Polynomials

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.
As -3 is the zero of given polynomial, p(x) = (k - 1) x² + kx + 1 is -3, Therefore

p (- 3) = 0

⇒ (k - 1)(- 3)² + k(- 3) + 1 = 0

⇒ (k - 1)(9) - 3k + 1 = 0

⇒ 9k - 3k - 8 = 0

⇒ 6k = 8

⇒ k = 4/3

- 3 and 4 are the zeroes of the polynomial p(x) = ax² + bx + c

∴ sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

- 3 + 4 = - b/a

1 = - b/a

b = - 1 and a = 1

Product of the zeroes = constant term ÷ coefficient of x²

c = (- 3)/4

⇒ c = 12

Putting the values of a, b and c in the polynomial p(x) = ax² + bx + c

∴ The polynomial is x² - x - 12

= x²/2 - x/2 - 6 (dividing by any constant will not change the polynomial)

OR

The equation of a quadratic polynomial is given by x² - (sum of the zeroes) x + (product of the zeroes) where,

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x² and

Product of the zeroes = constant term ÷ coefficient of x²

∴ Sum of the zeroes = - 3 + 4 = 1

and

product of the zeroes = (- 3)4 = - 12

⇒ x² - (1)x = (- 12)

= x² - x - 12

= x²/2 - x/2 - 6

(dividing by any constant will not change the polynomial)

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.

2 and - 3 are the zeroes of the polynomial p(x) = x² + (a + 1)x + b

i.e. p(2) = 0 and p(- 3) = 0

p(2) = (2)² + (a + 1)(2) + b = 0

= 4 + 2a + 2 + b = 0

= 6 + 2a + b = 0 (1)

P(- 3) = (- 3)² + 9 + (a + 1)(- 3) + b = 0

= 9 - 3a - 3 + b = 0

= 6 - 3a + b = 0 (2)

Equating (1) = (2), as both the equations are equal to zero. Hence both equations are equal to each other.

6 + 2a + b = 6 - 3a + b

= 5a = 0

⇒ a = 0

Putting the value of ‘a’ in (1)

6 + 2(0) + b = 0

⇒ b = - 6

OR

The equation of a quadratic polynomial is given by x² - (sum of the zeroes) x + (product of the zeroes)

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

⇒ sum of the zeroes = - (a + 1)

= 2 - 3 = - a - 1

= - 1 + 1 = - a

= - a = 0

⇒ a = 0

Product of the zeroes = constant term ÷ coefficient of x²

⇒ b = product of the zeroes

= 2(- 3)

= 6

let - 2 and 5 are the zeroes of the polynomials of the formp(x) = ax² + bx + c.

sum of the zeroes = - (coefficient of x) ÷ coefficient of x² i.e.

Sum of the zeroes = - b/a

- 2 + 5 = - b/a

3 = - b/a

⇒ b = - 3 and a = 1

Product of the zeroes = constant term ÷ coefficient of x² i.e.

Product of zeroes = c/a

(- 2)5 = c/a

- 10 = c

Put the values of a, b and c in the polynomial p(x) = ax² + bx + c.

∴ The equation is x² - 3x - 10

OR

The equation of a quadratic polynomial is given by x² - (sum of the zeroes)x + (product of the zeroes) where,

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x² and

Product of the zeroes = constant term ÷ coefficient of x²

∴ Sum of the zeroes = - 2 + 5 and product of the zeroes = (- 2)5

= 3 = - 10

∴ the equation is x² - 3x - 10

We know, the zeroes do not change if the polynomial is divided or multiplied by a constant

⇒ kx² - 3kx - 10k will also have - 2 and 5 as their zeroes.

As k can take any real value, there can be many polynomials having - 2 and 5 as their zeroes.

let α, β & γ be the zeroes of the polynomial ax³ + bx² + cx + d

And let α = 0(given)

sum of the product of two zeroes at a time = coefficient of x ÷ coefficient of x³ i.e.

c/a = sum of the product of two zeroes at a time

c/a = αβ + βγ + γα

c/a = β (0) + βγ + γ (0) (putting α = 0)

c/a = βγ

The product of the other two zeroes is c/a

let α, β & γ be the zeroes of the polynomial p(x) = x³ + ax² + bx + c and

α = - 1 (given)

Zeroes of a polynomial is all the values of x at which the polynomial is equal to zero.

i.e. p (α) = p(- 1) = 0

= (- 1)³ + (- 1)²a + (- 1)b + c = 0

= - 1 + a - b + c = 0

= c = 1 - a + b

Product of zeroes = - (constant term) ÷ coefficient of x³ i.e.

Product of zeroes = - c

αβγ = - c

= βγ = c

= βγ = 1 - a + b

let p(x) = x² + 99x + 127

x

=

=

=

= - 2.6/2, - 195.4/2

⇒ x = - 1.3, - 97.7

Both the zeroes are negative

OR

We know, in quadratic polynomial if the coefficients of the terms are of the same sign, then the zeroes of the polynomial are negative.

i.e. if either a > 0, b > 0 and c > 0 or a < 0, b < 0 and c < 0, then both zeroes are negative

So here, a = 1 > 0, b = 99 > 0 and c = 127 > 0

⇒ The zeroes are negative.

let p(x) = x² + kx + k

x =

=

=

⇒ k (k - 4) > 0 ⇒ k ∈ (- ∞, 0) ⋂ (4, ∞)

Here k lies in two intervals, therefore we need to consider both the intervals separately.

Case1:

When k (- ∞, 0)

i.e. k < 0

we know that in a quadratic equation p(x) = ax² + bx + c, if either a > 0, c < 0 or a < 0, c > 0, then the zeroes of the polynomial are of opposite signs.

Here a = 1 > 0, b = k < 0 and c = k < 0

⇒ both zeroes are of opposite signs

Case2:

When k ∈ (4, ∞)

i.e. k > 0

We know, in quadratic polynomial if the coefficients of the terms are of the same sign, then the zeroes of the polynomial are negative.

i.e. if either a > 0, b > 0 and c > 0 or a < 0, b < 0 and c < 0, then both zeroes are negative

Here a = 1 > 0, b = k > 0 and c = k > 0

⇒ both the zeroes are negative

Hence, by both cases, both the zeroes cannot be positive.

the zeroes of a quadratic polynomial are equal when the discriminant is equal to 0

i.e. D = 0

b² - 4ac = 0

4ac = b²

ac = b²/4 > 0 (square of any positive or negative number is positive )

ac > 0

For ac > 0, a and c must have the same sign

i.e., either a > 0 and c > 0 or a < 0 and c < 0

let p(x) = x² + ax + b

And let α be one of the zeroes

∴ - α is the other zero of the polynomial p(x)

Product of the zeroes = constant term ÷ coefficient of x²

Product of the zeroes = b

α(- α) = b

- α² = b

i.e. b is negative.

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

α - α = - a

0 = - a

⇒ a = 0

∴ The polynomials can be written as x² - α² (No linear term and negative constant term)

The shape of a quadratic polynomial is either upward or downward U - shaped curve i.e., an upward or downward parabola. Also, the graph of the quadratic equation cuts the X - axis at the most at two points, but in fig(D) it cuts the X - axis at three points.

∴ fig (D) is not the graph of a quadratic polynomial.

No, x² - 1 cannot be the quotient on division of x⁶ + 2x³ + x - 1 by a polynomial in x of degree 5. This is because when we divide a degree 6 polynomial with degree 5 polynomial we get the quotient to be of degree 1.

Let if possible (x² - 1) divides the polynomial with degree 6 and the quotient obtained is degree 5 polynomial (1)

i.e.,(degree6 polynomial) = (x² - 1)(degree 5 polynomial) + r(x) (a = bq + r)

= (degree 7 polynomial) + r(x) (x² term × x⁵ term = x⁷ term)

= (degree 7 polynomial)

Clearly, (degree6 polynomial) (degree 7 polynomial)

Hence, (1) is contradicted

∴ x² - 1 cannot be the quotient on division of x⁶ + 2x³ + x - 1 by a polynomial in x of degree 5

quotient = 0 and remainder = ax² + bx + c

Let p(x) = ax² + bx + c and g(x) = px³ + qx² + rx + s , p≠0

Clearly the degree of divisor is greater than the degree of the dividend

i.e. deg (p(x)) < deg(g(x))

∴ deg (p(x)) < deg (g(x)) × deg (q(x)) where deg (q(x)) can take any value …(1)

By division algorithm,

p(x) = g(x).q(x) + r(x)

Hence the degree in the L.H.S is equal to the degree of R.H.S

we know that the degree of r(x) will always be less than or equal to p(x)

⇒ deg (p(x)) = deg(g(x))× deg(q(x)) …(2)

(1) and (2) do not go hand in hand unless q(x) = 0

∴ p(x) = g(x).q(x) + r(x)

= p(x) = 0 + r(x)

⇒ p(x) = r(x)

∴ The quotient = 0 and the remainder is same as the dividend.

degree of p(x) is less than degree of g(x).

Given, the quotient = 0

i.e. q(x) = 0

by division algorithm,

p(x) = g(x).q(x) + r(x)

∴ the deg(p(x)) < deg(g(x))

g(x) is the factor of p(x) and degree of p(x)degree of g(x)

Here r(x) = 0

By division algorithm,

p(x) = g(x)q(x) + r(x)

p(x) = g(x) q(x)

∴ deg(p(x)) = deg(g(x))×deg(q(x))

⇒ deg(p(x))≥ deg(g(x))

Hence g(x) is the factor of p(x) and clearly degree of p(x) degree of g(x)

No, the quadratic polynomial x² + kx + k cannot have equal zeroes for some odd integer k > 1

Let suppose, if it has equal zeroes

the zeroes of a quadratic polynomial are equal when discriminate is equal to 0

i.e., D = 0

b² - 4ac = 0

b² = 4ac

k² = 4k

k = 0 or k = 4

But it is given that k > 1, so we reject the value k = 0 < 1

∴ k = 4

But 4 is not an odd number and only at k = 4 will the polynomial get equal roots.

Hence, the quadratic polynomial x² + kx + k cannot have equal zeroes for some odd integer k > 1

If the zeroes of a quadratic polynomial ax² + bx + c are both positive, then a, b and c all have the same sign: False.

Let α, β be the zeroes of the polynomial p(x) = ax² + bx + c

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

α + β = - b/a > 0 (∵ α > 0, β > 0 ⇒ α + β > 0)

∴ for – b/a > 0, b and a must have opposite signs.

Product of the zeroes = constant term ÷ coefficient of x²

αβ = c/a > 0 (∵ α,β > 0⇒ αβ > 0)

∴ for c/a > 0, c and a must have same signs.

Case 1: when a > 0

⇒ - b > 0 and c > 0

= b < 0 and c > 0

Case 2: when a < 0

⇒ - b < 0 and c < 0

= b > 0 and c < 0

Hence, the coefficients have different signs.

If the graph of a polynomial intersects the x - axis at only one point it need not be a quadratic polynomial: True.

The quadratic polynomial cuts the x - axis at most at two points i.e. it can either touch x - axis at two points or one point or doesn’t touch the axis at all.

Hence the polynomial intersecting the x - axis at only one point need not be a quadratic polynomial.

If the graph of a polynomial intersects the x - axis at exactly two points, it need not be a quadratic polynomial: True.

The polynomial with degree greater than 2 can also intersect the x - axis at two points. Hence, it need not be a quadratic polynomial.

If two of the zeroes of cubic polynomials are zero then it does not have linear and constant terms: True

Let α, β and γ be the zeroes of the polynomial p(x0 = ax³ + bx² + cx + d, where given α = β = 0

Clearly a≠ 0

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

αβγ = - d/a

0 = - d/a

⇒ d = 0 (no constant term)

∴ the polynomial is p(x) = ax³ + bx² + cx

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

0 + (0) γ + (0) γ = c/a

0 = c/a

⇒ c = 0 (no linear term)

∴ the polynomial is p(x) = ax³ + bx²

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

α + β + γ = - b/a

0 + 0 + γ = - b/a

⇒ b = - γ/a (there exists x² term ∵ b≠ 0)

∴ the polynomial is p(x) = ax³ - γx²/a

Hence, the polynomial does not have a linear and a constant term.

If all the zeroes of a cubic polynomial are negative, then all the coefficients and the constant term of the polynomial have the same sign: True

Let α, β and γ be the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, where α, β, γ < 0

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

αβγ = - d/a < 0 (∵ α, β, γ < 0 ⇒ αβγ < 0)

⇒ d/a > 0

⇒d and a have the same signs.

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a > 0 (∵ α, β, γ < 0 ⇒ αβ,βγ,αγ > 0 ⇒ αβ + βγ + αγ > 0 )

⇒c and a have the same signs.

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

α + β + γ = - b/a < 0 (∵ α, β, γ < 0 ⇒ α + β + γ < 0)

⇒ b/a > 0

⇒b and a have same signs.

⇒ a,b,c and d have same signs.

If all three zeroes of a cubic polynomial x³ + qx² - bx + c are positive, then at least one of a, b and c is non - negative: True.

Let α, β and γ be the zeroes of the polynomial p(x) = ax³ + bx² + cx + d, where α, β, γ > 0

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

αβγ = - d/a > 0 (∵ α, β, γ > 00 ⇒ αβγ > 0)

⇒ d/a < 0

⇒d and a have different signs.

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a > 0 (∵ α, β, γ > 0 ⇒ αβ, βγ, αγ > 0 ⇒ αβ + βγ + αγ > 0 )

⇒c and a have the same signs.

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

α + β + γ = - b/a > 0 (∵ α, β, γ > 0 ⇒ α + β + γ > 0)

⇒ b/a < 0

⇒b and a have different signs.

Case1: when a > 0 c > 0 , b < 0 and d < 0

Case2: when a < 0 c < 0 , b > 0 and d > 0

∴ in both cases two of the coefficients are non - negative.

The only value of k for which the quadratic polynomial kx² + x + k has equal zeroes is 1/2: False.

A quadratic polynomial kx² + x + k have equal roots when its discriminate is equal to

i.e. D = 0

b² - 4ac = 0

b² = 4ac

1² = 4(k) (k)

1 = 4k²

k² = 1/4

k = 1/2

Hence, k = 1/2 is not the only value for which the polynomial has equal roots.

By splitting the middle term

4x² - 3x - 1 = 0

4x² - (4x - x) - 1 = 0

4x² - 4x + x - 1 = 0

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

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

⇒ x = 1, - 1/4

Verification:

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

α + β = - b/a

1 - 1/4 = - (- 3)/4

= 3/4 = 3/4

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

1(- 1/4) = - 1/4

- 1/4 = - 1/4

By splitting the middle term

3x³ + 4x - 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

Verification:

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

α + β = - b/a

- 2 + (2/3) = - (4)/3

= - 4/3 = - 4/3

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

Product of the zeroes = (- 2) (2/3) = - 4/3

By splitting the middle term

5t² + 12t + 7 = 0

5t² + (5t + 7t) + 7 = 0

5t² + 5t + 7t + 7 = 0

5t (t + 1) + 7(t + 1) = 0

(t + 1)(5t + 7) = 0

(t + 1)(5t + 7) = 0

⇒ t = - 1. - 7/5

Verification:

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

α + β = - b/a

(- 1) + (- 7/5) = - (12)/5

= - 12/5 = - 12/5

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 1)(- 7/5) = - 7/5

- 7/5 = - 7/5

t³ - 2t - 15t = 0

Taking t common

t (t² - 2t - 15) = 0

By splitting the middle term

t {t² - (- 3t + 5t) - 15} = 0

t (t² + 3t - 5t - 15) = 0

t {t(t + 3) - 5(t + 3)} = 0

t (t + 3)(t - 5) = 0

⇒ t = - 3, 0, 5

Verification:

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

α + β + γ = - b/a

(0) + (- 3) + (5) = - (- 2)/1

= 2 = 2

Sum of the products of two zeroes at a time = coefficient of x ÷ coefficient of x³

αβ + βγ + αγ = c/a

(0)(- 3) + (- 3) (5) + (0) (5) = - 15/1

= - 15 = - 15

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

αβγ = - d/a

(0)(- 3)(5) = 0

0 = 0

By splitting the middle term

2x² + 7x/2 + 3/4 = 0

8x² + 14x + 3 = 0

8x² + (12x + 2x) + 3 = 0

8x² + 12x + 2x + 3 = 0

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

(2x + 3)(4x + 1) = 0

⇒ x = - 3/2, - 1/4

Verification:

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

α + β = - b/a

(- 3/2) + (- 1/4) = - (7)/4

= - 7/4 = - 7/4

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 3/2)(- 1/4) =

3/8 = 3/8

By splitting the middle term

4x² + 52x - 3 = 0

4x² + (6√ 2x - √2 x) - 3 = 0

4x² + (6√ 2x - √2 x) - 3 = 0

2√2 x (√2x + 3) - (√2x + 3) = 0

(√2x + 3)(2√2 - 1) = 0

⇒ x = -

Verification:

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

α + β = - b/a

() + () = -

=

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

=

- 3/4 = - 3/4

By splitting the middle term

2s² - (1 + 2√2) s + √2 = 0

2s² - s - 2√2s + √2 = 0

s (2s - 1) - √2(2s - 1) = 0

(2s - 1)(s - √2) = 0

⇒ s = 1/2, √2

Verification:

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

α + β = - b/a

(1/2) + (√2) = - {(1 + 2√2)}/2

= - (1 + 2√2)/2 = - (1 + 2√2)/2

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(1/2)(√2) = √2/2

1/√2 = 1/√2

By splitting the middle term

v² + 4√3 v - 15 = 0

v² + (5√3v - 3√3v) - 15 = 0

v² + 5√3v - 3√3v - 15 = 0

v (v + 5√3) - √3(v + 5√3) = 0

(v + 5√3)(v - √3) = 0

⇒ v = - 5√3, √3

Verification:

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

α + β = - b/a

(- 5√3) + (√3) = - 4√3

= - 4√3 = - 4√3

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 5√3)(√3) = - 15

- 15 = - 15

By splitting the middle term

y² + y - 5 = 0

2y² + 3√5y - 10 = 0

2y² + (4√5y - √5y) - 10 = 0

2y² + (4√5y - √5y) - 10 = 0

2y(y + 2√5) - √5(y + 2√5) = 0

(y + 2√5)(2y - √5) = 0

⇒ y = - 2√5, √5/2

Verification:

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

α + β = - b/a

(- 2√5) + (√5/2) = - (3√5)/2

= - 3√5/2 = - 3√5/2

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 2√5)(√5/2) = - 5

- 5 = - 5

By splitting the middle

7y² - 11y/3 - 2/3 = 0

21y² - 11y - 2 = 0

21y² + (3y - 14y) - 2 = 0

21y² + 3y - 14y - 2 = 0

3y (7y + 1) - 2(7y + 1) = 0

(7y + 1)(3y - 2) = 0

⇒ y = - 1/7, 2/3

Verification:

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

α + β = - b/a

(- 1/7) + (2/3) = - (- 11)/21

= 11/21 = 11/21

Product of the zeroes = constant term ÷ coefficient of x²

α β = c/a

(- 1/7)(2/3) =

- 2/21 = - 2/21

Sum of the zeroes = - 8/3

Product of the zeroes = 4/3

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

= x² - 8x/3 + 4/3

= 3x² - 8x + 4

By splitting the middle term

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 = 21/8

Product of the zeroes = 5/16

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

= x² - 21x/8 + 5/16

= 16x² - 42x + 5

By splitting the middle term

16x² - 42x + 5 = 0

16x² - (2x + 40x) + 5 = 0

16x² - 2x - 40x + 5 = 0

2x (8x - 1) - 5(8x - 1) = 0

(8x - 1)(2x - 5) = 0

⇒ x = 1/8, 5/2

Sum of the zeroes = - 2√3

Product of the zeroes = - 9

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

= x² - 2√3x 9

By splitting the middle term

x² - 2√3x - 9 = 0

x² - (- √3x + 3√3x) - 9 = 0

x² + √3x - 3√3x - 9 = 0

x(x + √3) - 3√3(x + √3) = 0

(x + √3)(x - 3√3) = 0

⇒ x = - √3, 3√3

Sum of the zeroes =

Product of the zeroes = - 1/2

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

= x² - x - 1/2

= 2√5x² - 3x - √5

By splitting the middle term

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 P(x) = x³ - 6x² + 3x + 10

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

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) = 6x³ + 2x² - 10x - 42

As √2 is one of the zeroes of P(x).

⇒ g(x) = (x - √2) is one of the factors of P(x).

By division algorithm

Dividend = (divisor) (quotient) + remainder

i.e. p(x) = g(x)q(x) + r(x)

clearly r(x) = 0 and q(x) = 6x² + 7√2 + 4

⇒ 6x³ + √2x² - 10x - 4√2 = (x - √2) (6x² + 7√2 + 4)

6x³ + √2x² - 10x - 4√2 = 0

= (x - √2) (6x² + 7√2 + 4) = 0

= (x - √2) {6x² + (3√2x + 4√2x) + 4} = 0 (by splitting the middle term)

= (x - √2) {6x² + 3√2x + 4√2x + 4} = 0

= (x - √2) {3√2x (√2x + 1) + 4(√2x + 1)} = 0

= (x - √2) (√2x + 1) (3√2x + 4) = 0

⇒ x = - 1/√2, 2√2/3, √2

let g(x) = x² + 2x + k and p(x) = 2x⁴ + x³ - 14x² + 5x + 6

Then if g(x) is a factor of p(x)

By division algorithm

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

Where q(x) is the quotient and

r(x) is the remainder which will be equal to zero.

i.e. r(x) = 0

Clearly q(x) = 2x² - 3x - 8 - 2k and r(x) = (21 + 7k)x + (2k² + 8k + 6) = 0

By comparing the coefficients, 21 + 7k = 0 and 2k² + 8k + 6 = 0

2k² + 8k + 6 = 0

k² + 4k + 3 = 0

k² + (k + 3k) + 3 = 0

k (k + 1) + 3(k + 1) = 0

(k + 1)(k + 3) = 0

⇒ k = - 1, - 3

Also, 21 + 7k = 0

Case1: k = - 1

21 + 7(- 1) = 0

= 21 - 7 = 14≠0

Hence k = - 1 is rejected.

Case2: k = - 3

21 + 7(- 3) = 0

= 21 - 21 = 0

∴ the value of k is - 3

g(x) = x² + 2x - 3

x² + 2x - 3 = 0

x² + (- x + 3x) - 3 = 0

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

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

⇒ x = 1, - 3

q(x) = 2x² - 3x - 8 - 2(- 3)

= 2x² - 3x - 2

2x² - 3x - 2 = 0

2x² - (4x - x) - 2 = 0

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

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

⇒ x = 2, - 1/2

Now, we know g(x) and q(x) are factors of p(x)

∴ the zeroes of g(x) and q(x) will be the zeroes of p(x)

Hence, the zeroes of p(x) = - 3, - 1/2, 1, 2

Let P(x) = x³ - 35x² + 13x - 35

As √5 is one of the zeroes of P(x).

⇒ (x - √5) is one of the factors of P(x).

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

⇒ x³ - 35x² + 13x - 35 = (x - √5) (x² - 2√5x + 3) + 0

x³ - 35x² + 13x - 35 = 0

= (x - √5) (x² - 2√5x + 3) = 0

= (x - √5) [x² - {(√5 + √2) x + (√5 - √2) x} + 3) = 0

= (x - √5) [x{x - (√5 + √2)} - (√5 - √2) {x - (√5 + √2)} + 3] = 0

= (x - √5) {x - (√5 + √2)} {x - (√5 - √2)} = 0

⇒ x = (√5 + √2), (√5 - √2), √5

Let P(x) = x⁵ - x⁴ - 4x³ + 3x + b and q (x) = x³ + 2x² + a

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

r(x) = - (a + 1)x² + 3(1 - a)x + b - 2a = 0 and g(x) = x² - 3x + 2

By comparing the coefficients

- (a + 1) = 0 , (1 - a) = 0

⇒ a = - 1

Also, b - 2a = 0

= b = 2a

= 2(- 1) = - 2

⇒ b = - 2

q (x) = x³ + 2x² - 1

x³ + 2x² - 1 = 0

x³ + x² + x² - 1 = 0

x²(x + 1) + (x² - 1) = 0

x²(x + 1) + (x + 1)(x - 1) = 0

(x + 1)(x² + x - 1) = 0

g(x) = x² - 3x + 2

x² - 3x + 2 = 0

x² - (x + 2x) + 2 = 0

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

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

Dividend = (divisor) (quotient) + remainder

⇒ p(x) = g(x).q(x) + r(x)

P(x) = x⁵ - x⁴ - 4x³ + 3x - 2 = q(x).g(x) + 0

= (x³ + 2x² - 1) (x2 - 3x + 2)

= (x + 1) (x2 + x - 1) (x - 1) (x - 2)

∴ 1 and 2 are the zeroes of p(x) that are not in q(x).