Polynomials

Let f(x) = x² + 7x + 12

Put f(x) = 0

x² + 7x + 12 = 0

x² + 4x + 3x + 12 = 0

3(x + 4) + x(x + 4) = 0

(3 + x) (x + 4) = 0

∴ x = -4 or x = -3

Now,

sum of zeroes = -3 + (-4) = -7=

Product of zeroes = (-3) × (-4)

Hence, relationship verified.

Let f(x) = x² + 2x - 8

Put f(x) = 0

x² + 2x - 8 = 0

x² + 2x - 4x - 8 = 0

x(x + 2) – 4 (x + 2) = 0

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

∴ x = 4 or x = -2

Now, sum of zeroes = -2 + 4 = 2 =

Product of zeroes = (4) × (-2)

Hence, relationship verified.

Let f(x) = x² + 3x - 10

Put f(x) = 0

x² + 3x - 10 = 0

x² + 5x - 2x - 10 = 0

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

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

∴ x = -5 or x = 2

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

Product of zeroes = (-5) × (2)

Hence, relationship verified.

Let f(x) = 4x² - 4x - 3

Put f(x) = 0

4x² - 4x - 3 = 0

4x² - 6x + 2x - 3 = 0

2x(2x - 3) + 1(2x - 3) = 0

(2x + 1) (2x - 3) = 0

Now, sum of zeroes = 1 =

Product of zeroes

Hence, relationship verified.

Let f(x) = 5x² - 4 – 8x

Put f(x) = 0

5x² - 8x - 4 = 0

5x² - 10x + 2x - 4 = 0

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

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

Now, sum of zeroes == =

Product of zeroes

Hence, relationship verified.

Let f(x)= 2√3 x² - 5x + √3

Put f(x) = 0

Now,

sum of zeroes

Product of zeroes

Hence, relationship verified.

Let f(x) = 2x² – 11x + 15

Put f(x) = 0

2x² - 11x + 15 = 0

2x² - 6x - 5x + 15 = 0

2x(x - 3) - 5(x - 3) = 0

(2x - 5) (x - 3) = 0

Now, sum of zeroes= =

Product of zeroes

Hence, relationship verified.

Let f(x) =4x² – 4x + 1 8x

Put f(x) = 0

4x² - 4x + 1 = 0

(2x)² – 2(2x)(1) + (1)² = 0

(2x - 1)² = 0

Now,

sum of zeroes

Product of zeroes

Hence, relationship verified.

Let f(x) = x² – 5

Put f(x) = 0

x² – 5 = 0

(x – √5) (x + √5) = 0

∴ x = √5 or x = -√5

Now, sum of zeroes

Product of zeroes

Hence, relationship verified.

Let f(x) = 8x² – 4

Put f(x) = 0

8x² – 4 = 0

(2√2x – 2) (2√2x + 2) = 0

Now, sum of zeroes

Product of zeroes

Hence, relationship verified.

Let f(x) = 5y² + 10y

Put f(x) = 0

5y² + 10y = 0

(5y) (y + 2) = 0

∴ x =0 or x = -2

Now, sum of zeroes

Product of zeroes

Hence, relationship verified.

Let f(x) =3x² – x – 4

Put f(x) = 0

3x² - x – 4 = 0

3x² - 4x + 3x - 4 = 0

x(3x - 4) + 1(3x - 4) = 0

(x + 1) (3x - 4) = 0

Now, sum of zeroes

Product of zeroes

Hence, relationship verified.

Let α = 2 and β = -6

Now, Sum of zeros, α + β = 2 – 6 = -4

And, product of zeroes, αβ = 2(-6) = -12

We know that,

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

= x²–(- 4)x + (-12)

= x² + 4x – 12

Now, sum of zeroes =

Product of zeroes =

Hence, relationship verified.

Now, Sum of zeros,

And, product of zeroes,

We know that,

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

=12x² – 5x – 2

Now, sum of zeroes

Product of zeroes

Hence, relationship verified.

Let the zero of the polynomial be α and β

According to the question,

α + β = 8

αβ = 12

∴f(x)= x² – (α + β)x + (αβ)

=x² - 8x + 12

∴required polynomial f(x) = x² - 8x + 12

Put f(x) = 0

x² - 8x + 12 = 0

x² - 6x -2x + 12 = 0

x(x – 6) -2(x – 6) = 0

(x - 6)(x - 2)=0

x = 6 or x = 2

Let the zero of the polynomial be α and β

According to the question,

α + β = 0

αβ = -1

∴f(x)= x² – (α + β)x + (αβ)

=x² - 0x - 1

∴required polynomial f(x) = x² - 1

Put f(x) = 0

x² - 1 = 0

(x - 1)(x + 1)=0

x = 1 or x = -1

Let the zero of the polynomial be α and β

According to the question:

αβ = 1

∴f(x)= x² – (α + β)x + (αβ)

∴ required polynomial f(x)

Put f(x) = 0

x² –(5/2)x + 1= 0

2x² – 5x + 2 = 0

2x² – 4x – x + 2 = 0

2x(x – 2) – 1(x - 2) =0

(x - 2)(2x - 1) = 0

x = 2 or x = 1/2

Let the zero of the polynomial be α and β

According to the question,

α + β = √2

∴ f(x)= x² – (α + β)x + (αβ)

∴ required polynomial f(x)= x² - √2 x + 1/3

Put f(x) = 0

x² - √2 x + 1/3 = 0

3x² – 3√2x + 1 = 0

We have, ax² + 7x + b = 0

Now, Sum of zeros

∴a = 3 (i)

Product of zeroes (From i)

∴b= -6

Since, x + a is a factor of 2x² + 2ax + 5x + 10

∴ x + a = 0

x = -a

Put x = -a in 2x² + 2ax + 5x + 10 = 0

2(-a)² + 2a(-a) + 5(-a) + 10 = 0

2a² – 2a² – 5a + 10 = 0

-5a = - 10

a = 2

It is given in the question that,

x = 2/3 is one of the zeros of the given polynomial 3x³ + 16x² + 15x – 18

We have, x = 2/3

x –2/3 = 0

To find the quotient we have to divide the given polynomial by x – 2/3

Quotient = 3x² + 18x + 27

∴ 3x² + 18x + 27 = 0

3x² + 9x + 9x + 27 = 0

3x (x + 3) + 9 (x + 3) = 0

(x + 3) (3x + 9) = 0

(x + 3) = 0 or (3x + 9) = 0

Hence, x = - 3 or x = - 3

It is given in the question that,

p (x) = x³ – 2x² – 5x + 6

Also, 3, -2 and 1 are the zeros of the given polynomial

∴ p (3) = (3)³ – 2 (3)² – 5 (3) + 6

= 27 – 18 – 15 + 6

= 33 – 33

= 0

p (-2) = (-2)³ – 2 (-2)² – 5 (-2) + 6

= -8 – 8 + 10 + 6

= - 16 + 16

= 0

And, p (1) = (1)³ – 2 (1)² – 5 (1) + 6

= 1 – 2 – 5 + 6

= 7 – 7

= 0

Verification of the relation is as follows:

Let us assume = 3, = - 2 and = 1

= 3 – 2 + 1

= 2

∴

Also, + + = 3 (-2) + (-2) (1) + 1 (3)

= - 6 – 2 + 3

= - 5

∴

And, = 3 × (-2) × 1

= - 6

∴

It is given in the question that,

p (x) = 3x³ – 10x² – 27x + 10

Also, 5, -2 and are the zeros of the given polynomial

∴ p (5) = 3 (5)³ – 10 (5)² – 27 (5) + 10

= 3 × 125 – 250 – 135 + 10

= 385 – 385

= 0

p (-2) = 3 (-2)³ – 10 (-2)² – 27 (-2) + 10

= - 24 – 40 + 54 + 10

= - 64 + 64

= 0

And, p () = 3 ()³ – 10 ()² – 21 () + 10

=

=

= 0

Verification of the relation is as follows:

Let us assume α = 5, β = - 2 and γ = 1/3

α + β + γ = 5 – 2 + 1/3 =

∴

Also, + + = 5 (-2) + (-2) () + () (5)

= - 27/3

= - 9

∴

And, αβγ =

Hence, verified

Let the zeros of the polynomial be a, b and c

Where a = 2, b = - 3 and c = 4

The cubic polynomial can be calculated as follows:

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

Putting the values of a, b and c in the above equation we get:

= x³ – (2 – 3 + 4) x² + (-6 – 12 + 8) x – (- 24)

= x³ – 3x² – 10x + 24

Let the zeros of the polynomial be a, b and c

Where a = 1/2, b = 1 and c = - 3

The cubic polynomial can be calculated as follows:

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

Putting the values of a, b and c in the above equation 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

The required cubic polynomial can be calculated as:

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

It is given that, sum of the product of its zeros taken two at a time, and the product of its zeros as 5, –2 and –24 respectively

Putting these values in the equation, we get:

x³ – 5x² – 2x + 24

It is given in the question that,

f (x) = x³ – 3x² + 5x – 3

And, g (x) = x² – 2

Hence,

Quotient q (x) = x – 3

Remainder r (x) = 7x – 9

It is given in the question that,

f (x) = x⁴ – 3x² + 4x + 5

And, g (x) = x² + 1 – x

Hence,

Quotient q (x) = x² + x - 3

Remainder r (x) = 8

It is given in the question that,

f (x) = x⁴ – 5x + 6 = x⁴ + 0x³ + 0x² – 5x + 6

And, g (x) = x² + 1 – x

Hence,

Quotient q (x) = x² – 2

Remainder r (x) = 5x + 10

It is given in the question that,

f (x) = 2x⁴ + 3x³ – 2x² – 9x – 12

And, g (x) = x³ – 3

Hence,

Quotient q (x) = 2x² + 3x + 4

Remainder r (x) = 0

As the remainder is 0

∴ x² – 3 is a factor of 2x⁴ + 3x³ – 2x² – 9x – 12

We know that,

According to the division rule, we have

Dividend = Quotient × Divisor + Remainder

Here, we have:

Dividend = 3x³ + x² + 2x + 5

Quotient = 3x – 5

Remainder = 9x + 10

Putting these values in the formula, we get

3x³ + x² + 2x + 5 = 3x – 5 × g (x) + 9x + 10

3x³ + x² + 2x + 5 – 9x – 10 = (3x – 5) × g(x)

3x³ + x² – 7x – 5 = (3x – 5) × g (x)

g (x) =

∴ g (x) = x² + 2x + 1

It is given in the question that,

f (x) = - 6x³ + x² + 20x + 8

g (x) = - 3x² + 5x + 2

∴ Quotient = 2x + 3

Remainder = x + 2

We know that, according to division rule:

Dividend = Quotient × Divisor + Remainder

Putting the values in the above formula, we get:

- 6x³ + x² + 20x + 8 = (- 3x² + 5x + 2) (2x + 3) + (x + 2)

-6x³ + x² + 20x + 8 = - 6x³ + 10x² + 4x – 9x² + 15x + 6 + x + 2

-6x³ + x² + 20x + 8 = -6x³ + x² + 20x + 8

Let us assume f (x) = x³ + 2x² ‒ 11x ‒ 12

It is given in the question that, -1 is a zero of the polynomial

∴ (x + 1) is a factor of f (x)

Now on dividing f (x) by (x + 1), we get

f (x) = x³ + 2x² ‒ 11x ‒ 12

= (x + 1) (x² + x – 12)

= (x + 1) {x² + 4x – 3x – 12}

= (x + 1) {x (x + 4) – 3 (x + 4)}

= (x + 1) (x – 3) (x + 4)

∴ f (x) = 0

(x + 1) (x – 3) (x + 4) = 0

(x + 1) = 0 0r (x – 3) = 0 or (x + 4) = 0

x = -1 or x = 3 or x = - 4

Hence, zeros of the polynomial are -1, 3 and -4

Let us assume f (x) = x³ – 4x² – 7x + 10

As 1 and – 2 are the zeros of the given polynomial therefore each one of (x – 1) and (x + 2) is a factor of f (x)

Consequently, (x – 1) (x + 2) = (x² + x – 2) is a factor of f (x)

Now, on dividing f (x) by (x² + x – 2) we get:

f (x) = 0

(x² + x – 2) (x – 5) = 0

(x – 1) (x + 2) (x – 5) = 0

∴ x = 1 or x = - 2 or x = 5

Hence, the third zero is 5

Let us assume f (x) = x⁴ + x³ ‒ 11x² ‒ 9x + 18

As 3 and – 3 are the zeros of the given polynomial therefore each one of (x + 3) and (x - 3) is a factor of f (x)

Consequently, (x – 3) (x + 3) = (x² – 9) is a factor of f (x)

Now, on dividing f (x) by (x² – 9) we get:

f (x) = 0

(x² + x – 2) (x² – 9) = 0

(x – 1) (x + 2) (x – 3) (x + 3) = 0

∴ x = 1 or x = - 2 or x = 3 or x = - 3

Hence, all the zeros of the given polynomial are 1, -2, 3 and -3

Let us assume f (x) = x⁴ + x³ ‒ 34x² ‒ 4x + 120

As 2 and – 2 are the zeros of the given polynomial therefore each one of (x - 2) and (x + 2) is a factor of f (x)

Consequently, (x – 2) (x + 2) = (x² – 4) is a factor of f (x)

Now, on dividing f (x) by (x² – 4) we get:

f (x) = 0

(x² + x – 30) (x² – 4) = 0

(x² + 6x – 5x – 30) (x – 2) (x + 2)

[x (x + 6) – 5 (x + 6)] (x – 2) (x + 2)

(x – 5) (x + 6) (x – 2) (x + 2) = 0

∴ x = 5 or x = - 6 or x = 2 or x = - 2

Hence, all the zeros of the given polynomial are 2, -2, 5 and -6

Let us assume f (x) = x⁴ + x³ ‒ 23x² ‒ 3x + 60

As √3 and –√3 are the zeros of the given polynomial therefore each one of (x - √3) and (x - √3) is a factor of f (x)

Consequently, (x – √3) (x + √3) = (x² – 3) is a factor of f(x)

Now, on dividing f (x) by (x² – 3) we get:

f (x) = 0

(x² + x – 20) (x² – 3) = 0

(x² + 5x – 4x – 20) (x² – 3)

[x (x + 5) – 4 (x + 5)] (x² – 3)

(x – 4) (x + 5) (x – ) (x + ) = 0

∴ x = 4 or x = - 5 or x = or x = -

Hence, all the zeros of the given polynomial are , -, 4 and - 5

Let us assume f (x) = 2x⁴ - 3x³ ‒ 5x² + 9x - 3

As and – are the zeros of the given polynomial therefore each one of (x - ) and (x + ) is a factor of f (x)

Consequently, (x – ) (x + ) = (x² – 3) is a factor of f (x)

Now, on dividing f (x) by (x² – 3) we get:

f (x) = 0

2x² - 3x² – 5x² + 9x - 3 = 0

(x² – 3) (2x² – 3x + 1) = 0

(x² - 3) (2x² – 2x – x + 1) (2x – 1) (x - 1) = 0

(x – ) (x + ) (2x – 1) (x - 1) = 0

∴ x = or x = - or x = or x = 1

Hence, all the zeros of the given polynomial are , -, and 1

Let us assume f (x) = x⁴ + 4x³ ‒ 2x² ‒ 20x - 15

As (x-√5) and (x-√5) are the zeros of the given polynomial therefore each one of (x-√5) and (x + √5) is a factor of f (x)

Consequently, (x-√5) (x + √5) = (x² – 5) is a factor of f (x)

Now, on dividing f (x) by (x² – 5) we get:

f (x) = 0

x⁴ + 4x³ – 7x² – 20x – 15 = 0

(x² – 5) (x² + 4x + 3) = 0

(x -√5) (x + √5) (x + 1) (x + 3) = 0

∴ x = √5 or x = - √5 or x = - 1 or x = - 3

Hence, all the zeros of the given polynomial are √5, -√5, - 1 and - 3

Let us assume f (x) = 2x⁴ – 11x³ + 7x² + 13x - 7

As (3 + √2) and (3- √2) are the zeros of the given polynomial therefore each one of (x + 3 + √2) and (x + 3 - √2) is a factor of f (x)

Consequently, [(x – (3 + √2)][(x – (3 -√2)

= [(x – 3) - √2] [(x – 3) + √2]

= [(x – 3)² – 2] = x² – 6x + 7 is a factor of f (x)

Now, on dividing f (x) by (x² – 6x + 7) we get:

f (x) = 0

2x⁴ - 11x³ + 7x² + 13x – 7 = 0

(x² - 6x + 7) (2x² + x – 1) = 0

(x + 3 + √2) (x + 3 -√2) (2x – 1) (x + 1) = 0

∴ x = - 3 - √2 or x = - 3 + √2 or x = 1/2 or x = - 1

Hence, all the zeros of the given polynomial are (-3 -√2), (-3 + √2), 1/2 and – 1

It is given in the question that, zero of the polynomial x² – 4x + 1 is (2 + )

Let the other zero of the polynomial be a

We know that,

Sum of zeros =

∴ 2 + √3 + a =

a = 2 - √3

Hence, the other zero of the given polynomial is (2 -√3)

We have,

f (x) = x² + x – p (p + 1)

f (x) = x² + (p + 1 - p)x – p (p + 1)

= x² + (p + 1) x – px – p (p + 1)

= x [x + (p + 1)] – p [x + (p + 1)]

= [x + (p + 1)] (x – p)

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

[x + (p + 1)] (x – p) = 0

[x + (p + 1)] = 0 or (x – p) = 0

x = p or x = - (p + 1)

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

We have,

f (x) = x² – 3x – m (m + 3)

Now, by adding and subtracting px, 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

Hence, the zeros of the given polynomial are - m and m + 3

It is given in the question that,

α + β= 6

And, αβ = 4

We know that,

If the zeros of the polynomial are then the quadratic polynomial can be found as x² – () x + (i)

Now substituting the values in (i), we get

x² – 6x + 4

It is given in the question that,

One of the zero of the polynomial kx² + 3x + k is 2

∴ It will satisfy the above polynomial

Now, we have

k (2)² + 3 (2) + k = 0

4k + 6 + k = 0

5k + 6 = 0

5k = - 6

∴ k = -6/5

It is given in the question that,

One of the zero of the polynomial 2x² + x + k is 3

∴ It will satisfy the above polynomial

Now, we have

2 (3)² + 3 + k = 0

21 + k = 0

k = - 21

It is given in the question that,

One of the zero of the polynomial x² – x- (2k + 2) is - 4

∴ It will satisfy the above polynomial

Now, we have

(- 4)² – (- 4) - (2k + 2) = 0

16 + 4 - 2k - 2 = 0

- 2k = - 18

k = 18/2

∴ k = 9

It is given in the question that,

One of the zero of the polynomial ax² – 3 (a – 1) x - 1 is 1

∴ It will satisfy the above polynomial

Now, we have

a (1)² – 3(a – 1) 1 – 1 = 0

a – 3a + 3 - 1 = 0

- 2a = - 2

∴ a = 1

It is given in the question that,

One of the zero of the polynomial 3x² + 4x + 2k is - 2

∴ It will satisfy the above polynomial

Now, we have

3 (- 2)² + 4 (- 2) + 2k = 0

12 - 8 + 2k = 0

4 + 2k = 0

2k = - 4

∴ k = - 2

We have,

f (x) = x² – x – 6

= x² – 3x + 2x – 6

= x (x – 3) + 2 (x – 3)

= (x – 3) (x + 2)

f (x) = 0

(x – 3) (x – 2) = 0

(x – 3) = 0 or (x – 2) = 0

x = 3 or x = 2

∴ The zeros of the given polynomial are 3 and – 2

It is given in the question that, zero of the polynomial kx² – 3x + 5 is 1

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

Sum of zeros =

⇒

⇒ k = 3

It is given in the question that, zero of the polynomial x² – 4x + k is 3

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

Sum of zeros =

3 = k/1

k = 3

It is given in the question that,

(x + 4) is a factor of 2x² + 2ax + 5x + 10

Now, we have

x + a = 0

x = - a

As (x + a) is a factor of 2x² + 2ax + 5x + 10

Thus, it will satisfy the given polynomial

∴ 2 (- a)² + 2a (- a) + 5 (- a) + 10 = 0

- 5a + 10 = 0

a = 2

It is given in the question that, zero of the polynomial 2x³ – 6x² + 5x - 7 is (a – b), a and (a + b)

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

Sum of zeros =

a – b + a + a + b = (- (-6))/2

3a = 3

a = 1

Firstly, equating x² – x to 0 to find the zeros we get:

x (x – 1) = 0

x = 0 or x – 1 = 0

x = 0 or x = 1

As x³ + x² – ax + b is divisible by x² – x

∴ The zeros o x² – x will satisfy x³ + x² – ax + b

Hence, (0)³ + 0² – a (0) + b = 0

b = 0

Also,

(1)³ + 1² – a (1) + 0 = 0

∴ a = 2

It is given in the question that, zeros of the polynomial 2x² + 7x + 5 are

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

Sum of zeros = and product of zeros =

= and

∴

The Division algorithm for polynomials is as follows:

If we have two polynomials f (x) and g (x) and the degree of f (x) is greater than the degree of g (x), where g (x) 0 than there exist two unique polynomials q (x) and r (x) such that:

f (x) = g (x)× q(x) + r(x)

where r (x) – 0 or degree of r (x) < degree of g (x)

We know that,

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula:

x² – (Sum of the zeroes)x + Product of zeros

∴ x² – (-1/2) x + (- 3)

∴ The required polynomial is x² + - 3

We know that,

To find the zeros of the quadratic polynomial we have to equate the f (x) to 0

∴ f (x) = 0

6x² – 3 = 0

3 (2x² – 1) = 0

2x² – 1 = 0

2x² = 1

x² = 1/2

x = � 1/√2

Hence, the zeros of the quadratic polynomial f (x) are 1/√2 and -1/√2

We know that,

To find the zeros of the quadratic polynomial we have to equate the f (x) to 0

∴ f (x) = 0

4√3 x² + 5x - 2√3 = 0

4√3 x² + 8x – 3x - 2√3 = 0

4x (√3 x + 2) - √3 (√3 x + 2) = 0

(√3 x + 2) = 0 or (4x - √3) = 0

x =- or x =

Hence, the zeros of the quadratic polynomial f (x) are - or

It is given in the question that,

Zeros of the polynomial x² – 5x + k are

Also,

α- β = 1

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

Sum of zeros = and product of zeros =

and

α + β=5 and α β = k/1

Now solving α- β=1 and α + β = 5, we get:

= 2

these values in we get:

It is given in the question that,

Zeros of the polynomial 6x² + x - 2 are

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

Sum of zeros = and product of zeros =

= and

Now we have:

=

=

=

=

=

It is given in the question that,

Zeros of the polynomial 5x² – 7x + 1 are

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

Sum of zeros = and product of zeros =

= and

= and

Now, we have:

=

=

= 7

It is given in the question that,

Zeros of the polynomial x² + x - 2 are

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

Sum of zeros = and product of zeros =

= and

= - 1 and

Now, we have:

= ()²

=

=

=

=

() =

It is given in the question that,

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 =

a – b + a + a + b =

3a = 3

a =

Now, we have product of zeros =

(a – b) (a) (a + b) =

(1 – b) (1) (1 + b) = - 1

1 – b² = - 1

b² = 2

∴ b =

We know that, Polynomial is an expression consist of constants, variables and exponents which are connected by addition, subtraction, multiplication and division but the exponent of variables cannot be in fraction and has to be a positive integer.

Hence, option D is correct.

We can observe that in the second expression the power of x is -1 but we know that the exponent of the variable in a polynomial has to be a positive term.

Hence, option D is correct.

We have, f(x) = x² – 2x – 3

Now, put f(x) = 0

x² – 2x – 3 = 0

x² – 3x + x – 3 = 0

x(x – 3) + 1(x – 3) = 0

(x – 3)(x + 1) = 0

Thus, x = 3, -1

Hence, option C is correct.

We have, f(x) = x² ‒ √2x – 12

Now, put f(x) = 0

x² ‒ √2x – 12 = 0

x(x ‒ 3√2) + 2√2(x – 3√2)=0

(x ‒ 3√2) + (x + 2√2) = 0

Thus, x= 3√2 , -2√2

Hence option B is correct.

We have, f(x) = 4x² + 5 x – 3

Now, put f(x) = 0

4x² + 5 x – 3 = 0

4x² + x -√2x – 3 = 0

2√2x(√2x + 3) -1(√2x + 3) = 0

(2√2x -1) (√2x + 3) = 0

Hence, option C is correct.

We have, f(x) = x² + x – 2

f(x) = 6x² + x – 12 =0

Now, put f(x) = 0

6x² + x – 12 = 0

6x² + 9x – 8x – 12 = 0

3x(2x + 3) -4(2x + 3) = 0

Hence, option B is correct.

f(x) = 21x² – 11x – 2 =0

Now, put f(x) = 0

21x² – 11x – 2 =0

21x² – 14x + 3x – 2 =0

7x(3x – 2) + 1 (3x – 2) = 0

(3x – 2) (7x + 1) = 0

Hence, option A is correct.

We have, sum of zeroes(α + β) = 3

Product of zeroes(αβ) = -10

We know that,

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

= x² – 3x – 10

Hence, option C is correct.

We have, α = 5, β = -3

Thus, sum of zeros(α + β) = 5 – 3 = 2

Product of zeros(αβ) = 5(-3) = -15

We know that,

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

= x² – 2x – 15

Hence, option C is correct.

Thus, sum of zeros(α + β)

We know that,

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

Hence, option D is correct.

Let the zeros be α and β

We know that, α + β = -88 (sum of zeros)

αβ = 125 (product of zeros)

This is only possible when both the zeroes are negative.

Since, α and β are the zeroes of polynomialx² + 5x + 8

And we know that,

x² – (α + β) + αβ is the polynomial with α and β as its zeros

Thus, (α + β) = -5

Hence, option B is correct.

Since, α and β are the zeroes of polynomial2x² + 5x – 9

And we know that,

x² – (α + β) + αβ is the polynomial with α and β as its zeros

Hence, option C is correct.

Given: 2 is the zero of the polynomial f(x) = kx² + 3x + k

Put f(2) = 0

k(2)² + 3(2) + k = 0

4k + 6 + k = 0

Hence, option D is correct.

Given: -4 is the zero of the polynomial f(x) = (k – 1) x² + kx + 1

Put f(-4) = 0

(k – 1)(-4)² + k(-4) + 1 = 0

16k -16 – 4k + 1 = 0

Hence, option B is correct.

It is given in the question that, -2 and 3 are the zeros of x² + (a + 1) x + b

∴ (- 2)² + (a + 1) × (- 2) + b = 0

4 – 2a – 2 + b = 0

b – 2a = - 2 (i)

Also, we have

3² + (a + 1) × 3 + b = 0

9 + 3a + 3 + b = 0

b + 3a = - 12 (ii)

Now, by subtracting the equation (i) from (ii) we get:

a = - 2

Also, b = - 2 – 4 = - 6

Thus, k = 3

Hence, option A is correct.

Let the zeros of the polynomial be α and β

According to the question,

α + β = αβ

Hence option D is correct.

Given: α and βare zeros of the given polynomial.

∴α = β= -6 and αβ = 2

Hence, option B is correct.

Given:α, β, γ are the zeros of the polynomial x³ – 6x² – x + 30

We know that,

Hence, option A is correct.

Given: α, β, γ are the zeros of the polynomial 2x³ + x² – 13x + 6

We know that,

Hence, option A is correct.

According to the question,

α, β, γ be the zeros of the polynomial p(x)such that (α + β + γ ) = 3, (αβ + βγ + γ α) = ‒10 and αβγ = ‒24

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

x³ – 3x² – 10x + 24

Hence, option C is correct.

Let5 us assume , -0 and 0 be the zeros of the given polynomial

∴ Sum of zeros =

+ 0 + 0 =

=

∴ The third zero is

Hence, option A is correct

Let us assume and 0 be the zeros of the given polynomial

∴ Sum of the products of zeros taking two at a time is given by:

() =

=

The product of other two zeros will be

Hence, option B is correct

It is given in the question that,

- 1 is the zero of the given polynomial x³ + ax² + bx + c

∴ (- 1)³ + a × (- 1)² + b × (- 1) + c = 0

a – b + c + 1 = 0

c = 1 – a + b

Also, the product of all zeros is:

× (- 1) = - c

= c

= 1 – a + b

Hence, option C is correct

It is given in the question that,

are the zeros of the given polynomial 2x² + 5x + k

= and =

,

=

∴ – =

(-)² – =

= 1

k = 2

Hence, option D is correct

As, we know that

According to the division algorithm on polynomials, we have

either r (x) = 0 or deg r (x) < deg g (x)

Hence, option C is correct

We know that,

Monomial is that which consist only one term and in the given option we have only one term i.e. 5x²

∴ Option D is correct

It is given in the question that,

p (x) = x² – 2x – 3

Let us assume x² – 2x – 3 = 0

x² – (3 – 1) x - 3 = 0

x² – 3x + x – 3 = 0

x (x – 3) + 1 (x – 3) = 0

(x – 3) (x + 1) = 0

x = 3, - 1

We have,

p (x) = x³ – 6x² – x + 3

Now we will compare the given polynomial with: x³ – () x² + () x –

By comparing we get:

() = - 1

We have,

p (x) = x² – 2x + 3k

Now by comparing the given polynomial with ax² + bx + c, we get:

a = 1, b = - 2 and c = 3k

In the question it is given that, are the roots of the given polynomial

∴ =

= -( )

= 2 (i)

:

=

=

= 3k (ii)

Hence, by using (i) and (ii), we have

=

2 = 3k

k =

Let us assume the zeros of the polynomial be

We have,

p (x) = 4x² – 8kx + 9

Now comparing the given polynomial with ax² + bx + c, we get:

a = 4, b = - 8kx + 9

Sum of roots =

=

2 + 4 = 2k

+ 2 = k

= (k – 2) (i)

, we have product of roots, =

+ 4) =

(k – 2) (k – 2 + 4) =

(k – 2) (k + 2) =

k² – 4 =

4k² – 16 = 9

4k² = 25

k² =

k =

k =

We have,

p (x) = x² + 2x – 195

Let us assume p (x) = 0

x² + (15 – 13) x – 195 = 0

x² + 15x – 13x – 195 = 0

x (x + 15) – 13 (x + 15) = 0

(x + 15) (x – 13) = 0

x = - 15, 13

Hence, the zeros of the polynomial are -15 and 13

We have,

(a + 9) x² – 13x + 6a = 0

Comparing with standard form of quadratic equation Ax² + Bx + C

A = (a² + 9), B = 13 and C = 6a

Let us assume be the zeros of the given polynomial

∴ Product =

=

1 =

a² + 9 = 6a

a² – 6a + 9 = 0

a² – 2 × a × 3 + 3² = 0

(a – 3)² = 0

a – 3 = 0

a = 3

It is given in the question that the two roots of the given polynomial are 2 and – 5

Let us assume and = - 5

We have:

Sum of Zeros = = 2 + (-5) = - 3

Product of Zeros = = 2 × (-5) = - 10

Hence,

Required polynomial = x² – (

= x² – (- 3) x + 10

= x² + 3x – 10

It is given in the question that the roots of the given polynomial are (a – b), a and (a + b)

Now by comparing the given polynomial with Ax³ + Bx² + Cx + D, we get:

A = 1, B = - 3, C = 1 and D = 1

Now,

(a – b) + a + (a + b) =

3a = -

a = 1

Also, we have:

(a – b) × a × (a + b) =

a (a² – b²) =

1 (1² – b²) = - 1

1 – b² = - 1

b² = 2

b =

Hence, a = 1 and b =

We have,

p (x) = x³ + 4x² – 3x – 18

Now,

p (2) = (2)³ + 4 × (2)² – 3 (2) – 18

= 8 + 16 – 6 – 18

= 24 – 24

= 0

∴ 2 is the zero of the given polynomial

It is given in the question that,

Sum of the zeros = - 5

And, product of the zeros = 6

∴ Required polynomial = x² – (Sum of the zeros) x + Product of the zeros

= x² – (- 5) x + 6

= x² + 5x + 6

Let us assume be the zeros of the required polynomial

We have,

= 3 + 5 + (- 2) = 6

= 3 × 5 + 5 × (- 2) + (- 2) × 3 = - 1

And, = 3 × 5 × - 2 = - 30

Now, we have:

p (x) = x³ – x² () + x () –

= x³ – x² × 6 + x × (- 1) – (- 30)

= x³ – 6x² – x + 30

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

Remainder theorem:
Let p(x) be any polynomial of degree greater than or equal to one and let a be any real number. If p(x) is divided by the linear polynomial x – a, then the remainder is p(a).
It is given in the question that,

p (x) = x³ + 3x² – 5x + 4

Now, p (2) = (2)³ + 3 (2)² – 5 (2) + 4

= 8 + 12 – 10 + 4

= 14

It is given in the question that,

f (x) = x³ + 4x² + x – 6

Now, we have

f (- 2) = (- 2)³ + 4 (- 2)² + (- 2) - 6

= - 8 + 16 – 2 - 6

= 0

Hence, (x + 2) is a factor of f (x)

It is given in the question that,

p (x) = 6x³ + 3x² – 5x + 1

= 6x³ – (- 3) x² + (- 5) x – 1

Now by comparing the polynomial with x³ – x² () + x ( – , we get:

= - 5 and,

= - 1

()

= ()

= ()

= 5

It is given in the question that,

f (x) = x² – 5x + k such that its coefficients are a = 1, b = - 5 and c = k

∴ =

= 5 (i)

, = 1 (ii)

, we get:

2α = 6

α = = 3

Now putting the value of α in (i), we get

3 + = 5

= 5 – 3 = 2

=

Hence, k = 6

Let us assume t = x²

∴ f (t) = t² + 4t + 6

Now first of all we have to equate f (t) = 0 in order to find the zeros

∴ t =

=

= - 2

As, t = x²

So, x² = - 2

x =

We know that, the zeros of the polynomial should be a real number and this is not a real number

∴ f (x) hs no zeros

It is given in the question that,

p (x) = x³ – 6x² + 11x – 6 having factor (x + 3)

Now, we have to divide p (x) by (x – 3)

x³ – 6x² + 11x – 6 = (x – 3) (x³ – 3x + 2)

= (x – 3) [(x² – (2 + 1) x + 2]

= (x – 3) (x² – 2x – x + 2)

= (x – 3) [x (x – 2) – 1 (x – 2)]

= (x – 3) (x - 1) (x – 2)

Hence, the two zeros of the polynomial are 1 and 2

It is given in the question that,

p (x) = 2x⁴ – 3x³ – 3x² + 6x – 2 having zeros and -

∴ The polynomial is (x + ) (x - ) = x² – 2

Let us now divide p (x) by (x² – 2)

2x⁴ – 3x³ – 3x² + 6x – 2 = (x² – 2) (2x² – 3x + 1)

= (x² – 2) [(2x² – (2 + 1) x + 1]

= (x² – 2) (2x² – 2x – x + 1)

= (x² – 2) [(2x (x – 1) – 1 (x – 1)]

= (x² – 2) (2x – 1) (x – 1)

Hence, the other two zeros are and 1

It is given in the question that,

p (x) = 3x⁴ + 5x³ – 7x² + 2x + 2

Now we have to divide p (x) by (x² + 3x + 1), we get:

Hence, the quotient is 3x² – 4x + 2

Given: when x³ + 2x² + kx + 3 is divided by (x – 3), then the remainder is 21.

To find: The value of k.

Solution:

If x-3 divides the equation and leaves 21 as remainder,
it means substituting x=3 in the equation and putting it equal to 21 will
value of k.

Let us assume,

p (x) = x³ + 2x² + kx + 3

Now, p (3) = (3)³ + 2 (3)² + 3k + 3

= 27 + 18 + 3k + 3

= 48 + 3k

It is given in the question that the remainder is 21

Hence, 3k + 48 = 21

3k = - 27

k = - 9