Polynomials

(i) A coefficient is a multiplicative factor in some term of a polynomial. It is the constant (-6 here) written before the variable (x⁵ here).

∴ The coefficient of x⁵ is -6.

(ii) Degree of p(x) is the highest power of x in p(x).

∴ The degree of p(x) is 7.

(iii) The term that is not attached to a variable (i.e., x) is -6.

∴ The constant term is -6.

(i) False

Since, Degree of a polynomial is the highest power of x in the polynomial which is 2 in.

(ii) False

A coefficient is a multiplicative factor in some term of a polynomial. It is the constant written before the variable.

∴ The coefficient of x² is -4.

(iii) True

Since, the power of x of a constant term is 0.

∴ the degree of a constant term is 0.

(iv) False

For an expression to be a polynomial term, any variables in the expression must have whole-number powers (i.e., x⁰, x¹, x²,……)

Since, the power in the expression has powers -2 and -1 which are not whole numbers, therefore, is not a polynomial.

(v) False

Degree of a polynomial is the highest power of x in the polynomial. The degree of polynomial is not related to the number of terms in the polynomial.

Therefore, the statement is false.

P(t) = t³ – 1

Therefore,

P(1) = 1³ – 1

⇒ P(1) = 1 – 1

⇒ P(1) = 0

P(-1) = (-1)³ – 1

⇒ P(-1) = -1 – 1

⇒ P(-1) = -2

P(0) = 0³ – 1

⇒ P(0) = 0 – 1

⇒ P(0) = -1

P(2) = 2³ – 1

⇒ P(2) = 8 – 1

⇒ P(2) = 7

P(-2) = (-2)³ – 1

⇒ P(2) = -8 – 1

⇒ P(2) = -9

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Now,

P(x) = x⁴ – 16

Therefore,

P(-2) = (-2)⁴ – 16

⇒ P(-2) = 16 – 16

⇒ P(-2) = 0

P(2) = 2⁴ – 16

⇒ P(2) = 16 – 16

⇒ P(2) = 0

Hence, Yes, -2 and -2 are zeroes of the polynomial x⁴ – 16.

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Now,

P(x) = x² – x – 6

Therefore,

P(-2) = (-2)² – (-2) – 6

⇒ P(-2) = 4 + 2 – 6

⇒ P(-2) = 0

P(3) = 3² – 3 – 6

⇒ P(3) = 9 – 3 – 6

-⇒ P(3) = 0

Hence, Yes, 3 and -2 are zeroes of the polynomial x² – x – 6.

(i) Since, the graph does not intersect with x-axis at any point therefore, it has no zeroes.

(ii) Since, the graph intersects with x-axis at only one point therefore, it has 1 number of zeroes.

(iii) Since, the graph intersects with x-axis three points therefore, it has 3 number of zeroes.

(iv) Since, the graph intersects with x-axis two points therefore, it has 2 number of zeroes.

(v) Since, the graph intersects with x-axis at four points therefore, it has 4 number of zeroes.

(vi) Since, the graph intersects with x-axis at three points therefore, it has 3 number of zeroes.

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, to find zeroes put p(x)=0.

p(x) = 0

⇒ 3x = 0

⇒ x = 0

Hence, x=0 is the zero of the polynomial.

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 and x = -2

Hence, x=-3 and x=-2 are the zeroes of the polynomial.

p(x) = 0

⇒ (x + 3)(x + 2)

⇒ x = -3 and x = -2

Hence, x=-3 and x=-2 are the zeroes of the polynomial.

p(x) = 0

⇒ x⁴ – 16 = 0

⇒ (x² – 4)(x² + 4)= 0

⇒ (x + 2)(x – 2)(x² + 4)= 0

⇒ x = -2, x = 2 and x² = 4

⇒ x = -2, x = 2 and x = ±2

Hence, x=2 and x=-2 are the zeroes of the polynomial.

p(x) = x² – x – 12

Clearly, the graph intersects with x – axis at x = -3 and x = 4

Hence, the zeroes of the polynomial x² – x – 12 are x = -3 and x=4.

p(x) = x² – 6x + 9

Clearly, the graph intersects with x – axis at x = 4

Hence, the zeroes of the polynomial x² – 6x + 9 are x=3.

p(x) = x² – 4x + 5

Clearly, the graph does not intersect with x – axis at any point.

Hence, there are no zeroes of the polynomial x² – 4x + 5.

p(x) = x² + 3x – 4

Clearly, the graph intersects with x – axis at x = -4 and x = 1.

Hence, the zeroes of the polynomial x² + 3x – 4 are x = -4 and x= 1

p(x) = x² – 1

Clearly, the graph intersects with x – axis at x = -1 and x = 1.

Hence, the zeroes of the polynomial x² – 1 are x = -1 and x= 1

A zero or root of a polynomial function is a number that, when plugged in for the variable, makes the function equal to zero.

Therefore, if p(x)=0, for a given x then the value of x is zero of polynomial.

p(x) = 4x² + 3x – 1

For 1/4,

p(1/4) = 4(1/4)² + 3(1/4) – 1

⇒ p(1/4) = 1/4 + 3/4 – 1

⇒ p(1/4) = 1 – 1

⇒ p(1/4) = 0

Therefore, x = 1/4 is a zero of the polynomial 4x² + 3x – 1.

For -1,

p(-1) = 4(-1)² + 3(-1) – 1

⇒ p(-1) = 4 – 3 – 1

⇒ p(-1) = 4 – 4

⇒ p(-1) = 0

Therefore, x = -1 is a zero of the polynomial 4x² + 3x – 1.

Hence, x = 1/4 and x = -1 are zeroes of the polynomial 4x² + 3x – 1.

p(x) = 0

⇒ x² – 2x – 8 = 0

⇒ x² + 2x – 4x – 8 = 0

⇒ x(x + 2) – 4(x + 2) = 0

⇒ (x + 2)(x – 4) = 0

⇒ x = -2 and x = 4

Hence, -2 and 4 are zeroes of the polynomial x² – 2x – 8.

Now,

Sum of zeroes = -2 + 4

⇒ Sum of zeroes = 2

Product of zeroes = -2 × 4

⇒ Product of zeroes = -8

p(s) = 0

⇒ 4s² – 4s + 1 = 0

⇒ 4s² – 2s – 2s + 1 = 0

⇒ 2s(2s – 1) – 1(2s – 1) = 0

⇒ (2s – 1) (2s – 1) = 0

⇒ x = 1/2 and x = 1/2

Hence, 1/2 and 1/2 are zeroes of the polynomial 4s² – 4s + 1.

Now,

Sum of zeroes = 1/2 + 1/2

⇒ Sum of zeroes = 1

Product of zeroes = 1/2 × 1/2

⇒ Product of zeroes = 1/4

p(x) = 0

⇒ 6x² – 3 – 7x = 0

⇒ 6x² – 7x – 3 = 0

⇒ 6x² – 9x + 2x – 3 = 0

⇒ 3x(2x – 3) + 1(2x – 3) = 0

⇒ (2x – 3) (3x + 1) = 0

⇒ and

Hence, and are zeroes of the polynomial 6x² – 3 – 7x.

Now,

⇒ Product of zeroes = -1/2

p(u) = 0

⇒ 4u² + 8u = 0

⇒ 4u(u + 2) = 0

⇒ u = 0 and u = -2

Hence, -2 and 0 are zeroes of the polynomial 4u² + 8u.

Now,

Sum of zeroes = -2 + 0

⇒ Sum of zeroes = -2

Product of zeroes = -2 × 0

⇒ Product of zeroes = 0

p(t) = 0

⇒ t² – 15 = 0

⇒ (t – √15) (t + √15) = 0

⇒ t = -√15 and t = √15

Hence, -√15 and √15 are zeroes of the polynomial t² – 15.

Now,

Sum of zeroes = -√15 + √15

⇒ Sum of zeroes = -√15 + √15

⇒ Sum of zeroes = 0

Product of zeroes = -√15 × √15

⇒ Product of zeroes = -15

p(x) = 0

⇒ 3x² – x – 4 = 0

⇒ 3x² + 3x – 4x – 4 = 0

⇒ 3x(x + 1) –4(x + 1) = 0

⇒ (x + 1)(3x – 4)= 0

Hence, -1 and are zeroes of the polynomial 3x² – x – 4.

Now,

Given: α + β = 1/4

αβ = -1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

…(2)

And,

If a = 4

⇒ c = -4 …(3)

From (2) and (3),

a = 4, b = -1 and c = -4

Hence, the polynomial is 4x² – x – 4

Given: α + β = √2

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 3

⇒ b = -3√2 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 3, b = -3 √2 and c = 1

Hence, the polynomial is 3x² – 3√2x + 1

Given: α + β = 0

αβ = √5

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

…(2)

And,

If a = 1

⇒c = √5 …(3)

From (2) and (3),

a = 1, b = 0 and c = √5

Hence, the polynomial is x² +√5

Given: α + β = 1

αβ = 1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 1

⇒ b = -1 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 1, b = -1 and c = 1

Hence, the polynomial is x² – x + 1

Given: α + β = -1/4

αβ = 1/4

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β .

Now we know that,

And,

If a = 4

⇒ b = 1 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 4, b = 1 and c = 1

Hence, the polynomial is 4x² + x + 1

Given: α + β = 4

αβ = 1

Let the quadratic polynomial be ax² + bx + c …(1)

Where, a≠0

And zeroes of the polynomial are α and β.

Now we know that,

And,

If a = 1

⇒ b = -4 …(2)

⇒c = 1 …(3)

From (2) and (3),

a = 1, b = -4 and c = 1

Hence, the polynomial is x² - 4x + 1

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = 2

β = -1

If a = 1,

⇒ b = -1

⇒ c = -2

Hence, the polynomial is x² – x – 2

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = √3

β = -√3

If a = 1,

⇒ b = 0

⇒ c = -3

Hence, the polynomial is x² – 3

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

α = 1/4

β = -1

If a = 4,

⇒ b = 3

⇒ c = -1

Hence, the polynomial is 4x² + 3x – 1.

Let the quadratic polynomial be ax² + bx + c

And, its zeroes be α and β

If a = 4,

⇒ b = -8

⇒ c = 3

Hence, the polynomial is 4x² – 8x + 3.

P(x) = x³ + 3x² – x – 3

For x = 1,

⇒ P(1) = 1³ + 3(1)² – 1 – 3

⇒ P(1) = 1+ 3 – 1 – 3

⇒ P(1) = 0

For x = -1,

⇒ P(-1) = (-1)³ + 3(-1)² – (-1) – 3

⇒ P(-1) = -1+ 3 + 1 – 3

⇒ P(-1) = 0

For x = -3,

⇒ P(-3) = (-3)³ + 3(-3)² – (-3) – 3

⇒ P(-3) = -27+ 27 + 3 – 3

⇒ P(-3) = 0

Now,

Sum of zeroes = 1 + (-1) + (-3) = -3

Hence,

And,

Product of zeroes = 1 × (-1) × (-3) = -3

p(x) = x³ – 3x² + 5x – 3

_{g(x) = x}² – 2

On dividing them,

The quotient is x – 3.

And,

The remainder is 7x – 9.

p(x) = x⁴ – 3x² + 4x + 5

g(x) = x² + 1 – x

On dividing them,

The quotient is x² + x - 3

And,

The remainder is 8

p(x) = x⁴ – 5x + 6

_{g(x) = 2 – x}²

On dividing them,

The quotient is –x² – 2

And,

The remainder is -5x + 10

p(x) = 2t⁴ + 3t³ – 2t² - 9t – 12

_{g(x) = t}² – 3

On dividing them,

The remainder is 0

Hence, yes t² – 3 is a factor of 2t⁴ + 3t³ – 2t² - 9t – 12

p(x) = 3x⁴ + 5x³ – 7x² + 2x + 2

_{g(x) = x}² + 3x + 1

On dividing them,

The remainder is 0

Hence, yes x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

p(x) = x⁵ – 4x³ + x² + 3x + 1

_{g(x) = x}³ – 3x + 1

On dividing them,

The remainder is 2 ≠ 0

_{Hence, no x}³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1

Two zeroes are and

is a factor

is a factor

is a factor

Now,

Therefore, 3x² + 6x +3 is also a factor

Dividing 3x² + 6x +3 by 3,

We get,

x² + 2x +1

Factorising x² + 2x +1,

x² + 2x +1 = 0

⇒ x² + x + x +1 = 0

⇒ x(x+ 1) + 1(x +1) = 0

⇒ (x+ 1)(x +1) = 0

⇒ x= -1, -1

Therefore, the other zeroes are -1 and -1.

Dividend = x³ – 3x² + x + 2

Quotient = x – 2

Remainder = -2x + 4

Dividend = Divisor × Quotient + Remainder

⇒ x³ – 3x² + x + 2 = Divisor ×(x – 2) + (-2x + 4)

⇒ x³ – 3x² + x + 2 – (-2x + 4) = Divisor ×(x – 2)

⇒ x³ – 3x² + x + 2 + 2x – 4 = Divisor ×(x – 2)

⇒ x³ – 3x² + 3x – 2 = Divisor ×(x – 2)

Therefore, g(x) = x² – x + 1

Let g(x) = 2

And p(x) = 2x² – 2x + 14

Then, dividing p(x) by g(x) gives.

q(x) = x² – x + 7

and,

r(x) = 0

Deg(p(x)) = Deg(q(x)) = 2

Hence,

g(x) = 2

p(x) = 2x² – 2x + 14

q(x) = x² – x + 7

r(x) = 0

Let g(x) = x + 1

And p(x) = x² + 3

Then, dividing p(x) by g(x) gives

q(x) = x

and,

r(x) = -x + 3

Deg(r(x)) = Deg(q(x)) = 1

Hence,

g(x) = x + 1

p(x) = x² + 3

q(x) = x

r(x) = -x + 3

Let g(x) = 3

And p(x) = 3x + 3

Then, dividing p(x) by g(x) gives

q(x) = x + 1

and,

r(x) = 0

Deg(r(x)) = 0

Hence,

g(x) = 3

p(x) = 3x + 3

q(x) = x + 1

r(x) = 0