10th Class Abhyasa Deepika Maths Solutions 3 Polynomials Telangana SSC (English & Telugu Medium)

Are you searching for 10th class abhyasa deepika maths solutions 3 Polynomials Telangana SSC? Chapter 3 – Polynomials is one of the most important chapters in SSC Mathematics. It forms the base for many upcoming concepts and carries direct questions in the public exam.

In this guide, you will find clear explanations, important formulas, key answer highlights, and smart preparation tips for both English and Telugu medium students.

10 Abyasa Deepika Solutions

1. If the sum of the zeroes of the polynomial P(x) = 2x³ – 3kx² + 4x – 5 is 6, then the value of ‘k’ is
a) 2
b) 4
c) -2
d) -4

View Answer

Solution: For a cubic polynomial $ax^3 + bx^2 + cx + d$

, the sum of the zeroes is given by $-\frac{b}{a}$

.
Given: $a = 2, b = -3k$

.
Sum of zeroes = $-\frac{(-3k)}{2} = \frac{3k}{2}$

.
According to the problem, $\frac{3k}{2} = 6$

.
$3k = 12 \implies k = 4$

.
Correct Option: b

2. Among the following, is not a polynomial
a) $\sqrt{3}x^2 - 2\sqrt{3}x + 3$
b) $\frac{3}{2}x^3 - 5x^2 - \frac{1}{\sqrt{2}}x - 1$
c) $x + \frac{1}{x} (x \neq 0)$
d) $5x^2 - 3x + \sqrt{2}$

View Answer

Solution: An algebraic expression is not a polynomial if the variable has a negative or fractional exponent in any term. In option (c), $\frac{1}{x}$

is $x^{-1}$

, which has a negative exponent.
Correct Option: c

3. If P(x) = 3x⁴ – 5x³ + x² + 8, then the value of P(-1) is
a) 2
b) 15
c) 17
d) -17

View Answer

Solution: Substitute $x = -1$

into the polynomial:
$P(-1) = 3(-1)^4 - 5(-1)^3 + (-1)^2 + 8$

$P(-1) = 3(1) - 5(-1) + 1 + 8$

$P(-1) = 3 + 5 + 1 + 8 = 17$

.
Correct Option: c

4. Degree of the polynomial P(x) = x³ – 2x² – \sqrt{3}x + \frac{1}{2} is
a) $\frac{1}{2}$
b) 2
c) 3
d) 4

View Answer

Solution: The degree of a polynomial is the highest power of the variable $x$

in the expression. Here, the highest power is 3.
Correct Option: c

5. If the sum and product of the zeroes of a quadratic polynomial are ‘2’ and ‘-15’ respectively, then one of the form of the quadratic polynomial is
a) $x^2 - 2x + 15$
b) $x^2 - 2x - 15$
c) $x^2 + 2x - 15$
d) $x^2 + 2x + 15$

View Answer

Solution: A quadratic polynomial is given by $x^2 - (\text{sum of zeroes})x + (\text{product of zeroes})$

.
Given: Sum = 2, Product = -15.
Polynomial: $x^2 - (2)x + (-15) = x^2 - 2x - 15$

.
Correct Option: b

6. If one zero of 2x² − 8x − k is 5/2, find k.

View Answer

Solution:

If 5/2 is zero, substitute in polynomial:
2(5/2)² − 8(5/2) − k = 0
= 2(25/4) − 20 − k
= 25/2 − 20 − k
Convert 20 into fraction:
20 = 40/2
So,
25/2 − 40/2 − k = 0
−15/2 − k = 0
⇒ k = −15/2

Answer: −15/2

7. Find coefficient of x² in P(x) = 2x³ − 5x² − 3x + 7.

View Answer

Solution:

Coefficient of x² = −5

Answer: −5

8. Find zeroes of P(x) = x² − 9.

View Answer

Solution:

x² − 9 = 0
Using identity:
a² − b² = (a − b)(a + b)
= (x − 3)(x + 3) = 0
So,
x = 3
x = −3

Zeroes are 3 and −3.

9. Find zeroes of P(x) = x² − 7 and verify relation.

View Answer

Solution:

x² − 7 = 0
x² = 7
x = ±√7
Zeroes are √7 and −√7
Now verify:
For ax² + bx + c,
a = 1, b = 0, c = −7
Sum of zeroes = −b/a = 0
Actual sum:
√7 + (−√7) = 0 ✔
Product of zeroes = c/a = −7
Actual product:
√7 × (−√7) = −7 ✔
Verified.

10. If P(x) = x² − 5x + 6, find P(−1), P(0), P(1).

View Answer

Solution:

P(−1) = (−1)² − 5(−1) + 6
= 1 + 5 + 6
= 12

P(0) = 6
P(1) = 1 − 5 + 6
= 2

Answer: 12, 6, 2

11. Write quadratic polynomial with given sum and product of zeroes.

i) 1/4, −1
ii) 0, 5

View Answer

Solution:

Formula:
x² − (Sum)x + Product
i) x² − (1/4)x − 1

To remove fraction multiply by 4:
4x² − x − 4
ii) x² − 0x + 5
= x² + 5

12. Write quadratic polynomials with given zeroes.

View Answer

Solution:

If zeroes are α and β, polynomial:
(x − α)(x − β)

i) 2, 3
(x − 2)(x − 3)
= x² − 5x + 6

ii) 1/4, 1

(x − 1/4)(x − 1)
= x² − (5/4)x + 1/4
Multiply by 4:
4x² − 5x + 1

iii) −√2, √2

(x + √2)(x − √2)
= x² − 2

iv) 4+√5, 4−√5

= (x − 4 − √5)(x − 4 + √5)
= (x − 4)² − 5
= x² − 8x + 11

v) −3, 3

(x + 3)(x − 3)
= x² − 9

13. Verify that 3, −1 and −1/3 are zeroes of P(x) = 3x³ − 5x² − 11x − 3.

View Answer

Solution:

Substitute x = 3:
P(3) = 81 − 45 − 33 − 3
= 0
Substitute x = −1:
P(−1) = −3 − 5 + 11 − 3
= 0

Substitute x = −1/3:
P(−1/3) = 0

Hence verified.

14. Complete table for P(x) = 6x² − 13x + 6.

View Answer

Solution:

Substitute values:

x = −2 → 56
x = −1 → 25
x = 0 → 6
x = 1 → −1
x = 2 → 4

These points are used to draw graph.

15. Find zeroes from graph.

View Answer

Solution:

i) x² + 4x + 4 = (x + 2)² → zero = −2

ii) 5x − x² − 6 = 0
x² − 5x + 6 = 0
(x − 2)(x − 3)
Zeroes: 2, 3

iii) x² − 3x + 2
= (x − 1)(x − 2)
Zeroes: 1, 2

iv) x² + x + 1
Discriminant < 0
No real zero

v) x³ − 3x
= x(x² − 3)
Zeroes: 0, √3, −√3

vi) x² − 16
= (x − 4)(x + 4)
Zeroes: −4, 4

16. Prove that 2x⁴ − 3x³ − 3x² + 6x − 2 is divisible by x² − 2.

View Answer

Solution:

On dividing polynomial by x² − 2 using long division, remainder = 0.

Hence exactly divisible.

17. Find remainder when 3x³ + x² + 3x + 5 is divided by x² + 2x + 1.

View Answer

Solution:
Divide polynomial:
Remainder = 6x + 4

18. If volume is x³ + 9x² + 26x + 24 and one dimension is x+3, find remaining dimensions.

View Answer

Solution:

Factor polynomial:
x³ + 9x² + 26x + 24
= (x + 3)(x + 2)(x + 4)
Remaining dimensions:
x + 2
x + 4

19. Find other zeroes of 2x⁴ − 3x³ − 5x² + 9x − 3 if two zeroes are −√3 and √3.

View Answer

Solution:

Since ±√3 are zeroes,
(x² − 3) is factor.
Divide polynomial by (x² − 3):
Remaining quadratic:
2x² − 3x + 1
Factor:

(2x − 1)(x − 1)
Other zeroes:
1/2 and 1

20. If zeroes of x³ − 9x² + 23x − 15 are in arithmetic progression, find them.

View Answer

Solution:
Factor polynomial:
= (x − 1)(x − 3)(x − 5)

Zeroes:
1, 3, 5
These are in AP.

10 Abyasa Deepika Solutions

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10th Class Abhyasa Deepika Maths Solutions 3 Polynomials Telangana SSC (English & Telugu Medium) – Complete Key & Exam Guide

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