Polynomials

(i) In general, we know that for a linear polynomial ax + b, a≠0, the graph of y = ax + b is a straight line which intersects the x – axis at exactly one point.

And here, we can see that the graph of y = p(x) is a straight line and intersects the x – axis at exactly one point. Therefore, the given graph is of a Linear Polynomial.

(ii) Here, the graph of y = p(x) is a straight line and parallel to the x – axis . Therefore, the given graph is of a Linear Polynomial.

(iii) For any quadratic polynomial ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = a x² + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0 or a < 0. (These curves are called parabolas.)

Here, we can see that the shape of the graph is a parabola. Therefore, the given graph is of a Quadratic Polynomial.

(iv) For any quadratic polynomial ax² + bx + c, a ≠ 0, the graph of the corresponding equation y = ax² + bx + c has one of the two shapes either open upwards like or open downwards like depending on whether a > 0or a < 0. (These curves are called parabolas.)

Here, we can see that the shape of the graph is parabola. Therefore, the given graph is of a Quadratic Polynomial.

(v) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial nor a quadratic polynomial.

(vi) The given graph have a straight line but it doesn’t intersect at x – axis and the shape of the graph is also not a parabola. So, it is not a graph of a quadratic polynomial. Therefore, it is not a graph of linear polynomial or quadratic polynomial.

(vii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.

(viii) The shape of the graph is neither a straight line nor a parabola. So, the graph is not of a linear polynomial or a quadratic polynomial.

(i) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(ii) Here, the graph of y = p(x) intersects the x – axis at three points. So, the number of zeroes is 3.

(iii) Here, the graph of y = p(x) intersects the x – axis at one point only. So, the number of zeroes is 1.

(iv) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(vi) Here, the graph of y = p(x) intersects the x – axis at exactly one point. So, the number of zeroes is 1.

(i) Here, the graph of y = p(x) intersect the x – axis at zero points. So, the number of zeroes is 0.

(ii) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(iii) Here, the graph of y = p(x) intersects the x – axis at four points. So, the number of zeroes is 4.

(iv) Here, the graph of y = p(x) does not intersects the x – axis. So, the number of zeroes is 0.

(v) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

(vi) Here, the graph of y = p(x) intersects the x – axis at two points. So, the number of zeroes is 2.

Let f (x) = x² – 3

Now, if we recall the identity

(a² – b²) = (a – b)(a + b)

Using this identity, we can write

x² – 3 = (x – √3) (x + √3)

So, the value of x² – 3 is zero when x = √3 or x = – √3

Therefore, the zeroes of x² – 3 are √3 and – √3.

Verification

Now,

Sum of zeroes = α + β = √3 + ( – √3) = 0 or

Product of zeroes = αβ = (√3)( – √3) = – 3 or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 2x² – 8x + 6

By splitting the middle term, we get

f(x) = 2 x² – (2 + 6)x + 6 [∵ – 8 = – (2 + 6) and 2×6 = 12]

= 2 x² – 2x – 6x + 6

= 2x(x – 1) – 6(x – 1)

= (2x – 6) (x – 1)

On putting f(x) = 0, we get

(2x – 6) (x – 1) = 0

⇒2x – 6 = 0 or x – 1 = 0

⇒x = 3 or x = 1

Thus, the zeroes of the given polynomial 2 x² – 8x + 6 are 1 and 3

Verification

Sum of zeroes = α + β = 3 + 1 = 4 or

Product of zeroes = αβ = (3)(1) = 3 or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = x² – 2x – 8

By splitting the middle term, we get

f(x) = x² – 4x + 2x – 8 [∵ – 2 = 2 – 4 and 2×4 = 8]

= x(x – 4) + 2(x – 4)

= (x + 2) (x – 4)

On putting f(x) = 0, we get

(x + 2) (x – 4) = 0

⇒ x + 2 = 0 or x – 4 = 0

⇒x = – 2 or x = 4

Thus, the zeroes of the given polynomial x² – 2x – 8 are – 2 and 4

Verification

Sum of zeroes = α + β = – 2 + 4 = 2 or

Product of zeroes = αβ = ( – 2)(4) = – 8 or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 3x² + 5x – 2

By splitting the middle term, we get

f(x) = 3x² + (6 – 1)x – 2 [∵ 5 = 6 – 1 and 2×3 = 6]

= 3x² + 6x – x – 2

= 3x(x + 2) – 1(x + 2)

= (3x – 1) (x + 2)

On putting f(x) = 0 , we get

(3x – 1) (x + 2) = 0

⇒ 3x – 1 = 0 or x + 2 = 0

or x = – 2

Thus, the zeroes of the given polynomial 3x² + 5x – 2 are – 2 and

Verification

Sum of zeroes or

Product of zeroes or

=

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 3x² – x – 4

By splitting the middle term, we get

f(x) = 3x² – (4 – 3)x – 4 [∵ – 1 = 3 – 4 and 4×3 = 12]

= 3x² + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (3x – 4) (x + 1)

On putting f(x) = 0, we get

(3x – 4) (x + 1) = 0

⇒ 3x – 4 = 0 or x + 1 = 0

or x = – 1

Thus, the zeroes of the given polynomial 3x² – x – 4 are – 1 and

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = x² + 7x + 10

By splitting the middle term, we get

f(x) = x² + 5x + 2x + 10 [∵ 7 = 2 + 5 and 2×5 = 10]

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

= (x + 2) (x + 5)

On putting f(x) = 0 , we get

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

⇒ x + 2 = 0 or x + 5 = 0

⇒x = – 2 or x = – 5

Thus, the zeroes of the given polynomial x² + 7x + 10 are – 2 and – 5

Verification

Sum of zeroes = α + β = – 2 + ( – 5) = – 7 or

Product of zeroes = αβ = ( – 2)( – 5) = 10 or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = t² – 15

Now, if we recall the identity

(a² – b²) = (a – b)(a + b)

Using this identity, we can write

t² – 15 = (t – √15) (x + √15)

So, the value of t²– 15 is zero when t = √15 or t = – √15

Therefore, the zeroes of t²– 15 are √15 and – √15.

Verification

Now,

Sum of zeroes = α + β = √15 + ( – √15) = 0 or

Product of zeroes = αβ = (√15)( – √15) = – 15 or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 4 s² – 4s + 1

By splitting the middle term, we get

f(x) = 4 s² – (2 – 2)s + 1 [∵ – 4 = – (2 + 2) and 2×2 = 4]

= 4 s² – 2s – 2s + 1

= 2s(2s – 1) – 1(2s – 1)

= (2s – 1) (2s – 1)

On putting f(x) = 0 , we get

(2s – 1) (2s – 1) = 0

⇒ 2s – 1 = 0 or 2s – 1 = 0

or

Thus, the zeroes of the given polynomial 4 s² – 4s + 1 are and

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 8x² – 22x – 21

By splitting the middle term, we get

f(x) = 8x² – 28x + 6x – 21

= 4x(2x – 7) + 3(2x – 7)

= (4x + 3) (2x – 7)

On putting f(x) = 0 , we get

(4x + 3) (2x – 7) = 0

⇒ 4x + 3 = 0 or 2x – 7 = 0

or

Thus, the zeroes of the given polynomial 8x² – 22x – 21 are and

Verification

Sum of zeroes or

The product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 2 x² – 7x

In this the constant term is zero.

f(x) = 2 x² – 7x

= x(2x – 7)

On putting f(x) = 0 , we get

x(2x – 7) = 0

⇒ 2x – 7 = 0 or x = 0

or x = 0

Thus, the zeroes of the given polynomial 2x² – 7x are 0 and

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 10x² + 3x – 1

By splitting the middle term, we get

f(x) = 10x² – 2x + 5x – 1

= 2x(5x – 1) + 1(5x – 1)

= (2x + 1) (5x – 1)

On putting f(x) = 0, we get

(2x + 1) (5x – 1) = 0

⇒ 2x + 1 = 0 or 5x – 1 = 0

= or =

Thus, the zeroes of the given polynomial10x² + 3x – 1 are and

Verification

Sum of zeroes

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = px² + (2q – p²)x – 2pq

f(x) = px² + 2qx – p² x – 2pq

= x(px + 2q) – p(px + 2q)

= (x – p) (px + 2q)

On putting f(x) = 0, we get

(x – p) (px + 2q) = 0

⇒ x – p = 0 or px + 2q = 0

⇒x = p or =

Thus, the zeroes of the given polynomial px² + (2q – p²)x – 2pq are p and

Verification

Sum of zeroes

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = x² – (2a + b)x + 2ab

f(x) = x² – 2ax – bx + 2ab

= x(x – 2a) – b(x – 2a)

= (x – 2a) (x – b)

On putting f(x) = 0 , we get

(x – 2a) (x – b) = 0

⇒ x – 2a = 0 or x – b = 0

⇒x = 2a or x = b

Thus, the zeroes of the given polynomial x² – (2a + b)x + 2ab are 2a and b

Verification

Sum of zeroes = α + β = 2a + b or

Product of zeroes = αβ = 2a × b = 2ab or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = r²s²x² + 6rstx + 9t²

Now, if we recall the identity

(a + b)² = a² + b² + 2ab

Using this identity, we can write

r²s²x² + 6rstx + 9t² = (rsx + 3t)²

On putting f(x) = 0 , we get

(rsx + 3t)² = 0

Thus, the zeroes of the given polynomial r²s²x² + 6rstx + 9t² are and

Verification

Sum of zeroes or

or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 5x² – 8x – 4

By splitting the middle term, we get

f(x) = 5x² – 10x + 2x – 4

= 5x(x – 2) + 2(x – 2)

= (5x + 2) (x – 2)

On putting f(x) = 0 we get

(5x + 2) (x – 2) = 0

⇒ 5x + 2 = 0 or x – 2 = 0

or x = 2

Thus, the zeroes of the given polynomial5x² – 8x – 4 are and 2

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = 4 x² – 4x – 3

By splitting the middle term, we get

f(x) = 4 x² – 6x + 2x – 3

= 2x(2x – 3) + 1(2x – 3)

= (2x + 1) (2x – 3)

On putting f(x) = 0, we get

(2x + 1) (2x – 3) = 0

⇒ 2x + 1 = 0 or 2x – 3 = 0

or

Thus, the zeroes of the given polynomial4x² – 4x – 3 are and

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let f(x) = √3 x² – 8x + 4√3

By splitting the middle term, we get

(x) = √3 x² – 6x – 2x + 4√3

= √3 x(x – 2√3) – 2(x – 2√3)

= (√3 x – 2) (x – 2√3)

On putting f (x) = 0, we get

(√3 x – 2) (x – 2√3) = 0

⇒ √3 x – 2 = 0 or x – 2√(3 = 0)

or

Thus, the zeroes of the given polynomial√3 x² – 8x + 4√3 are and 2

Verification

Sum of zeroes or

Product of zeroes or

So, the relationship between the zeroes and the coefficients is verified.

Let the quadratic polynomial be 2 x² + 3x – 6, and its zeroes are α and β.

We have

Here, a = 2 , b = 3 and c = – 6

….(1)

….(2)

(i) α² + β²

We have to find the value of α² + β²

Now, if we recall the identity

(a + b)² = a² + b² + 2ab

Using the identity, we get

(α + β)² = α² + β² + 2αβ

{from eqⁿ (1) & (2)}

(ii) α² + β² + αβ

{ from part (i)}

and, we have αβ = – 3

So,

=

=

(iii) α²β + α β²

Firstly, take common, we get

αβ(α + β)

and we already know the value of and .

So, α²β + α β² = αβ(α + β)

{from eqⁿ (1) and (2)}

(iv)

Let’s take the LCM first then we get,

(v)

Let’s take the LCM first then we get,

{from part(i) and eqⁿ (2)}

(vi)

Now, recall the identity

(a – b)² = a² + b² – 2ab

Using the identity , we get

(α – β)² = α² + β² – 2 αβ

{from part(i) and eqⁿ (2)}

= =

(vii)

Now, recall the identity

(a + b)³ = a³ + b³ + 3a² b + 3ab²

Using the identity, we get

⇒(α + β)³ = α³ + β³ + 3α² β + 3αβ²

(viii)

Let’s take the LCM first then we get,

{from part(vii) and eqⁿ (2)}

Let the quadratic poynomial be ax² + bx + c , and its zeroes be α and β.

We have

α + β = and α β =

(i)

We have to find the value of

Now, if we recall the identity

(a + b)² = a² + b² + 2ab

Using the identity, we get

{from eqⁿ (1) & (2)}

(ii)

Let’s take the LCM first then we get,

{}

(iii)

Now, recall the identity

(a + b)³ = a³ + b³ + 3a² b + 3ab²

Using the identity, we get

⇒(α + β)³ = α³ + β³ + 3α² β + 3αβ²

The given quadratic polynomial is x² + kx = 12 and – = 1

If we rearrange the polynomial then we get

p(x) = x² + kx – 12

We have,

and

So,

…(1)

…(2)

Now, if we recall the identities

(a + b)² = a² + b² + 2ab

Using the identity, we get

(α + β)² = α² + β² + 2 αβ

( – k)² = α² + β² + 2( – 12)

⇒ α² + β² = k² + 24 …(3)

Again, using the identity

(a – b)² = a² + b² – 2ab

Using the identity, we get
(α – β)² = α² + β² – 2 αβ

(1)² = α² + β² – 2( – 12) {∵ (α – β) = 1}

⇒ α² + β² = 1 – 24

⇒ α² + β² = – 23 …(4)

From eqⁿ (3) and (4), we get

k² + 24 = – 23

⇒ k² = – 23 – 24

⇒ k² = – 47

Now the square can never be negative, so the value of k is imaginary.

Given : p(x) = x² – 8x + k

α² + β² = 40

We have,

and

So,

…(1)

…(2)

Now, if we recall the identities

(a + b)² = a² + b² + 2ab

Using the identity, we get

(α + β)² = α² + β² + 2 αβ

(8)² = + 2(k)

⇒2k = 64 – 40

= = 12

Let one zero of the given polynomial is

According to the given condition,

The other zero of the polynomial is

We have,

Product of zeroes,

⇒ α² – 6α + 9 = 0

⇒ α² – 3α – 3α + 9 = 0

⇒ α(α – 3) – 3(α – 3) = 0

⇒(α – 3)(α – 3) = 0

⇒(α – 3) = 0 & (α – 3) = 0

⇒ α = 3, 3

Given Product of zeroes, α β = 4

p(x) = α x² – 6x – 6

to find: value of α

We know,

Product of zeroes ,

Given x + a is a factor of

So,g(x) = x + a

x + a = 0

⇒x = – a

Putting the value x = – a in the given polynomial, we get

2( – a)² + 2a( – a) + 5( – a) + 10 = 0

2a² – 2a² – 5a + 10 = 0

– 5a + 10 = 0

a = 2

Given: Sum of zeroes = α + β = 1

Product of zeroes = αβ = 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (1)x + 1

= x² – x + 1

Given: Sum of zeroes = α + β = 0

Product of zeroes = αβ = 3

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (0)x + 3

= x² + 3

Given: Sum of zeroes = α + β = 1/4

Product of zeroes = αβ = – 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

=

=

=

We can consider 4x² – x – 4 as required quadratic polynomial because it will also satisfy the given conditions.

Given: Sum of zeroes = = 4

Product of zeroes = = 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (4)x + 1

= x² – 4x + 1

Given: Sum of zeroes

Product of zeroes = αβ = – 1

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

=

=

=

We can consider 3x² – 10x – 3 as required quadratic polynomial because it will also satisfy the given conditions.

Given: Sum of zeroes

Product of zeroes

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

=

=

=

We can consider 2x² + x – 1 as required quadratic polynomial because it will also satisfy the given conditions.

Let the zeroes of the quadratic polynomial be

α = 3 , β = – 3

Then, α + β = 3 + ( – 3) = 0

αβ = 3 × ( – 3) = – 9

Sum of zeroes = α + β = 0

Product of zeroes = αβ = – 9

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (0)x + ( – 9)

= x² – 9

Let the zeroes of the quadratic polynomial be

Then,

Sum of zeroes = α + β = 2

Product of zeroes = =

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

=

=

=

We can consider 4x² – 8x – 1 as required quadratic polynomial because it will also satisfy the given conditions.

Let the zeroes of the quadratic polynomial be

α = 3 + √7, β = 3 – √7

Then, α + β = 3 + √7 + 3 – √7 = 6

αβ = (3 + √(7)) × (3 – √7) = 9 – 7 = 2

Sum of zeroes = α + β = 6

Product of zeroes = αβ = 2

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (6)x + 2

= x² – 6x + 2

Let the zeroes of the quadratic polynomial be

α = 1 + 2√3, β = 1 – 2√3

Then, α + β = 1 + 2√3 + 1 – 2√3 = 2

αβ = (1 + 2√3) × (1 – 2√3) = 1 – 12 = – 11

Sum of zeroes = α + β = 2

Product of zeroes = αβ = – 11

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (2)x + ( – 11)

= x² – 2x – 11

Let the zeroes of the quadratic polynomial be

Then,

Sum of zeroes

Product of zeroes

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

=

=

=

We can consider 9x² – 12x + 1 as required quadratic polynomial because it will also satisfy the given conditions.

Let the zeroes of the quadratic polynomial be

α = √2, β = 2√2

Then, α + β = √2 + 2√2 = √2 (1 + 2) = 3√2

αβ = √2× 2√2 = 4

Sum of zeroes = α + β = 3√2

Product of zeroes = αβ = 4

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (3√2)x + 4

= x² – 3√2 x + 4

et the zeroes of the polynomial x² – x – 1 be α and β

We have,

and

So,

Now, according to the given condition,

α² β² = ( – 1)² = 1

& (α + β)² = α² + β² + 2 α β

α² + β² = (α + β)² – 2αβ

α² + β² = (1)² – 2( – 1)

α² + β² = 3

So, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (3)x + 1

= x² – 3x + 1

Given : p(x) = x² + 10x + 30

So, Sum of zeroes …(1)

Product of zeroes …(2)

Now,

Let the zeroes of the quadratic polynomial be

α^' = α + 2β , β' = 2α + β

Then, α’ + β’ = α + 2β + 2α + β = 3α + 3β = 3(α + β)

α'β’ = (α + 2β) ×(2α + β) = 2α² + 2β² + 5αβ

Sum of zeroes = 3(α + β)

Product of zeroes = 2α² + 2β² + 5αβ

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (3(α + β))x + 2α² + 2β² + 5αβ

= x² – 3( – 10)x + 2 (α² + β²) + 5(30) {from eqⁿ (1) & (2)}

= x² + 30x + 2(α² + β² + 2αβ – 2αβ) + 150

= x² + 30x + 2 (α + β)² – 4αβ + 150

= x² + 30x + 2( – 10)² – 4(30) + 150

= x² + 30x + 200 – 120 + 150

= x² + 30x + 230

So, the required quadratic polynomial is x² + 30x + 230

Given : p(x) = x² + 4x + 3

So, Sum of zeroes …(1)

Product of zeroes …(2)

Now,

Let the zeroes of the quadratic polynomial be

Then,

Sum of zeroes

Product of zeroes

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= +

= {from eqⁿ (1) & (2)}

= +

=

So, the required quadratic polynomial is 3x² – 16x + 16

Let the zeroes of the quadratic polynomial be

α = 1 , β = – 3

Then, α + β = 1 + ( – 3) = – 2

αβ = 1 × ( – 3) = – 3

Sum of zeroes = α + β = – 2

Product of zeroes = αβ = – 3

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – ( – 2)x + ( – 3)

= x² + 2x – 3

Verification

Sum of zeroes = α + β = 1 + ( – 3) = – 2 or

Product of zeroes = αβ = (1)( – 3) = – 3 or

So, the relationship between the zeroes and the coefficients is verified.

Given: Sum of zeroes = α + β = 8

Product of zeroes = αβ = 12

Then, the quadratic polynomial

= x² – (sum of zeroes)x + product of zeroes

= x² – (8)x + 12

= x² – 8x + 12

Now, we have

and

So,

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = 2x² – 4x + 11

Remainder = – 21

No, 2x² – 4x + 11 is not a factor of 2x³ + 3x + 1 because remainder ≠ 0

Dividend = 3x³ + x² + 2x + 5

Divisor = x² + 2x + 1

Now, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = 3x – 5

Remainder = 9x + 10

No, x² + 2x + 1 is not a factor of 3x³ + x² + 2x + 5 because remainder ≠ 0

p(x) = x³ – 3x² + 4x + 2,g(x) = x – 1

Dividend = x³ – 3x² + 4x + 2

Divisor = x – 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = x² – 2x + 2

Remainder = 4

p(x) = 4x³ – 3x² + 2x + 3,g(x) = x + 4

Dividend = 4x³ – 3x² + 2x + 3

Divisor = x + 4

Here, dividend and divisor both are in the standard form.

Now, on dividing by we get the following division process

Quotient = 4x² – 13x + 54

Remainder = – 213

p(x) = 2x⁴ + 3x³ + 4x² + 19x + 45, g(x) = x – 2

Dividend = 2x⁴ + 3x³ + 4x² + 19x + 45

Divisor = x – 2

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = 2x³+7x²+18x+55

r(x) = 155

p(x) = x⁴ + 2x³ – 3x² + x – 1, g(x) = x – 2

Dividend = x⁴ + 2x³ – 3x² + x – 1

Divisor = x – 2

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³+4x²+5x+11

r(x) = 21

p(x) = x³ – 3x²– x + 3, g(x) = x² – 4x + 3

Dividend = x³ – 3x²– x + 3

Divisor = x² – 4x + 3

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process we get,

q(x)= x + 1

r(x) = 0

p(x) = x⁶ + x⁴ + x³ + x² + 2x + 2, g(x) = x³ + 1

Dividend = x⁶ + x⁴ + x³ + x² + 2x + 2

Divisor = x³ + 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³ + x

r(x) = x² + x + 2

p(x) = x⁶ + 3x² + 10 , g(x) = x³ + 1

Dividend = x⁶ + 3x² + 10

Divisor = x³ + 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process we get,

q(x) = x³ – 1

r(x) = 3x² + 11

p(x) = x⁴ + 1, g(x) = x + 1

Dividend = x⁴ + 1,

Divisor = x + 1

q(x) = x³ – x² + x – 1

r(x) = 0

On dividing x³ – 6x² + 11x + k by x – 1 we get,

Since x – 1 is a factor of x³ – 6x² + 11x + k,

This means x – 1 divides the given polynomial completely.

→ 6 x + k = 0

→ k = – 6x

On dividing 4x⁴ – 3x² + 3x + c by 2x + 1 we get,

Since 2x + 1 is a factor of 4x⁴ – 3x² + 3x + c,

This means 2x + 1 divides the given polynomial completely,

→ 4x + c = o

→ c = – 4x

p(x) = 2x² + 3x + 1, g(x) = x + 2

Dividend = 2x² + 3x + 1,

Divisor = x + 2

Apply the division algorithm we get,

q(x) = 2x – 1

r(x) = 3

p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2

Dividend = x³ – 3x² + 5x – 3,

Divisor = x² – 2

On applying division algorithm, we get,

q(x) = x – 3

r(x) = 7x – 9

p(x) = x⁴ – 1,g(x) = x + 1

Dividend = x⁴ – 1

Divisor = x + 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = x³ – x² + x – 1

Remainder = 0

p(x) = x³ – 3x² + 4x + 2,g(x) = x – 1

Dividend = x³ – 3x² + 4x + 2

Divisor = x – 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = x – 5

Remainder = 13x + 7

p(x) = x³ – 6x² + 11x – 6,g(x) = x² – 5x + 6

Dividend = x³ – 6x² + 11x – 6

Divisor = x² – 5x + 6

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = x – 1

Remainder = 6x

p(x) = 6x³ + 13x² + x – 2,g(x) = 2x + 1

Dividend = 6x³ + 13x² + x – 2

Divisor = 2x + 1

Here, dividend and divisor both are in the standard form.

Now, on dividing p(x) by g(x) we get the following division process

Quotient = 3x² + 5x – 2

Remainder = 0

Let us divide x³ + 3x² – 12x + 4 by x – 2

The division process is

Here, the remainder is 0, therefore x – 2 is a factor of x³ + 3x² – 12x + 4

Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1

The division process is

Here, the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2

Let us divide 2x⁴ – 11x³ + 29x² – 30x + 29 by x² – 3x + 4

The division process is

Here, the remainder is 58x – 115 , therefore x² – 3x + 4 is not a factor of 2x⁴ – 11x³ + 29x² – 30x + 29

Let us divide x³ – 3x² – x + 3 by x² – 4x + 3

The division process is

Here, the remainder is 0, therefore x² – 4x + 3 is a factor of x³ – 3x² – x + 3

Let us divide t³ + t² – 2t + 1 by t – 1

The division process is

Here, the remainder is 1, therefore t – 1 is not a factor of t³ + t² – 2t + 1

Let us divide t² + 11t – 6 by t² – 5t + 6

The division process is

Here, the remainder is 16t – 12,

Therefore, t² – 5t + 6 is not a factor of t² + 11t – 6

(i)Let p(x) = 12x² + 8x + 25, g(x) = 4,

q(x) = 3x² + 2x + 6 , r(x) = 0

Here, degree p(x) = degree q(x) = 2

Now, g(x).q(x) + r(x) = (3x² + 2x + 6)×4 + 1

= 12x² + 8x + 24 + 1

= 12x² + 8x + 25

(ii) Let p(x) = t³ + t² – 2t, g(x) = t² + 2t,

q(x) = t – 1 , r(x) = 0

Here, degree q(x) = 1

Now, g(x).q(x) + r(x) = (t² + 2t)×(t – 1) + 0

= t³ – t² + 2t² – 2t

= t³ + t² – 2t

(iii) Let p(x) = x³ + x² + x + 1 , g(x) = x² – 1,

q(x) = x + 1 , r(x) = 2x + 2

Here, degree q(x) = degree r(x) + 1 = 1

Now, g(x).q(x) + r(x) = (x² – 1)×(x + 1) + 2x + 2

= x³ + x² – x – 1 + 2x + 2

= x³ + x² + x + 1

Given zeroes is 3

So, (x – 3) is the factor of x³ – 6x² + 11x – 6

Let us divide x³ – 6x² + 11x – 6 by x – 3

The division process is

Here, quotient = x² – 3x + 2

= x² – 2x – x + 2

= x(x – 2) – 1(x – 2)

= (x – 1)(x – 2)

So, the zeroes are 1 and 2

Hence, all the zeroes of the given polynomial are 1, 2 and 3.

Given zeroes are 1 and 2

So, (x – 1)and (x – 2) are the factors of x⁴ – 8x³ + 23x² – 28x + 12

⟹ (x – 1)(x – 2) = x² – 3x + 2 is a factor of given polynomial.

Consequently, x² – 3x + 2 is also a factor of the given polynomial.

Now, let us divide x⁴ – 8x³ + 23x² – 28x + 12 by x² – 3x + 2

The division process is

Here, quotient = x² – 5x + 6

= x² – 2x – 3x + 6

= x(x – 2) – 3(x – 2)

= (x – 3)(x – 2)

So, the zeroes are 3 and 2

Hence, all the zeroes of the given polynomial are 1, 2 ,2 and 3.

Given zeroes is – 2

So, (x + 2) is the factor of x³ + 2x² – x – 2

Let us divide x³ + 2x² – x – 2 by x + 2

The division process is

Here, quotient = x² – 1

= (x – 1)(x + 1)

So, the zeroes are – 1 and 1

Hence, all the zeroes of the given polynomial are – 1, – 2 and 1.

Given zeroes is – 3

So, (x + 3) is the factor of x³ + 5x² + 7x + 3

Let us divide x³ + 5x² + 7x + 3 by x + 3

The division process is

Here, quotient = x² + 2x + 1

= (x + 1)²

So, the zeroes are – 1 and – 1

Hence, all the zeroes of the given polynomial are – 1, – 1 and – 3.

x⁴ – 6x³ – 26x² + 138x – 35;2±√3

Given zeroes are 2 + √3 and 2 – √3

So, (x – 2 – √3)and (x – 2 + √3) are the factors of x⁴ – 6x³ – 26x² + 138x – 35

⟹ (x – 2 – √3)(x – 2 + √3)

= x² – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3

= x² – 4x + 1 is a factor of given polynomial.

Consequently, x² – 4x + 1 is also a factor of the given polynomial.

Now, let us divide x⁴ – 6x³ – 26x² + 138x – 35 by x² – 4x + 1

The division process is

Here, quotient = x² – 2x – 35

= x² – 7x + 5x – 35

= x(x – 7) + 5(x – 7)

= (x + 5)(x – 7)

So, the zeroes are – 5 and 7

Hence, all the zeroes of the given polynomial are – 5, 7, 2 + √3 and

x⁴ + x³ – 34x² – 4x + 120;2, – 2.

Given zeroes are – 2 and 2

So, (x + 2)and (x – 2) are the factors of x⁴ + x³ – 34x² – 4x + 120

⟹ (x + 2)(x – 2) = x² – 4 is a factor of given polynomial.

Consequently, x² – 4 is also a factor of the given polynomial.

Now, let us divide x⁴ + x³ – 34x² – 4x + 120 by x² – 4

The division process is

Here, quotient = x² + x – 30

= x² + 6x – 5x – 30

= x(x + 6) – 5(x + 6)

= (x + 6)(x – 5)

So, the zeroes are – 6 and 5

Hence, all the zeroes of the given polynomial are – 2 , – 6, 2 and 5.

2x⁴ + 7x³ – 19x² – 14x + 30;√(2, – √2)

Given zeroes are √2 and – √2

So, (x – √2)and (x + √2) are the factors of 2x⁴ + 7x³ – 19x² – 14x + 30

⟹ (x – √2)(x + √2) = x² – 2 is a factor of given polynomial.

Consequently, x² – 2 is also a factor of the given polynomial.

Now, let us divide 2x⁴ + 7x³ – 19x² – 14x + 30 by x² – 2

The division process is

Here, quotient = 2x² + 7x – 15

= 2x² + 10x – 3x – 15

= 2x(x + 5) – 3(x + 5)

= (2x – 3)(x + 5)

So, the zeroes are – 5 and

Hence, all the zeroes of the given polynomial are – 5, – √2, √2 and

2x⁴ – 9x³ + 5x² + 3x – 1;2±√3

Given zeroes are 2 + √3 and 2 – √3

So, (x – 2 – √3)and (x – 2 + √3) are the factors of 2x⁴ – 9x³ + 5x² + 3x – 1

⟹ (x – 2 – √3)(x – 2 + √3)

= x² – 2x + √3 x – 2x + 4 – 2√3 – √3 x + 2√3 – 3

= x² – 4x + 1 is a factor of given polynomial.

Consequently, x² – 4x + 1 is also a factor of the given polynomial.

Now, let us divide 2x⁴ – 9x³ + 5x² + 3x – 1 by x² – 4x + 1

The division process is

Here, quotient = 2x² – x – 1

= 2x² – 2x + x – 1

= 2x(x – 1) + 1(x – 1)

= (2x + 1)(x – 1)

So, the zeroes are and 1

Hence, all the zeroes of the given polynomial are , 1, 2 + √3 and 2 – √3

2x³ – 4x – x² + 2;√2, – √2

Given zeroes are √2 and – √2

So, (x – √2) and (x + √2) are the factors of 2x³ – 4x – x² + 2

⟹ (x – √2)(x + √2) = x² – 2 is a factor of given polynomial.

Consequently, x² – 2 is also a factor of the given polynomial.

Now, let us divide 2x³ – 4x – x² + 2by x² – 2

The division process is

Here, quotient = 2x – 1

So, the zeroes is

Hence, all the zeroes of the given polynomial are , and

Let p(x) = 3 x³ – 5 x² – 11x – 3

Then, p( – 1) = 3( – 1)³ – 5( – 1)² – 11( – 1) – 3

= – 3 – 5 + 11 – 3

= 0

= 0

p(3) = 3(3)³ – 5(3)² – 11(3) – 3

= 81 – 45 – 33 – 3

= 0

Hence, we verified that 3, – 1 and are the zeroes of the given polynomial.

So, we take α = 3, β = – 1,

Verification

α + β + γ

αβ + βγ + γα

and αβγ

= 1

Thus, the relationship between the zeroes and the coefficients is verified.

Let p(x) = x³ – 4x² + 5x – 2

Then, p(2) = (2)³ – 4(2)² + 5(2) – 2

= 8 – 16 + 10 – 2

= 0

p(1) = (1)³ – 4(1)² + 5(1) – 2

= 1 – 4 + 5 – 2

= 0

Hence, 2, 1 and 1 are the zeroes of the given polynomial x³ – 4x² + 5x – 2.

Now, Let α = 2 , β = 1 and γ = 1

Then, α + β + γ = 2 + 1 + 1 = 4

αβ + βγ + γα = (2)(1) + (1)(1) + (1)(2)

= 2 + 1 + 2

= 5

and αβγ = 2 × 1 × 1

= 2

Thus, the relationship between the zeroes and the coefficients is verified.

Let p(x) = x³ – 6x² + 11x – 6

Then, p(1) = (1)³ – 6(1)² + 11(1) – 6

= 1 – 6 + 11 – 6

= 0

p(2) = (2)³ – 6(2)² + 11(2) – 6

= 8 – 24 + 22 – 6

= 0
p(3) = (3)³ – 6(3)² + 11(3) – 6

= 27 – 54 + 33 – 6

= 0

Hence, 1, 2 and 3 are the zeroes of the given polynomial x³ – 6x² + 11x – 6.

Now, Let α = 1 , β = 2 and γ = 3

Then, α + β + γ = 1 + 2 + 3 = 6

αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)

= 2 + 6 + 3

= 11

and αβγ = 1 × 2 × 3

= 6

Thus, the relationship between the zeroes and the coefficients is verified.

Let p(x) = x³ + 2x² – x – 2

Then, p( – 2) = ( – 2)³ + 2( – 2)² – ( – 2) – 2

= – 8 + 8 + 2 – 2

= 0
p(1) = (1)³ + 2(1)² – (1) – 2

= 1 + 2 – 1 – 2

= 0

Hence, – 2, – 2 and 1 are the zeroes of the given polynomial x³ + 2x² – x – 2.

Now, Let α = – 2 , β = – 2 and γ = 1

Then, α + β + γ = – 2 + ( – 2) + 1 = – 3

αβ + βγ + γα = ( – 2)( – 2) + ( – 2)(1) + (1)( – 2)

= 4 – 2 – 2

= 0

and αβγ = ( – 2) × ( – 2) × 1

= 4

Thus, the relationship between the zeroes and the coefficients is not verified.

Let p(x) = x³ + 5x² + 7x + 3.

Then, p( – 1) = ( – 1)³ + 5( – 1)² + 7( – 1) + 3

= – 1 + 5 – 7 + 3

= 0
p( – 3) = ( – 3)³ + 5( – 3)² + 7( – 3) + 3

= – 27 + 45 – 21 + 3

= 0

Hence, – 1, – 1 and – 3 are the zeroes of the given polynomial x³ + 5x² + 7x + 3.

Now, Let α = – 1 , β = – 1 and γ = – 3

Then, α + β + γ = – 1 + ( – 1) + ( – 3) = – 5

αβ + βγ + γα = ( – 1)( – 1) + ( – 1)( – 3) + ( – 3)( – 1)

= 1 + 3 + 3

= 7

and αβγ = ( – 1) × ( – 1) × ( – 3)

= – 3

Thus, the relationship between the zeroes and the coefficients is verified.

Let the zeroes of the cubic polynomial be

α = 1, β = 2 and γ = 3

Then, α + β + γ = 1 + 2 + 3 = 6

αβ + βγ + γα = (1)(2) + (2)(3) + (3)(1)

= 2 + 6 + 3

= 11

and αβγ = 1 × 2 × 3

= 6

Now, required cubic polynomial

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

= x³ – (6) x² + (11)x – 6

= x³ – 6 x² + 11x – 6

So, x³ – 6 x² + 11x – 6 is the required cubic polynomial which satisfy the given conditions.

Let the zeroes of the cubic polynomial be

α = – 3, β = – 2 and γ = 2

Then, α + β + γ = – 3 + ( – 2) + 2

= – 3 – 2 + 2

= – 3

αβ + βγ + γα = ( – 3)( – 2) + ( – 2)(2) + (2)( – 3)

= 6 – 4 – 6

= – 4

and αβγ = ( – 3) × ( – 2) × 2

= 6 × 2

= 12

Now, required cubic polynomial

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

= x³ – ( – 3) x² + ( – 4)x – 12

= x³ + 3 x² – 4x – 12

So, x³ + 3x^2 – 4x – 12 is the required cubic polynomial which satisfy the given conditions.

Let the zeroes be α, β and γ.

Then, we have

α + β + γ = 0

αβ + βγ + γα = – 7

and αβγ = – 6

Now, required cubic polynomial

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

= x³ – (0) x² + ( – 7)x – ( – 6)

= x³ – 7x + 6

So, x³ – 7x + 6 is the required cubic polynomial.

Let the zeroes be α, β and γ.

Then, we have

α + β + γ = 2

αβ + βγ + γα = – 7

and αβγ = – 14

Now, required cubic polynomial

Sox³ – 2x² – 7x + 14 is the required cubic polynomial.

Let the zeroes be α, β and γ.

Then, we have

Now, required cubic polynomial

So, 6x³ + 24x² + 3x – 2 is the required cubic polynomial which satisfy the given conditions.

Let the zeroes be α, β and γ.

Then, we have

Now, required cubic polynomial

So, 7x³ – 5x² + x – 1 is the required cubic polynomial which satisfy the given conditions.

Let the zeroes be α, β and γ.

Then, we have

Now, required cubic polynomial

So, 10x³ – 4x² + x – 5 is the required cubic polynomial which satisfy the given conditions.