial. 

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

= 9 x 2 – 9 x 2 + 1 

= 18 – 18 + 1

 ∴ p( 3√2 ) = 1

p(m) = m³ + 2m² – m + 10 

Put m = a in the given polynomial. 

∴ p(a) = a³ + 2a² – a + 10 …(i) 

Put m = -a in the given polynomial.

 p(-a) = (-a)³ + 2(-a)² – (-a) +10 

∴ p (-a) = -a³ + 2a² + a + 10 …(ii) 

Adding (i) and (ii), 

p(a) + p(-a) = (a³ + 2a² – a + 10) + (-a³ + 2a² + a + 10) 

= a³ – a³ + 2a² + 2a² – a + a + 10 + 10 

∴ p(a) + p(-a) = 4a² + 20

p(y) = 2y³ – 6y² – 5y + 7 

Put y = 2 in the given polynomial. 

∴ p(2) = 2(2)³ – 6(2)² – 5(2) + 7

= 2 x 8 – 6 x 4 – 10 + 7 

 16 – 24 – 10 + 7 

∴ P(2) = -11

p(x) = 2x – 2x³ + 7

 Put x = 3 in the given polynomial. 

∴ p(3) = 2(3) – 2(3)³ + 7 

= 6 – 2 x 27 + 7 

= 6 – 54 + 7 

∴ P(3) = – 41

 p(x) = 2x – 2x³ + 7

Put x = -1 in the given polynomial

∴  p(- 1) = 2(- 1) – 2(-1)³ + 7 

= – 2 – 2(-1) + 7 

= -2 + 2 + 7 

∴ p(-1) = 7

The value of p(x) at x = 1 is 4(1)² – 3(1) + 7 = 8

p(x) = 2x – 2x³+ 7

 Put x = 0 in the given polynomial. 

∴ p(0) = 2(0) – 2(0)³ + 7 

= 0 – 0 + 7 

∴ P(0) = 7

a₁₄ p^₁₄ + a₁₃ p¹³ + a₁₂ p¹² + …………….+ a₁ p + a₀

polynomial with 2 terms in variable x is 2x + 3x²

p(x) = x³ 

∴ p(1) = 1³ = 1

 p(x) = x³

∴ p(0) = 0³ = 0 

p(x) = x³ 

∴ p(-2) = (-2)³ = -8

i. Number of coefficients = 5 

∴ Degree = 5 – 1 = 4 

∴Index form = 3x⁴ – 2x³ + 0x² + 7x + 18 

ii. Number of coefficients = 4 

∴Degree = 4 – 1 = 3 

∴ Index form = 6x³ + x² + 0x + 7

iii. Number of coefficients = 4 

∴ Degree = 4 – 1 = 3 

∴ Index form = 4x³ + 5x² – 3x + 0

i. 2m⁴ – 3m² + 7 

= 2m⁴ + 0m³ – 3m² + 0m + 7 

∴ Coefficient form of the given polynomial = (2, 0, -3, 0, 7)

ii.  –\(\frac2{3}\) 

∴Coefficient form of the given polynomial = (-\(\frac2{3}\))

p(y) = y² – 2y + 5 

∴ p(1) = 1² – 2(1) + 5 

= 1 – 2 + 5

∴ P(1) = 4

 p(y) = y² – 2y + 5 

∴ p(0) = 0² – 2(0) + 5

 = 0 – 0 + 5 

∴ p(0) = 5

p(y) = y² – 2y + 5 

∴ p(- 2) = (- 2)² – 2(- 2) + 5 

= 4 + 4 + 5 

∴ p(-2) = 13

i. x⁴ + 16 

Index form = x⁴ + 0x³ + 0x² + 0x + 16 

∴ Coefficient form of the polynomial = (1,0,0,0,16) 

ii. m⁵ + 2m² + 3m + 15

Index form = m⁵ + 0m⁴+ 0m³ + 2m²+ 3m + 15 

∴ Coefficient form of the polynomial = (1, 0, 0, 2, 3, 15)