These TS 10th Class Maths Chapter Wise Important Questions Chapter 3 Polynomials given here will help you to solve different types of questions.

## TS 10th Class Maths Important Questions Chapter 3 Polynomials

Previous Exams Questions

Question 1.

Find p(3) if p(x) = x^{2} – 5x + 6 is given.

Solution:

p(x) = x^{2} – 5x + 6 (given)

then p(3) = 3^{2} – 5(3) + 6

= 9 – 15 + 6 = 15 – 15 = 0

∴ p(3) = 0.

Question 2.

Find p(3) if p (x) = x^{2} – 5x – 6 is given. (A.P.Mar.’16)

Solution:

p(x) = x^{2} – 5x + 6 (given)

p (3) = 3^{2} – 5(3) + 6

= 9 – 15 + 6 = 15 – 15 = 0

∴ p(3) = 0.

Question 3.

Find the quotient \(\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}\) when x ≠ 1.

Solution:

\(\frac{x^5+x^4+x^3+x^2}{x^3+x^2+x+1}\)

= \(\frac{x^2\left(x^3+x^2+x+1\right)}{\left(x^3+x^2+x+1\right)}\)

= x^{2}

So the quotient is [x^{2}]

Question 4.

We can write a trinomial having degree 7^{th}. Justify the given statement by giving one example. (T.S. Mar.15)

Solution:

Since the required is a trinomial, it should have 3 terms. As its degree is 7 the highest exponent (power) of the variable is 7. So the required will be

1) x^{7} – x^{6} + x^{5} (or) x^{7} – x + 10 (or) x^{7} – x^{5} – x (or) x^{7} + 5x^{6} + 6x^{5} or …

Question 5.

If we multiply or divide both sides of a linear equation by a non zero number, then the roots of linear equation will remain the same. Is it true ? If so justify with an example. (T.S. Mar. ’15)

Solution:

Let linear equation be 2x + 8 = 11

Its solution is 2x = 11 – 8 ⇒ 2x = 3

So x = \(\frac{3}{2}\)

Then 2x + 8 = 11 linear equation is divided by 10 on both sides.

\(\frac{2 x+8}{10}\) = \(\frac{11}{10}\) = 20x + 80 = 110

⇒ 20x = 110 – 80 = 30

then x = \(\frac{30}{2}\) = \(\frac{3}{2}\)

So its solution doesn’t change.

So the given statement is true.

Question 6.

Laxmi does not want to disclose the l, b, h of a cuboid of her project. She has constructed a polynomial x^{3} – 6x^{2} + 11x – 6 by taking the values of l, b, h as its zeros. Can you open the secret ? (T.S. Mar. 15)

Solution:

P(x) = x^{3} – 6x^{2} + 11x – 6

P(1) = (1)^{3} – 6(1)^{2} + 11 (1) – 6

= 1 – 6 + 11 – 6

= 12 – 12 = 0

∴ for P(x), (x – 1) is a factor

x^{2} – 5x + 6 = x^{2} – 3x – 2x + 6

= x (x – 3) – 2(x – 3)

= (x – 3) (x – 2)

P(x) = (x – 1) (x – 2) (x – 3)

P(x) zeroes of the polynomial are 1, 2,3

∴ Measurements of the cuboid are 1,2 and 3 units.

Its solution doesn’t change

So it is true.

Question 7.

Draw a graph for the polynomial p(x) = x^{2} + 3x – 4 and find its zeros from the graph.

Solution:

p(x) = x^{2} + 3x – 4

∴ Zeroes of polynomial are -4 and 1

Question 8.

Given an example for a quadratic polynomial which has no zeroes.

Solution:

A quadratic polynomial is in the form of ax^{2} + bx + c. As this has no zeroes. Its discriminant will not be a real number.

So b^{2} – 4ac < 0

So we can choose certain a, b, c values where

b^{2} – 4ac < 0

For examples a = 1, b = 4 and c = 9

So ax^{2} + bx + c = 0

⇒ x^{2} + 4x + 9 = 0 will not have zeroes.

Question 9.

The length of a rectangle is 5 more than its breadth. So express its perimeter in the form of polynomial.

Solution:

Let the breadth of rectangle = xm and its

length = x + 5m.

So the perimeter = 2(l + b)

= 2(x + 5 + x)

= 2(2x + 5)

= 4x + 10m

4x + 10 is the polynomial which represents the perimeter of above rectangle.

Question 10.

Draw the graph of polynomial

p(x) = x^{2} – 3x + 2 and find its zeros.

Sol, let y = p(x) = x^{2} – 3x + 2

If x = 0 then p(0) = 0 – 0 + 2

= 2 So (0, 2)

x = 1 then p(1) = 1^{2} – 3(1) + 2

= 1 – 3 + 2

= 0 So (1, 0)

x = 2 then p(2) = 2^{2} – 3(2) + 2

= 4 – 6 + 2

= 0 So (2, 0)

x = 3 then p(3) = 3^{2} – 3(3) + 2

= 9 – 9 + 2

= 2 So (3, 2)

and if x = -1 then p(-1)

= (-1)^{2} – 3(-1) + 2

= 1 + 3 + 2

= 6 So (-1, 6)

x = -2 then p(-2) = (-2)^{2} – 3(2) + 2

= 4 + 6 + 2 = 12 So (-2, 12)

that means the graph of the polynomial

p(x) = x^{2} – 3x + 2 passes through the points. (0, 2), (1, 0) (2, 0) (3, 2) (-1, 6) and (-2, 12)

So 1 and 2 are zeroes of the given polynomial.

Question 11.

Draw the graphs of the following two linear equations and find their solution.

Solution:

First we have to recognise the points through which the line 3x – 2y = 2 passes then after 2x + y = 6 pass. Let us find the points

3x – 2y = 2

So y = \(\frac{3 x-2}{2}\) —- (1)

Put x = 0 in above equation

∴ y = \(\frac{3(0)-2}{2}\) = \(\frac{0-2}{2}\) = 1

So (0, 1)

Now x = 1 then

y = \(\frac{3(1)-2}{2}\) = \(\frac{3-2}{2}\) = \(\frac{1}{2}\) So(1, \(\frac{1}{2}\))

Now x = 2 then

y = \(\frac{3(2)-2}{2}\) = \(\frac{6-2}{2}\) = \(\frac{4}{2}\) So(2, 2)

that means the line 3x – 2y = 2 passes through the points (0, -1), (1, \(\frac{1}{2}\)) and (2, 2)

Similarly

2x + y = 6

⇒ y = 6 – 2x —– (2)

Put x = 0 in the above equation (2) we get

y = 6 – 2(0) = 6 – 0 = 6 So(0, 6)

and x = 1 ⇒ y = 6 – 2(1) = 6 – 2 = 4

So(1, 4)

and x = 2 ⇒ y = 6 – 2(2) = 6 – 4 = 2

So (2, 2)

So the line 2x + y = 6 passes through the points (0, 6) (1, 4) and (2, 2)

Here we observe (2, 2) is the common point.

That means they intersert at (2, 2)

So x = 2 and y = 2 will be the solution of above the equations.

Additional Questions

Question 1.

If p(x) = 6x^{7} – 2x^{5} + 4x – 8, find

(i) coefficient of x^{5}

(ii) degree of p(x)

(iii) constant term. (Each 1 Mark)

Solution:

If p(x) = 6x^{7} – 2x^{5} + 4x – 8,

i) coefficient of x^{5} = – 2

ii) degree of p(x) = highest degree of x – 7

iii) constant term = – 8

Question 2.

If p(t) = 3t^{3} + 4t – 5, find the values of p(1), p(-1), P(0), p(2), p(-2).

Solution:

given p(t) = 3t^{3} + 4t – 5

∴ p(1) = 3(1)^{3} + 4(1) – 5 = 3 + 4 – 5 = 2

p(-1) = 3(-1)^{3} + 4(-1) – 5 = -3 – 4 – 5 = -12

p(0) = 3(0)^{3} + 4(0) – 5 = 0 + 0 – 5 = -5

p(2) = 3(2)^{3} + 4(2) – 5 = 3(8) + 8 – 5

= 24 + 8 – 5 = 27

p(-2) = 3(-2)^{3} + 4 (-2) – 5

= – 24 – 8 – 5 = – 37

Question 3.

Check whether – 4 and 4 are the zeroes of the polynomial x^{4} – 256.

Solution:

given p(x) = x^{4} – 256

p(-4) = (-4)^{4} – 256 = 256 – 256 = 0

p(4) = (4)^{4} – 256 = 256 – 256 = 0

yes, -4 and 4 are zeroes of the given polynomial

Question 4.

Check whether 5 and -4 are the zeroes of the polynomial p(x) when p(x) = x^{2} – x – 20

Solution:

given p(x) = x^{2} – x – 20

p(5) = (5)^{2} – 5 – 20 = 25 – 5 – 20 = 25 – 25 = 0

P(-4) = (-4)^{2} – (-4) – 20

= 16 + 4 – 20 = 20 – 20 = 0

yes, 5 and – 4 are the zeroes of the polynormial

p(x) = x^{2} – x – 20.

Question 5.

Check whether 3 and – 7 are the zeroes of the polynomial p(x) = x^{2} + 4x – 21.

Solution:

given p(x) = x^{2} + 4x – 21

p(3) = 3^{2} + 4(3) – 21

= 9 + 12 – 21 = 21 – 21 = 0

p(-7) = (-7)^{2} + 4(-7) – 21

= 49 – 28 – 21 = 49 – 49 = 0

yes, 3 and – 7 are the zeroes of the polynomial p(x) = x^{2} + 4x – 21.

Question 6.

Find the zeroes of the given polynomials.

i) p(x) = 4x

ii) p(x) = x^{2} – 9x + 20

iii) p(x) = (x + 5) (x + 6)

iv) p(x) = x^{4} – 81

Solution:

i) p(x) = 4x

p(0) = 4 × 0 = 0

∴ The zero of p(x) = 4x is 0.

ii) p(x) = x^{2} – 9x + 20

= x^{2} – 4x – 5x + 20

= x (x – 4) -5 (x – 4)

= (x – 4) (x – 5)

To find zeroes, let p(x) = 0

⇒ (x – 4) (x – 5) = 0

⇒ x – 4 = 0 or x – 5 = 0

⇒ x = 4 or x = 5

∴ The zeroes of x^{2} – 9x + 20 are 4 nd 5.

iii) p(x) = (x + 5) (x + 6)

To find zeroes, let p(x) = 0

⇒ (x + 5) (x + 6) = 0

⇒ x + 5 = 0 or x + 6 = 0

⇒ x = -5 or x = -6

∴ The zeroes of (x + 5) (x + 6) are – 5 and – 6

iv) p(x) = x^{4} – 81.

To find zeroes, let p(x) = 0

⇒ x^{4} – 81 = 0

⇒ (x^{2})^{2} – (9)^{2} = 0

⇒ (x^{2} + 9) (x^{2} – 9) = 0

⇒ x^{2} + 9 = 0 or x^{2} – 9 = 0

x^{2} = -9 or (x + 3) (x – 3) = 0

x = ±\(\sqrt{-9}\) = 9 or x + 3 = 0 or x – 3 = 0

x = -3 or x = 3

∴ The zeroes of the polynomial x^{4} – 81 are -3, 3 and ± \(\sqrt{-9}\).

Question 7.

Find the zeroes of the given polynomials.

i) p(x) = x^{2} – x – 12

ii) p(x) = x^{2} – 6x + 9

iii) x^{2} p(x) = x^{2} – 4x + 5

iv) p(x) = x^{2} + 3x – 4

Solution:

i) p(x) = x^{2} – x – 12

To find zeroes, let p(x) = 0

⇒ x^{2} – x – 12 = 0

x^{2} – 4x + 3x – 12 = 0

x(x – 4) + 3 (x – 4) = 0

(x – 4) (x + 3) = 0

⇒ x – 4 = 0 or x + 3 = 0

∴ The zeroes of x^{2} – x – 12 are 4 and – 3

ii) p(x) = x^{2} – 6x + 9

To find zeroes, let p(x) = 0

x^{2} – 6x + 9 = 0

x^{2} – 3x – 3x + 9 = 0

x(x – 3) – 3 (x – 3) = 0

⇒ (x – 3) (x – 3) = 0

⇒ x – 3 = 0 or x – 3 = 0

⇒ x = 3 or x = 3

∴ The zeroes of x^{2} – 6x + 9 is 3.

iii) p(x) = x^{2} – 4x – 5

To find zeroes, let p(x) = 0

x^{2} – 4x – 5 = 0

x^{2} – 5x + x – 5 = 0

⇒ x (x – 5) + 1(x – 5) = 0

⇒ (x – 5) (x + 1) = 0

⇒ x – 5 = 0 or x + 1 = 0

⇒ x = 5 or x = -1

∴ The zeroes of x^{2} – 4x – 5 are 5 and – 1

iv) p(x) = x^{2} + 3x – 4

To find zeroes, let p(x) = 0

x^{2} + 3x – 4 = 0

⇒ x^{2} + 4x – x – 4 = 0

⇒ x(x + 4) – 1(x + 4) = 0

⇒ (x + 4) (x – 1) = 0

⇒ x + 4 = 0 or x – 1 = 0

⇒ x = -4 or x = 1

∴ The zeroes of x^{2} + 3x – 4 are -4 and 1.

Question 8.

Why are \(\frac{1}{3}\) and -2 zeroes of polynomial

p(x) = 3x^{2} + 5x – 2 ?

Solution:

p(x) = 3x^{2} + 5x – 2

Now p(\(\frac{1}{3}\)) = 3(\(\frac{1}{3}\))^{2} + 5(\(\frac{1}{3}\)) – 2

= \(\frac{1}{3}\) + \(\frac{5}{3}\) – 2

= \(\frac{1+5-6}{3}\) = \(\frac{6-6}{3}\) = \(\frac{0}{3}\) = 0

p(-2) = 3 (-2)^{2} + 5(-2) – 2

= 3(4) – 10 – 2

= 12 – 12 = 0

since p(\(\frac{1}{3}\)) and p(-2) are equal to zero

–\(\frac{1}{3}\) and – 2 are zeroes of the polynomial.

Question 9.

Find the zeroes of the following quadratic polynomial and verify the relationship between the zeroes and coefficients.

i) x^{2} – 5x + 6

ii) 3x^{2} – 7x + 4

iii) 5x^{2} + 10x

iv) t^{2} – 12

Solution:

i) Let p(x) = x^{2} – 5x + 6

= x^{2} – 3x – 2x + 6

= x(x – 3) -2 (x – 3)

= (x – 3) (x – 2)

To find zeroes, let p(x) = 0

Hence the zeroes of p(x) are 3 and 2

sum of the zeroes = 3 + 2 = 5 = –\(\frac{(-5)}{1}\)

Product of zeroes = 3 × 2 = 6 = \(\frac{6}{1}\) constant term

ii) 3x^{2} – 7x + 4

Let p(x) = 3x^{2} – 7x + 4

= 3x^{2} – 3x – 4x + 4

= 3x (x – 1) -4 (x – 1)

= (x – 1) (3x – 4)

To find zeroes, let p(x) = 0

⇒ (x – 1) (3x – 4) = 0

x = 1 or x = 4/3

Hence, the zeroes of p(x) are 1 and \(\frac{4}{3}\).

sum of the zeroes = 1 + \(\frac{4}{3}\) = \(\frac{3+4}{3}\) = \(\frac{7}{3}\)

product of zeroes = 1 × \(\frac{4}{3}\) = \(\frac{4}{3}\)

iii) Let p(x) = 5x^{2} + 10x

= 5x (x + 2)

To find zeroes, Let p(x) = 0

⇒ 5x (x + 2) = 0

⇒ x = 0 or x + 2 = 0

⇒ x = 0 or x = -2

Hence the zeroes of p(x) are 0 and -2.

sum of the zeroes = 0 + (-2) = -2

coefficient of x Product of zeroes = 0 x – 2 = 0

iv) t^{2} – 12

let p(t) = t^{2} – 12

to find zeroes, let p(t) = 0

⇒ t^{2} – 12 = 0

⇒ t^{2} = 12 ⇒ t = ±\(\sqrt{12}\) = \(\sqrt{12}\) and –\(\sqrt{12}\)

Hence the zeroes of p(t) are \(\sqrt{12}\) and –\(\sqrt{12}\)

sum of zeroes = \(\sqrt{12}\) + (-\(\sqrt{12}\))

= \(\sqrt{12}\) – \(\sqrt{12}\) = 0 = \(\frac{0}{1}\)

product of zeroes = \(\sqrt{12}\) × (-\(\sqrt{12}\))

= -12 = \(-\frac{12}{1}\)

Question 10.

Find the quadratic polynomial in each case, with the given numbers as the sum and product of its zeroes.

i) \(\frac{1}{3}\), -2

ii) 0, \(\sqrt{3}\)

iii) 3, 5

iv) –\(\frac{1}{2}\), \(\frac{1}{2}\)

Solution:

i) \(\frac{1}{3}\), -2

Let α, β be the zeroes of the quadratic polynomial sum of zeroes = α + β = \(\frac{1}{3}\)

product of zeroes = αβ = -2

The required quadratic polynomial will be

k [x^{2} – x(α + β) + αβ] when k is a constant

⇒ k[x^{2} – \(\frac{x}{3}\) – 2]

If k = 3, then the polynomial will be

3[x^{2} – \(\frac{x}{3}\) – 2] = 3x^{2} – x – 6

ii) 0, \(\sqrt{3}\)

Let α, β be the zeroes of quadratic polynomial.

sum of the zeroes = α + β = 0

product of the zeroes = αβ = \(\sqrt{3}\)

The required quadratic polynomial will be k[x^{2} – x(α + β) + αβ] where k is a constant.

⇒ k [x^{2} – x(0) + \(\sqrt{3}\))

⇒ k [x^{2} – 0 + \(\sqrt{3}\)] ⇒ k[x^{2} + \(\sqrt{3}\)

If k = 1, then the polynomial will be 1[x^{2} + \(\sqrt{3}\)] = x^{2} + \(\sqrt{3}\).

iii) 3, 5

Let α, β be the zeroes of quadratic polynomial.

sum of the zeroes = α + β = 3

product of the zeroes = αβ = 5

The required quadratic polynomial will be k[x^{2} – x(α + β) + αβ] where k is a constant

⇒ k [x^{2} – x(3) + 5]

⇒ k [x^{2} – 3x + 5]

If k = 1, then the polynomial will be

1[x^{2} – 3x + 5] = x^{2} – 3x + 5

iv) –\(\frac{1}{2}\), \(\frac{1}{2}\)

Let α, β be the zeroes of quadratic polynomial sum of the zeroes = α + β

= –\(\frac{1}{2}\)

product of the zeroes αβ = \(\frac{1}{2}\)

The required quadratic polynomial will be k[x^{2} – x(α + β) + αβ] when k is a constant

k[x^{2} – x(-\(\frac{1}{2}\)) + \(\frac{1}{2}\)]

⇒ k[x^{2} + \(\frac{x}{2}\) + \(\frac{1}{2}\)]

If k = 2, then the polynomial will be 2[x^{2} + \(\frac{x}{2}\) + \(\frac{1}{2}\)] = 2x^{2} + x + 1

Question 11.

Find the quadratic polynomial the zeroes α, β given in each case

i) 4, -3

ii) \(\sqrt{5}\) – \(\sqrt{5}\)

iii) \(\frac{1}{3}\), \(\frac{5}{3}\)

Solution:

i) 4, -3

Let the quadratic polynomial be ax^{2} + bx + c, a ≠ 0,

and its zeroes be a and b.

then α = 4, β = – 3

sum of the zeroes = a + b = 4 – 3 = 1

product of the zeroes = a . b = 4 × -3

= -12

The requiral quadratic polynomial is k [x^{2} – x (α + β) + αβ] where k is a constant

⇒ k[x^{2} – x(1) + (-12)]

⇒ k[x^{2} – x – 12]

When k = 1, the quadratic polynomial will be x^{2} – x – 12.

ii) \(\sqrt{5}\) – \(\sqrt{5}\)

Let the quadratic polynomial be

ax^{2} + bx + c, a ≠ 0 and its zeroes be a and b, Here α = \(\sqrt{5}\), β = –\(\sqrt{5}\)

sum of the zeroes = α + β = \(\sqrt{5}\) – \(\sqrt{5}\) = 0

product of the zeroes = α . β

= \(\sqrt{5}\) × –\(\sqrt{5}\) = -5

The required quadratic polynomial is

k^{2} [x – x (α + β) + αβ] where k is a constant

⇒ k^{2} [x^{2} – x(0) + (-5)]

⇒ k[x^{2} – 5]

wherek k = 1, the quadratic polynomial will be x^{2} – 5

iii) \(\frac{1}{3}\), \(\frac{5}{3}\)

Let the quadrate polynomial be

ax^{2} + bx + c, a ≠ 0 and its zeroes be α and β.

Then α = \(\frac{1}{3}\) and β = \(\frac{5}{3}\)

sum of the zeroes = α + β = \(\frac{1}{3}\) + \(\frac{5}{3}\)

= \(\frac{6}{3}\) = 2

Product of the zeroes = αβ = \(\frac{1}{3}\) . \(\frac{5}{3}\)

= \(\frac{5}{9}\)

The required quadratic polynomial is

k[x^{2} – x(α + β) + αβ] where k is a constant

⇒ k[x^{2} – x(2) + \(\frac{5}{9}\)]

⇒ k[x^{2} – 2x + \(\frac{5}{9}\)]

Where k = 9, the quadratic polynomial will be a [x^{2} – 2x + \(\frac{5}{9}\)] = 9x^{2} – 18x + 5

Question 12.

Verify that 1, -2 and -3 are the zeroes of the cubic polynomial x^{3} + 4x^{2} + x – 6 and check the relationship between zeroes and the coefficients.

Solution:

The given polynomial is x^{3} + 4x^{2} + x – 6.

Comparing the given polynomial with

ax^{3} + bx^{2} + cx + d, we get

a = 1, b = 4, c = 1, d = – 6

Let p(x) = x^{3} + 4x^{2} + x – 6

p(1) = (1)^{3} + 4(1)^{2} + 1 – 6

= 1 + 4 + 1 – 6 = 6 – 6 = 0

p(-2) = (-2)^{3} + 4(-2)^{2} + (-2) – 6

= – 8 + 4(4) – 2 – 6

= – 8 + 16 – 2 – 6 = 16 – 16 = 0

p(-3) =(-3)^{3} + 4(-3)^{2} + (-3) – 6

= – 27 + 4(9) – 3 – 6

= 36 – 36 = 0

∴ 1, -2 and – 3 are the zeroes of given polynomial. So, α = 1, β = -2, ∝= – 3

Question 13.

Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following.

i) p(x) = x^{3} + 4x^{2} + 5x + 6, g(x) = x^{2} + 3

ii) p(x) = x^{4} + 3x^{2} – 4x + 8

g(x) = x^{2} + 2 – x

Solution:

i) p(x) = x^{3} + 4x^{2} + 5x + 6,

g(x) = x^{2} + 3

The degree of x^{2} + 3 is 2. The degree of 2x – 6 is 1

Since degree of (2x – 6)^{2} < degree of (x^{2} + 3)

∴ we stop here.

So, the quotient is x + 4 and the remainder is 2x – 6.

ii) p(x) = x^{4} + 3x^{2} – 4x + 8

g(x) = x^{2} + 2 – x = x^{2} – x + 2

(∴ writing it in standard form)

since degree of (-4x + 4) < degree of x^{2} – x + 2

∴ we stop here

So, the quotient is x^{2} + x + 2 and the remainder is (-4x + 4).

Question 14.

Check in each case the first polynomial is a factor of the second polynomial.

i) x^{2} – 2, 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2

ii) x^{2} – x + 1, x^{4} – 3x^{2} + 4x – 3

Solution:

Hint : If the remainder is zero, then the first polynomial is a factor of the second one

i) x^{2} – 2, 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2

The given polynomials are in standard form.

Since the remainder is zero, x^{2} – 2 is the factor of 2x^{4} – 3x^{3} – 3x^{2} + 6x – 2.

ii) x^{2} – x + 1, x^{4} – 3x^{2} + 4x – 3

The given polynomials is standard form.

Since the remainder is zero, x^{2} – x + 2 is the factor of x^{4} – 3x^{2} + 4x – 3.

Question 15.

Draw the graph of the polynomial x^{2} + x – 6 and mark the zeroes by the …….

Solution:

given that y = x^{2} + x – 6

= x^{2} + x – 6

zeroes of the polynomial x^{2} + x – 6 are the x – co-ordinates of the points on the – 3 and 2 are the zeroes of the given polynomial on graph.

Question 16.

Check whether \(\frac{1}{2}\) is the zero of the polynomial 2x^{2} + x – 1 or not. (AP-SA-II : 2016)

Solution:

Let p(x) = 2x^{2} + x – 1

we have p(\(\frac{1}{2}\)) = 2(\(\frac{1}{2}\))^{2} + \(\frac{1}{2}\) – 1

= 2 × \(\frac{1}{4}\) + \(\frac{1}{2}\) – 1

= \(\frac{1}{2}\) + \(\frac{1}{2}\) – 1 = 1 – 1

= 0

∴\(\frac{1}{2}\) is the zero of the polynomial.

Question 17.

Verify the relationship between the zeroes and the coefficient of x^{2} – 25 by finding its zeroes. (AP-SA-II : 2016)

Solution:

Given polynomial is x^{2} – 25

we have x^{2} – 25 = 0

⇒ (x + 5) (x – 5) = 0

⇒ x = -5 or x = 5

∴ The zeroes of x^{2} – 25 are – 5 and 5

∴ The sum of zeroes = -5 + 5 = 0 coefficient of x

= –\(\frac{0}{1}\) = 0

And product of the zeroes = (-5) × (5) = – 25

= –\(\frac{25}{1}\) = -25

Hence verified.

Question 18.

Give two examples of the polynomials p(x) and g(x) satisfying division algori-tham. (AP-SA-II : 2016)

Solution:

Using division alogaritham,

We have p(x) = q(x) × g(x) + r(x)

Example (1) : on dividing 12x^{2} + 8x + 24 by 3x^{2} + 2x + 6, we get

Here p(x) = 12x^{2} + 8x + 24

g(x) = 3x^{2} + 2x + 6

g(x) = 4

r(x) = 0

Example – 2 : an dividing 3x^{3} + 6x^{2} + 18x by x^{2} + 2x + 6

Here p(x) = 3x^{2} + 6x^{2} + 18x

g(x) = x^{2} + 2x + 6

g(x) = 3x

r(x) = 0

Question 19.

Verify that 4, -1, –\(\frac{1}{4}\) are the zeroes of the cubic polynomial 4x^{3} – 11x^{2} – 19x – 4 and check. (AP-SA-II: 2016)

Solution:

Given polynomial is 4x^{3} – 11x^{2} – 19x – 4

p(x) comparing the given polynomial with

ax^{3} + bx^{2} + cx + d, we get

a = 4, b = – 11, c = -19, d = – 4

Further p(4) = 4(4)^{3} – 11(4)^{2} – 19(4) – 4

= 4 × 64 – 11(16) – 19(4) – 4

= 256 – 176 – 76 – 4

= 256 – 256 = 0

p(-1) = 4(-1)^{3} – 11(-1)^{2} – 19(-1) – 4

= 4(-1) – 11(1) + 19-4

= -4 – 11 + 19 – 4

= -19 + 19 = 0

Therefore 4, -1, –\(\frac{1}{4}\) are the zeroes of

4x^{3} – 11x^{2} – 19x – 4.

We take α = 4, β = – 1 and γ = –\(\frac{1}{4}\)

now α + β + γ = 4 – 1 – \(\frac{1}{4}\) = 3 – \(\frac{1}{4}\) = \(\frac{11}{4}\)

= \(\frac{-(-11)}{4}\) = –\(\frac{b}{a}\)

Question 20.

Draw the graph of the polynomial x^{2} + x – 6 and mark the zeroes of the polynomial

Solution:

Given that y = x^{2} + x – 6

Scale : On X-axis :1 cm = 1 unit

On Y-axis :1 cm = 1 unit

Zeroes of the polynomial x^{2} + x – 6 are the x – coordinates of the points on the x – axis.

-3 and 2 are the zeroes of the given polynomial on graph.

Question 21.

If α, β are zeroes of the polynomial 2x^{2} + 7x + 5, find the value of α + β + αβ? (AP New SCERT Model Paper) Solution:

2x^{2} + 7x + 5

α + β = \(\frac{-b}{a}\) = \(\frac{-7}{2}\)

α.β = \(\frac{\mathrm{c}}{\mathrm{a}}\) = \(\frac{5}{2}\)

∴ α + β + α.β = \(\frac{-7}{2}\) + \(\frac{5}{2}\) = \(\frac{-2}{2}\) = -1

Question 22.

If a, b and c are the zeroes of a polynomial of degree 3, then give the relations between the zeroes and the coefficients of the polynomial. (AP New SCERT Model Paper)

Solution:

α + β + γ = \(\frac{-b}{a}\)

αβ + βγ + γα = \(\frac{c}{a}\)

αβγ = \(\frac{-d}{a}\)

Question 23.

Solve the quadratic polynomial x^{2} – 3x – 4 graphically. (AP New SCERT Model Paper)

Solution:

Let y = x^{2} – 3x – 4

o.p. = (-2, 6), (-1, 0), (0, -4), (1, -6), (2, -6) (3, -4), (4, 0), (5, 6)

Scale : On X-axis 1 cm = 1 unit

On Y-axis 1 cm = 1 unit

The graph of y = x^{2} – 3x – 4 intersects X-axis at (-1, 0) and (4, 0).

Hence the zeroes of p(x) are -1 and 4.