Polynomials

(1) Here p (x) = x² — 5x + 6

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(2) Here p(x) = x² — x³ + x + 1

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

(3) Here p(x) = 5x² + 8x + 3

Degree of the polynomial is 2. Hence it is a quadratic polynomial in x.

(4) Here p(x) = x³

Degree of the polynomial is 3. Hence it is a cubic polynomial in x.

(1) Here p(x) = 3x —x⁴ + x² + 2x³ + 7

Highest power of x in the above expression is 4. Hence, the degree of the polynomial is 4.

(2) Here p(x) = x³ — 3x —x² + 6

Highest power of x in the above expression is 3. Hence, the degree of the polynomial is 3.

(3) Here p(x) = 3x — 9

Highest power of x in the above expression is 1. Hence, the degree of the polynomial is 1.

(4) Here p(x) = 2x² — x + 1

(5) Highest power of x in the above expression is 2. Hence, the degree of the polynomial is 2.

(1) p(x) = 10x³+ 7 x² — 3x + 5

Coefficient of the underlined term is 10.

(2) p(x) = 7 — 5x⁵+ 3x⁴ + x² — x

Coefficient of the underlined term is –5.

(3) p(x) = 25 — 125x

Coefficient of the underlined term is –125.

(4) p(x) = x³ — x² + x + 7

Coefficient of the underlined term x is –1.

(1) Given, p(x) = 2x³ + 3x² + 7x + 9

p(0) = 2(0)³ + 3(0)² + 7(0) + 9

p(0) = 0 + 0 + 9 = 9

p(1) = 2(1)³ + 3(1)² + 7(1) + 9

p(1) = 2 + 3 + 7 + 9 = 21

(2) Given, p(x) = 3x² + 10x + 7

p(–3) = 3(–3)² + 10(–3) + 7

p(–3) = 27 – 30 + 7 = 4

p(1) = 3(1)² + 10(1) + 7

p(1) = 3 + 10 + 7 = 20

(3) Given, p(x) = x² — 2x + 5

p(–1) = (–1)² – 2(–1) + 5

p(–1) = 1 + 2 + 5 = 8

p(5) = (5)² – 2(5) + 5

p(5) = 25 – 10 + 5 = 20

(4) Given, p(x) = 2x⁴ — 3x³ + 7x + 5

p(–2) = 2(–2)⁴ – 3(–2)³ + 7(–2) + 5

p(–2) = 32 + 24 – 14 + 5 = 47

p(2) = 2(2)⁴ – 3(2)³ + 7(2) + 5

p(2) = 32– 24 + 14 + 5 = 27

Given, p(x) = 3x³ + 2x² + 7 x + 8

For (x + 1) to be a factor of p(x), the condition: p(–1) = 0; should be satisfied.

p(–1) = 3(–1)³ + 2(–1)² + 7(–1) + 8

= 3(–1) + 2(1) + 7(–1) + 8

= –3 + 2 – 7 + 8

= 0

Thus, (x + 1) is a factor of the given polynomial.

⇒ The given statement is valid.

Given, p(x) = x³ + x² + x + 2

For (x + 2) to be a factor of p(x), the condition: p(–2) = 0; should be satisfied.

p(–2) = (–2)³ + (–2)² + (–2) + 2

= (–8) + (4) + (–2) + 2

= – 8 + 4 – 2 + 2

= –4 ≠ 0

Thus, (x + 2) is not a factor of the given polynomial.

⇒ The given statement is invalid.

Given, p(x) = x⁴ — 2x³ + 3x —2

For (x – 1) to be a factor of p(x), the condition: p(1) = 0; should be satisfied.

p(1) = (1)⁴ – 2(1)³ + 3(1) – 2

= 1 – 2 + 3 – 2

= 0

Thus, (x – 1) is a factor of the given polynomial.

⇒ The given statement is valid.

Given, p(x) = x² — 2x — 3

For (x – 3) to be a factor of p(x), the condition: p(3) = 0; should be satisfied.

p(3) = (3)² – 2(3) – 3

= 9 – 6 – 3

= 0

Thus, (x –3) is a factor of the given polynomial.

⇒ The given statement is valid.

p(x) = x³ – x² – x + 1

= x² (x – 1) – 1(x – 1)

= (x – 1)(x² – 1)

= (x – 1)(x – 1)(x + 1)

Using the identity: (a² – b²) = (a – b) (a + b)

= (x – 1)² (x + 1)

p(x) = 5x² + 11x + 6

= 5x² + 5x + 6x + 6

(Splitting 11x as 5x + 6x)

= 5x (x + 1) + 6(x + 1)

= (x + 1) (5x + 6)

p(x) = x³ – 3x² + 9x – 27

= x² (x – 3) + 9 (x – 3)

= (x – 3) (x² + 9)

p(x) = x³ + 2x² + 3x + 2

= x³ + x² + x² + x + 2x + 2 (2x² = x² + x² and 3x = x + 2x)

= x² (x + 1) + x (x + 1) + 2 (x + 1)

= (x + 1) (x² + x + 2)

Given, p(x) = x³ — 2x²

For (x – 2) to be a factor of p(x), the condition: p(2) = 0 ;should be satisfied.

p(2) = (2)³ – 2(2)²

= 8 – 2(4)

= 8 – 8

= 0

Hence proved that, x — 2 is a factor of p(x) = x³ — 2x²

Here, p(x) = x² — x

To find the zeros of p(x), consider p(x) = 0

∴ x² – x = 0

∴ x (x – 1) = 0

∴ x = 0 or x = 1

∴ 0 and 1 are zeros of p(x).

Hence, the number of zeros of the given polynomial is 2.

Here, p(x) = x —x² —1

To find the zeros of p(x), consider p(x) = 0

∴ x – x² – 1 = 0

Taking negative common

x² – x + 1 = 0

x² – x + + 1 – = 0

(Adding and subtracting in order to do factorization)

+ = 0

Using the identity: (a– b) ² = (a² – 2ab + b²)

= –

The above equation cannot be true as the square of a real number cannot be negative.

⇒ No zero exists for p(x).

Hence, the number of zeros of the given polynomial is 0.

Here, p(x) = 3x —2

To find the zeros of p(x), consider p(x) = 0

∴ 3x – 2 = 0

∴ x =

∴ is a zero of p(x).

Hence, the number of zeros of the given polynomial is 1.

Here, p(x) = x³ — x

To find the zeros of p(x), consider p(x) = 0

∴ x³ – x = 0

∴ x (x² – 1) = 0

∴ x (x – 1) (x + 1) = 0 Using the identity: (a² – b²) = (a – b) (a + b)

∴ x = 0 or x = –1 or x = 1

∴ 0, –1 and 1 are the zeros of p(x).

Thus, the number of zeros of the given polynomial is 3.

Given, p(x) = x³ + 1

To find the zeros of p(x), consider p(x) = 0

∴ x³ + 1 = 0

∴ (x + 1) (x² – x + 1) = 0 Using the identity (a + b)³ = (a + b) (a² – ab + b²)

∴ x + 1 = 0 or x² – x + 1 = 0

∴ x = –1 ; real zeros of x² – x + 1 are not possible.

∴ –1 is the only zero of p(x).

Since –1 is real, number of real zeros is 1.

The given polynomial has degree 3, so it can have 3 zeros at most.

⇒ No. of zeros = 3 ; No. of real zeros = 1

For p(x) = x³ + 1, taking x = –2, –1, 0 and 2.

We obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

It can be seen that the graph intersects the X-axis at (–1, 0), so –1 is the zero of p(x).

Taking values as x = –3, –2, –1, 0, 1, 2, 3 in p(x), we obtain the following table:

Plotting these points on graph paper, we obtain the following figure:

The graph does not intersect the X -axis at any point.

∴ p(x) has no real zero.

(i) The graph of y = p(x) intersects X-axis at one point. So, the number of real zeros of p(x) is one.

(ii) The graph of y = p(x) does not intersect X-axis at any point. So, p(x) has no real zeros.

(iii) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

(iv) The graph of y = p(x) intersects X-axis at two points. So, the number of real zeros of p(x) is two.

(v) The graph of y = p(x) intersects X-axis at four points. So, the number of real zeros of p(x) is four.

(vi) The graph of y = p(x) intersects X-axis at three points. So, the number of real zeros of p(x) is three.

Given, p(x) = x² – 4

To find the zeros of p(x), consider p(x) = 0

∴ x² – 4 = 0

(x – 2)(x + 2) = 0

Using the identity: (a² – b²) = (a – b) (a + b)

x = 2 or x = –2

⇒ The real zeros of p(x) are 2 and –2, i.e. p(x) has two zeros.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

Plotting these points on a graph paper as shown in the figure:

It can be observed that the graph of p(x) intersects the X-axis at two distinct points (–2, 0) and (2, 0).

The X–coordinates of these point are considered as zeros of p(x).

Thus, –2 and 2 are the zeros of p(x).

Here p(x) = x² – 5x + 4

To prove that 4 and 1 are the zeros of p(x), we need to prove that p(4) = p(1) = 0.

p(4) = (4)² – 5(4) + 4

= 16 – 20 + 4

= 0

p(1) = (1)² – 5(1) + 4

= 1– 5 + 4

= 0

Hence proved that 4 and 1 are the zeros of the quadratic polynomial p(x) = x² – 5x + 4.

In p(x) = x² – 5x + 4

a = 1, b = –5, c = 4

Sum of zeros = 4 + 1 = 5

= – (–5)

= – = – = –

Product of zeros = 4×1 = 4

= = =

Hence verified the relationship between the zeros and the coefficients of p(x).

Here p(x) = x² + 4x – 21

To find the zeros of p(x), let p(x) = 0

∴ x² + 4x – 21 = 0

∴ x² + 7x – 3x – 21 = 0

∴ x(x + 7) – 3(x + 7) = 0

∴ (x + 7)(x – 3) = 0

∴ x = –7 or x = 3

⇒ –7 and 3 are the zeros of the quadratic polynomial p(x).

Here p(x) = 6x² — 11x + 5

To find the zeros of p(x), let p(x) = 0

∴6x² — 11x + 5 = 0

∴ 6x² – 6x – 5x + 5 = 0

∴ 6x(x – 1) – 5(x – 5) = 0

∴ (x – 1)(6x – 5) = 0

∴ x = 1 or x =

⇒ 1 and are the zeros of the quadratic polynomial p(x).

Here p(x) = 4x² + 9x + 5

To find the zeros of p(x), let p(x) = 0

∴4x² + 9x + 5 = 0

∴4x² + 4x + 5x + 5 = 0

∴4x(x + 1) + 5(x + 1) = 0

∴ (x + 1)(4x + 5) = 0
∴ x = –1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

Here p(x) = 3x² + 5x — 8

To find the zeros of p(x), let p(x) = 0

∴3x² + 5x — 8 = 0

∴ 3x² + 8x – 3x – 8 = 0

∴ x(3x + 8) – 1(3x + 8) = 0

∴ (3x + 8)(x – 1) = 0
∴ x = 1 or x = –

⇒ 1 and – are the zeros of the quadratic polynomial p(x).

Here p(x) = x² — 81

To find the zeros of p(x), let p(x) = 0

∴ x² – 81 = 0

∴ (x)² – (9)² = 0

∴ (x – 9) (x + 9) = 0

[Using the identity: (a² – b²) = (a – b) (a + b)]

∴ x = 9 or x = –9

–9 and 9 are the zeros of the quadratic polynomial p(x).

Here p(x) = x² — x — 6

To find the zeros of p(x), let p(x) = 0

∴ x² — x — 6 = 0

∴ x² — 3x + 2x — 6 = 0

∴ x(x – 3) + 2 (x – 3) = 0

∴ (x – 3) (x + 2) = 0

∴ x = 3 or x = –2

3 and –2 are the zeros of the quadratic polynomial p(x).

Here p(x) = 3x² — x — 4

To find the zeros of p(x), let p(x) = 0

∴3x² — x — 4 = 0

∴ 3x² + 3x – 4x – 4 = 0

∴ 3x(x + 1) – 4(x + 1) = 0

∴ (3x – 4)(x + 1) = 0

∴ x = or x = –1

⇒ and –1are the zeros of the quadratic polynomial p(x).

Sum of zeros = + (–1) =

Product of zeros = × (–1) = –

Let α and β be the zeros of the polynomial p(x) = ax² + bx + c.

Given, α + β = 2 ; αβ = –3

We know that α + β = – = – . So, – = = k, say

Thus, b = –2k, a = k

And αβ = = – . So, c = –3a = –3(k) = –3k

p(x) = ax² + bx + c

∴ p(x) = (k)x² + (–2k)x – 3k

∴ p(x) = k (x² – 2x – 3)

Let α and β be the zeros of the polynomial p(x) = ax² + bx + c.

Given, α + β = –3 ; αβ = –4

We know that α + β = – = – . So, – = = k, say

Thus, b = –3k, a = k

And αβ = = – . So, c = –4a = –4(k) = –4k

p(x) = ax² + bx + c

∴ p(x) = (k)x² + (–3k)x – 4k

∴ p(x) = k (x² – 3x– 4)

Let α and β be the zeros of the polynomial p(x) = ax² + bx + c.

Given, α + β = ; αβ =

We know that α + β = – = . So, – = = k, say

Thus, b = –k, a = 3k

And αβ = = . So, c = =

p(x) = ax² + bx + c

∴ p(x) = (3k)x² + (–k)x +

∴ p(x) = k (3x² – x + )

∴ p(x) = (6x² – 2x + 3) [Taking common]

(1) The quadratic polynomial is,

p(x) = ax² + bx + c; a ≠ 0, a, b, c ε R.

∴ Substituting a = 6, b = 17 and c = 11, we get the required quadratic polynomial

p(x) = 6x² + 17x + 11.

(2) The cubic polynomial is,

p(x) = ax³ + bx² + cx + d; a ≠ 0, a, b, c, d ε R.

∴ Substituting a = 1, b = —1, c = —1 and d = 1, we get the required cubic polynomial

p(x) = x³ – x² – x + 1.

(3) The quadratic polynomial is,

p(x) = ax + bx + c; a ≠ 0, a, b, c ε R.

∴ Substituting a = 5, b = 7 and c = 2, we get the required quadratic polynomial

p(x) = 5x² + 7x + 2.

(4) The cubic polynomial is,

p(x) = ax + bx + cx + d; a ≠ 0, a, b, c, d ε R.

∴ Substituting a = 1, b = –3, c = –1 and d = 3, we get the required cubic polynomial

p(x) = x³ – 3x² – x + 3.

(5) The cubic polynomial is,

p(x) = ax³ + bx² + cx + d; a ≠ 0, a, b, c, d ε R.

∴ Substituting a = 3, b = –5, c = –11 and d = –3, we get the required cubic polynomial

p(x) = 3x³ – 5x² – 11x – 3.

Here, dividend polynomial = p(x) = 2x³ – 13x² + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x² – 5x + 4 and the reminder polynomial r(x) = 0.

Here, dividend polynomial = p(x) = x² + 5x + 6

and divisor polynomial = s(x) = x + 6

Thus, the quotient polynomial q(x) = x + 1 and the reminder polynomial r(x) = 0.

Here, dividend polynomial = p(x) = 40x² + 11x — 63

and divisor polynomial = s(x) = 8x — 9

Thus, the quotient polynomial q(x) = 5x + 7 and the reminder polynomial r(x) = 0.

Here, dividend polynomial = p(x) = 2x³ + 9x² + 13x + 6

and divisor polynomial = s(x) = 2x² + 5x + 3

Thus, the quotient polynomial q(x) = x + 2 and the reminder polynomial r(x) = 0.

Here, dividend polynomial = p(x) = 2x³ – 13x² + 23x – 12

and divisor polynomial = s(x) = 2x — 3

Thus, the quotient polynomial q(x) = x² + 4x + 6 and the reminder polynomial r(x) = –3x + 3.

Here the dividend polynomial p(x) = x³ — 3x² + 4x + 5 and the divisor polynomial s(x) = x – 2

Coefficients of x³, x², x and x⁰ in the dividend polynomial are 1, –3, 4 and 5 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x³ — 3x² + 4x + 5 is divided by x–2 is r(x) = 9.

Given, 3 is a zero of p(x).

⇒ p(3) = 0

∴ 3(3)³ — (3)² — a(3) — 45 = 0

∴ 81 – 9 – 3a – 45 = 0

∴ 3a = 27

∴ a = 9

Here p(x) = the dividend polynomial = 6x³ + 29x² + 44x + 21

s(x) = the divisor polynomial = 3x + 7

Now,

∴ The quotient polynomial q(x) = 3x + 7 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 2x² + 5x + 3

Here, divisor polynomial s(x) = x² + 3x + 5

quotient polynomial q(x) = 2x² + x + 1

remainder polynomial r(x) = x – 3

We know that,

Dividend = Divisor × Quotient + Remainder

∴ p(x) = s(x) × q(x) + r(x)

∴ p(x) = (x² + 3x + 5) × (2x² + x + 1) + (x – 3)

∴ p(x) = 2x⁴ + x³ + x² + 6x³ + 3x² + 3x + 10x² + 5x + 5 + x – 3

∴ p(x) = 2x⁴ + 7x³ + 14x² + 9x + 2

Here the dividend polynomial p(x) = x³ — 4x² + 5x — 2

and the divisor polynomial s(x) = x – 2

Coefficients of x³, x², x and x⁰ in the dividend polynomial are 1, –4, 5 and –2 respectively.

Equating divisor polynomial x – 2 to 0, we get x = 2.

The reminder obtained when the cubic polynomial x³ — 4x² + 5x — 2 is divided by x–2 is r(x) = 0.

Here, dividend polynomial = p(x) = x⁴ + 57x + 15

and divisor polynomial = s(x) = x² + 4x + 2

Thus, the quotient polynomial q(x) = x² – 4x + 14 and the reminder polynomial r(x) = 9x – 13.

⇒ Each student will get (x² – 4x + 14) pens and number of pens left undistributed = 9x – 13

Cost of 2x² — x + 2 TV sets = Rs. 8x⁴ + 7x — 6

Here, dividend polynomial = p(x) = 8x⁴ + 7x — 6

and divisor polynomial = s(x) = 2x² — x + 2

⇒ Cost of one TV set = 4x² + 2x – 3.

Given, – √2 and √2 are zeros of p(x).

⇒ (x + √2) and (x – √2) are the factors of p(x).

Thus (x + √2) (x – √2) = (x² – 2) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8

and divisor polynomial = s(x) = x² – 2

⇒ p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8 = (x² – 2)(2x² + 7x — 4)

On factorising 2x² + 7x — 4, we get

2x² + 7x — 4 = 2x² + 8x – x — 4

= 2x (x + 4) –1 (x + 4)

= (2x –1) (x + 4)

Hence the other two zeros of p(x) are and –4.

(1) p(x) = 5x + 7

∴ p = 5 + 7 = 14

⇒ p ≠0

⇒ is not a zero of p(x).

So the statement is False.

(2) Given, p(x) = x² + 2x + 1

To find the zeros of p(x), let p(x) = 0

∴ x² + 2x + 1 = 0

∴ (x + 1)² = 0 Using the identity: (a+ b) ² = (a² + 2ab + b²)

∴ x + 1 = 0 or x = –1

∴ x = –1 or x = –1

Here, both the zeros are equal, i.e. –1, and hence not distinct.

So the statement is False.

(3) Given, p(x) = x³ + x² — x — 1

To find the zeros of p(x), let p(x) = 0

∴ x³ + x² — x — 1 = 0

∴ x² (x + 1) – 1 (x + 1) = 0

∴ (x² – 1)(x + 1) = 0

∴ (x – 1) (x + 1)(x + 1) = 0 Using the identity: (a² – b²) = (a – b) (a + b)

∴ x – 1 = 0 or x + 1 = 0 or x + 1 = 0

∴ x = 1 or x = –1 or x = –1

∴ Two distinct zeros of p(x) are 1 and –1.

Hence, p(x) has at the most two distinct zeros.

So the statement is True.

(4) Given, p(x) = x³

To find the zeros of p(x), let p(x) = 0

∴x³ = 0

⇒ x = 0.

⇒ The graph of p(x) = x³ meets the X-axis at only one point i.e. (0, 0).

So the statement is True.

(5) If the graph of the quadratic polynomial p(x) does not intersect the x-axis at any point, then the quadratic polynomial does not have any real zero.

So the statement is False.

Here, p(x) = x² + 9x + 18

To find the zeros of p(x), let p(x) = 0.

∴ x² + 9x + 18 = 0

∴ x² + 6x + 3x + 18 = 0

∴ x(x + 6) + 3(x + 6) = 0

∴ (x + 6)(x + 3) = 0

∴ x = –6 or x = –3

Thus, the number of zeros of p(x) = 2.

To draw the graph of this polynomial, consider the following values of x and corresponding values of p(x):

On plotting the above points on a graph paper we obtain the following shape of the graph:

Given, p(x) = 4x² + 12x + 5

= 4x² + 10x + 2x + 5

= 2x(2x + 5) + 1(2x + 5)

= (2x + 5)(2x + 1)

To find the zeros of p(x), put p(x) = 0

∴ (2x + 5)(2x + 1) = 0

∴ x = – or x = –

The zeros of p(x) are – and – .

Now, sum of zeros = = = –3

and, product of zeros = – × – =

Let α and β be the zeros of the polynomial p(x) = ax² + bx + c.

Given, α + β = –4 ; αβ = 9

Putting the values in this equation we get,

p(x) = x² - (- 4) x + 9

p(x) = x² + 4x + 9

Here, dividend polynomial = p(x) = 11x — 21 + 2x² = 2x² + 11x — 21

and divisor polynomial = s(x) = 1 + 2x = 2x + 1

The quotient polynomial q(x) = x + 5 and the reminder polynomial r(x) = –26.

Here, dividend polynomial = p(x) = 2x³ + 3x² — 11x — 6

and divisor polynomial = s(x) = x² + x — 6

Thus, the quotient polynomial q(x) = 2x + 1 and the reminder polynomial r(x) = 0.

Given, 4 is a zero of polynomial p(x).

So (x – 4) is the factor of p(x).

Here, dividend polynomial = p(x) = x³ — 3x² — 6x + 8

and divisor polynomial = s(x) = x – 4.

Coefficients of x³, x², x and x° are 1, –3, –6 and 8 respectively.

Taking x – 4 = 0 we get x = 4

∴ p(x) = x³ — 3x² — 6x + 8

= (x – 4) (x² + x – 2)

= (x – 4) (x² – x + 2x – 2)

= (x – 4) (x(x – 1) + 2(x –1))

= (x – 4)(x + 2)(x – 1)

To find the remaining zeros, let p(x) = 0

i.e. (x – 4)(x + 2)(x – 1) = 0

∴ The remaining zeros of p(x) are –2 and 1.

Here p(x) = the dividend polynomial = 3x⁴ + 5x³ — 21x² — 53x — 30

s(x) = the divisor polynomial = x² — x — 6

Now,

∴ The quotient polynomial q(x) = 3x² + 8x + 5 and the remainder polynomial r(x) = 0.

∴ The other polynomial is 3x² + 8x + 5.

Given, 2 + √3 and 2–√3 are zeros of p(x).

⇒ and are the factors of p(x).

Thus = (x² – 4x + 4 –3) = (x² – 4x + 1) is also a factor of p(x).

To find the remaining zeros, we find the remaining factors using the division process.

Here, dividend polynomial = p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8

and divisor polynomial = s(x) = x² – 2

⇒ p(x) = 2x⁴ + 7x³ — 8x² — 14x + 8 = (x² – 4x + 1)(x² – 2x — 35)

On factorising x² – 2x — 35, we get

x² – 2x — 35 = x² –7x + 5x— 35

= x (x – 7) + 5 (x – 7)

= (x – 7) (x + 5)

Hence the other two zeros of p(x) are 7 and –5.

Here, p(x) = 7x — 3

To find the zeros of p(x), consider p(x) = 0

∴ 7x — 3 = 0

∴ x =

Hence, the zero of the given polynomial is .

The correct option is B.

Here, p(x) = x³ — x

To find the zeros of p(x), consider p(x) = 0

∴x³ — x = 0

∴ x (x² – 1) = 0

∴ x(x – 1)(x + 1) = 0 Using the identity: (a² – b²) = (a – b) (a + b)

∴ x = 0, x = 1, x = –1

∴ The cubic polynomial has 3 zeros.

The correct option is D.

The zeros of p(x) are the intersections points of the equation, p(x) = 3x – 2 – x² with the x-axis.

∴ The number of distinct zeros of p(x) gives the number of distinct intersection points on the X-axis.

To find the zeros of p(x), consider p(x) = 0

∴3x – 2 – x² = 0

∴ x² – 3x+ 2 = 0

∴ x² – x – 2x+ 2 = 0

∴ x(x – 1) – 2(x – 1) = 0

∴ (x – 1)(x – 2) = 0

∴ x = 1 or x = 2 are the zeros of p(x).

⇒The graph of p(x) intersects X-axis at two points.

The correct option is C.

Given, p(x) = 3x² + 5x — 2

Here a = 3, b = 5, c = –2

Sum of zeros, α + β = – = –

The correct option is D.

Given, p(x) = 3x+ 5

Since p(x) is a linear polynomial, so its graph is a straight line.

The correct option is A.

The number of intersection points of a quadratic polynomial at X-axis is the number of its real zeros.

Given, the quadratic polynomial has no zero.

⇒ Its graph does not intersect X-axis or its graph lies in any one half plane of X-axis.

The correct option is D.

The graph intersects the X-axis at four distinct points.

⇒ p(x) has four zeros.

The correct option is D.

Given, p(x) = x² — 4x + 3

Here a = 1, b = –4, c = 3

Product of zeros, αβ = = – = 3

The correct option is B.

The standard form of a cubic polynomial is p(x) = ax³ + bx² + cx + d

∴ For a = 3, b = 5, c = 7 and d = 11,

p(x) = 3x³ + 5x² + 7x + 11.

The correct option is D.