AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.3 Textbook Questions and Answers.

## AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.3

### 10th Class Maths 3rd Lesson Polynomials Ex 3.3 Textbook Questions and Answers

Question 1.

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

i) x^{2} – 2x – 8

ii) 4s^{2} – 4s + 1

iii) 6x^{2} – 3 – 7x

iv) 4u^{2} + 8u

v) t^{2} – 15

vi) 3x^{2} – x – 4

Answer:

i) Given polynomial is x^{2} – 2x – 8

We have x^{2} – 2x – 8 = x^{2} – 4x + 2x – 8

= x(x – 4) + 2(x – 4)

= (x – 4) (x + 2)

So, the value of x^{2} – 2x – 8 is zero

when x – 4 = 0 or x + 2 = 0 i.e.,

when x = 4 or x = -2

So, the zeroes of x^{2} – 2x – 8 are 4 and -2.

Sum of the zeroes = 4 – 2 = 2 Coefficient of ,x -(-2)

= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-2)}{1}\) = 2

And product of the zeroes = 4 × (-2) = -8

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-8}{1}\) = -8

ii) Given polynomial is 4s^{2} – 4s + 1

We have, 4s^{2} – 4s + 1

= 4s^{2} – 2s – 2s + 1

= 2s (2s – 1) – 1(2s – 1)

= (2s – 1) (2s – 1)

= (2s – 1)^{2}

So, the value of 4s2 – 4s + 1 is zero

when 2s-1 = 0 or s = \(\frac{1}{2}\)

∴ Zeroes of the polynomial are \(\frac{1}{2}\) and \(\frac{1}{2}\)

∴ Sum of the zeroes = \(\frac{1}{2}\) + \(\frac{1}{2}\) = 1.

= – \(\frac{\text { Coefficient of } s}{\text { Coefficient of } s^{2}}\) = –\(\frac{-4}{4}\) = 1

And product of the zeroes = \(\left(\frac{1}{2}\right) \times\left(\frac{1}{2}\right)\) = \(\frac{1}{4}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{1}{4}\)

iii) Given polynomial is 6x^{2} – 3 – 7x

We have, 6x^{2} – 3 – 7x = 6x^{2} – 7x – 3

= 6x^{2} – 9x + 2x – 3

= 3x(2x – 3) + 1(2x – 3)

= (2x – 3) (3x + 1)

The value of 6x^{2} – 3 – 7x is zero, when the value of (3x +1) (2x – 3) is 0

i.e., when 3x + 1 = 0 and 2x – 3 = 0

3x = -1 and 2x = 3

x = \(\frac{-1}{3}\) and x = \(\frac{3}{2}\)

∴ The zeroes of 6x^{2} – 3 – 7x = \(\frac{-1}{3}\) and \(\frac{3}{2}\)

∴ Sum of the zeroes = \(\frac{1}{3}\) + \(\frac{3}{2}\) = \(\frac{7}{6}\).

= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-7)}{6}\) = \(\frac{7}{6}\)

And product of the zeroes = \(\left(\frac{-1}{3}\right) \times\left(\frac{3}{2}\right)\) = \(\frac{-1}{2}\)

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-3}{6}\) = \(\frac{-1}{2}\)

iv) Given polynomial is 4u^{2} + 8u

We have, 4u^{2} + 8u = 4u (u + 2)

The value of 4u^{2} + 8u is 0,

when the value of 4u(u + 2) = 0, i.e.,

when u = 0 or u + 2 = 0, i.e.,

when u = 0 (or) u = – 2

∴ The zeroes of 4u^{2} + 8u are 0 and – 2.

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

= – \(\frac{\text { Coefficient of } u}{\text { Coefficient of } u^{2}}\) = \(\frac{-8}{4}\) = -2

And product of the zeroes 0 . (-2) = 0

= \(\frac{\text { Constant term }}{\text { Coefficient of } u^{2}}\) = \(\frac{0}{4}\) = 0

v) Given polynomial is t^{2} – 15.

We have, t^{2} – 15 = (t – √15 ) (t + √l5)

The value of t^{2} – 15 is 0,

when the value of (t – √15 ) (t + √l5) = 0, i.e.,

when t – √15 = 0 or t + √15 = 0, i.e.,

when t = √15 (or) t = -√15

∴ The zeroes of t^{2} – 15 are √15 and -√15.

Therefore, sum of the zeroes = √15 + (-√15) = 0

= – \(\frac{\text { Coefficient of } t}{\text { Coefficient of } t^{2}}\) = –\(\frac{0}{1}\) = 0

And product of the zeroes √15 × (-√15) = -15

= \(\frac{\text { Constant term }}{\text { Coefficient of } t^{2}}\) = \(\frac{-15}{1}\) = -15

vi) Given polynomial is 3x^{2} – x – 4

we have, 3x^{2} – x – 4 = 3x^{2} + 3x – 4x – 4

= 3x(x + 1) – 4(x + 1)

= (x + 1) (3x – 4)

The value of 3x^{2} – x – 4 is 0 when the value of (x + 1) (3x – 4) is 0.

i.e., when x + 1 = 0 or 3x – 4 = 0

i.e., when x = -1 or x = \(\frac{4}{3}\)

∴ The zeroes of 3x^{2} – x – 4 are -1 and \(\frac{4}{3}\)

Therefore, sum of the zeroes = -1 + \(\frac{4}{3}\) = \(\frac{-3+4}{3}\) = \(\frac{1}{3}\)

= – \(\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{2}}\) = \(\frac{-(-1)}{3}\) = \(\frac{1}{3}\)

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

= \(\frac{\text { Constant term }}{\text { Coefficient of } x^{2}}\) = \(\frac{-4}{3}\)

Question 2.

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

i) \(\frac{1}{4}\), -1

ii) √2, \(\frac{1}{3}\)

iii) 0, √5

iv) 1, 1

v) –\(\frac{1}{4}\), \(\frac{1}{4}\)

vi) 4, 1

Answer:

Let the polynomial be ax^{2} + bx + c

and its zeroes be α and β.

i) Here, α + β = \(\frac{1}{4}\) and αβ = -1

Thus, the polynomial formed = x^{2} – (sum of the zeroes)x + product of the zeroes

= x^{2} – (\(\frac{1}{4}\))x – 1

= x^{2} – \(\frac{x}{4}\) – 1

The other polynomials are (x^{2} – \(\frac{x}{4}\) – 1)

then the polynomial is 4x^{2} – x – 4.

ii) Here, α + β = √2 and αβ = \(\frac{1}{3}\)

Thus, the polynomial formed = x^{2} – (sum of the zeroes)x + product of the zeroes

= x^{2} – (√2)x + \(\frac{1}{3}\)

= x^{2} – √2x + \(\frac{1}{3}\)

The other polynomials are (x^{2} – √2x + \(\frac{1}{3}\))

then the polynomial is 3x^{2} – 3√2x + 1.

iii) Here, α + β = 0 and αβ = √5

Thus, the polynomial formed = x^{2} – (sum of the zeroes)x + product of the zeroes

= x^{2} – (0)x + √5

= x^{2} + √5

iv) Let the polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then α + β = 1 = \(\frac{-(-1)}{1}\) = \(\frac{-b}{a}\) and

αβ = 1 = latex]\frac{1}{1}[/latex] = \(\frac{c}{a}\)

If a = 4, then b = 1 and c = 1

∴ One quadratic polynomial which satisfies the given conditions is 4x^{2} + x + 1.

v) Let the polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then α + β = \(\frac{-1}{4}\) = \(\frac{-b}{a}\) and

αβ = \(\frac{1}{4}\) = \(\frac{c}{a}\)

If a = 4, then b = 1 and c = 1

∴ One quadratic polynomial which satisfies the given conditions is 4x^{2} + x + 1.

vi) Let the polynomial be ax^{2} + bx + c and its zeroes be α and β.

Then α + β = 4 = \(\frac{-(-4)}{1}\) = \(\frac{-b}{a}\) and

αβ = 1 = \(\frac{1}{1}\) = \(\frac{c}{a}\)

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

∴ One quadratic polynomial which satisfies the given conditions is x^{2} – 4x + 1.

Question 3.

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

i) 2, -1

ii) √3, -√3

iii) \(\frac{1}{4}\), -1

iv) \(\frac{1}{2}\), \(\frac{3}{2}\)

Answer:

i) Let the polynomial be ax^{2} + bx + c, a ≠ 0 and its zeroes be α and β.

Here α = 2 and β = – 1

Sum of the zeroes = α + β = 2 + (-l) = 1

Product of the zeroes = αβ = 2 × (-1) = -2

Therefore the quadratic polynomial ax^{2} + bx + c is x^{2} – (α + β)x + αβ = [x^{2} – x – 2]

the quadratic polynomial will be x^{2} – x – 2.

ii) Let the zeroes be α = √3 and β = -√3

Sum of the zeroes = α + β

= √3 + (-√3) = 0

Product of the zeroes = αβ

= √3 × (-√3) = -3

∴ The quadratic polynomial

ax^{2} + bx + c is [x^{2} – (α + β)x + αβ]

= [x^{2} – 0.x + (-3)] = [x^{2} – 3]

the quadratic polynomial will be x^{2} – 3.

iii) Let the zeroes be α = \(\frac{1}{4}\) and β = -1

Sum of the zeroes = α + β

= \(\frac{1}{4}\) + (-1) = \(\frac{1+(-4)}{4}\) = \(\frac{-3}{4}\)

Product of the zeroes = αβ

= \(\frac{1}{4}\) × (-1) = \(\frac{-1}{4}\)

∴ The quadratic polynomial

ax^{2} + bx + c is [x^{2} – (α + β)x + αβ]

= [x^{2} – \(\left(\frac{-3}{4}\right)\).x + (\(\frac{-1}{4}\))]

the quadratic polynomial will be 4x^{2} + 3x – 1.

iv) Let the zeroes be α = \(\frac{1}{2}\) and β = \(\frac{3}{2}\)

Sum of the zeroes = α + β

= \(\frac{1}{2}\) + \(\frac{3}{2}\) = \(\frac{1+3}{2}\) = \(\frac{4}{2}\) = 2

Product of the zeroes = αβ

= \(\frac{1}{2}\) × \(\frac{3}{2}\) = \(\frac{3}{4}\)

∴ The quadratic polynomial

ax^{2} + bx + c is [x^{2} – (α + β)x + αβ]

= [x^{2} – 2x + (\(\frac{3}{4}\))]

the quadratic polynomial will be 4x^{2} – 8x + 3.

Question 4.

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

Answer:

Given cubic polynomial

p(x) = x^{3} + 3x^{2} – x – 3

Comparing the given polynomial with ax^{3} + bx^{2} + cx + d, we get a = 1, b = 3, c = -1, d = -3

Futher given zeroes are 1,-1 and – 3

p(1) = (1)^{3} + 3(1)^{2} – 1 – 3

= 1 + 3 – 1 – 3 = 0

p(-1) = (-1)^{3} + 3(-1)^{2} – 1 – 3

= -1 + 3 + 1 – 3 = 0

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

= -27 + 27 + 3 – 3 = 0

Therefore, 1, -1 and -3 are the zeroes of x^{3} + 3x^{2} – x – 3.

So, we take α = 1, β = -1 and γ = -3 Now,

α + β + γ = 1 + (-1) + (-3) = -3

αβ + βγ + γα = 1(-l) + (-1) (-3) + (-3)1

= -1 + 3 – 3 = -1

= \(\frac{c}{a}\) = \(\frac{-1}{1}\) = -1

αβγ = 1 (-1) (-3) = 3 = \(\frac{-d}{a}\) = \(\frac{-(-3)}{1}\) = 3