Polynomial Problems and Solutions

1-Mark Questions

If x ≠ –1, then find the quotient of (x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) (M'15)

Step 1: Factor numerator and denominator
Numerator: x⁵ + x⁴ + x³ + x² = x²(x³ + x² + x + 1)
Denominator: x³ + x² + x + 1

Step 2: Simplify the expression
(x⁵ + x⁴ + x³ + x²) / (x³ + x² + x + 1) = [x²(x³ + x² + x + 1)] / (x³ + x² + x + 1)

Step 3: Cancel common factors
Since x ≠ -1, (x³ + x² + x + 1) ≠ 0, so we can cancel:
= x²

∴ Quotient = x²

"We can write a trinomial having degree 7". Justify the above statement by giving one example. (M'15)

Step 1: Understand the terms
- A trinomial is a polynomial with exactly three terms.
- Degree of a polynomial is the highest power of the variable.

Step 2: Create an example
Example: 3x⁷ + 2x³ + 5
This has:
- Three terms: 3x⁷, 2x³, and 5
- Highest power: 7
- Therefore, it's a trinomial of degree 7.

∴ Example: 3x⁷ + 2x³ + 5
This is a trinomial with degree 7, which justifies the statement.

Write an example for a quadratic Polynomial that has no zeros. (M'16)

Step 1: Condition for no zeros
A quadratic polynomial has no real zeros when its discriminant (D = b² - 4ac) is negative.

Step 2: Create an example
Let's take p(x) = x² + 2x + 3
Discriminant D = b² - 4ac = (2)² - 4(1)(3) = 4 - 12 = -8 < 0
Since D < 0, this quadratic has no real zeros.

∴ Example: x² + 2x + 3
This quadratic polynomial has no real zeros as its discriminant is negative.

If p(x) = x³ – 3x² + 2x – 3 is a polynomial, then find the value of p(1). (J'16)

Step 1: Substitute x = 1
p(1) = (1)³ - 3(1)² + 2(1) - 3

Step 2: Calculate
p(1) = 1 - 3 + 2 - 3

Step 3: Simplify
p(1) = (1 + 2) + (-3 - 3) = 3 - 6 = -3

∴ p(1) = -3

Srikar says that the order of the polynomial (x² – 5)(x³ + 1) is 6. Do you agree with him? (J'17)

Step 1: Expand the polynomial
(x² - 5)(x³ + 1) = x²(x³ + 1) - 5(x³ + 1) = x⁵ + x² - 5x³ - 5

Step 2: Rearrange in standard form
= x⁵ - 5x³ + x² - 5

Step 3: Find the degree
The highest power of x is 5, so the degree (order) is 5.

∴ I don't agree with Srikar.
The order of the polynomial is 5, not 6.

Find zeros of the polynomial P(x) = x² – 4. (J'17)

Step 1: Set P(x) = 0
x² - 4 = 0

Step 2: Solve for x
x² = 4
x = ±√4 = ±2

∴ Zeros are x = 2 and x = -2

Verify the relation between zeros and coefficients of the quadratic polynomial x² – 4. (M'18)

Step 1: Identify coefficients
For x² - 4:
a = 1, b = 0, c = -4

Step 2: Find zeros
Zeros: α = 2, β = -2

Step 3: Verify sum of zeros
α + β = 2 + (-2) = 0
-b/a = -0/1 = 0
∴ α + β = -b/a

Step 4: Verify product of zeros
αβ = (2)(-2) = -4
c/a = -4/1 = -4
∴ αβ = c/a

∴ The relation between zeros and coefficients is verified:
Sum of zeros = -b/a = 0
Product of zeros = c/a = -4

Whether 1/2 and 1 are zeros of the polynomial p(x) = 2x² – 3x + 1 or not? Justify. (J'18)

Step 1: Check if 1/2 is a zero
p(1/2) = 2(1/2)² - 3(1/2) + 1 = 2(1/4) - 3/2 + 1 = 1/2 - 3/2 + 1 = -1 + 1 = 0
∴ 1/2 is a zero.

Step 2: Check if 1 is a zero
p(1) = 2(1)² - 3(1) + 1 = 2 - 3 + 1 = 0
∴ 1 is a zero.

∴ Both 1/2 and 1 are zeros of the polynomial p(x) = 2x² - 3x + 1.

If P(x) = x⁴ + 1, then find P(2) – P(-2). (M'19)

Step 1: Calculate P(2)
P(2) = (2)⁴ + 1 = 16 + 1 = 17

Step 2: Calculate P(-2)
P(-2) = (-2)⁴ + 1 = 16 + 1 = 17

Step 3: Find P(2) - P(-2)
P(2) - P(-2) = 17 - 17 = 0

∴ P(2) - P(-2) = 0

–3, 0 and 2 are the zeroes of the polynomial p(x) = x³ + (a – 1)x² + bx + c. Find a and c. (J'19)

Step 1: Use the zero x = 0
p(0) = (0)³ + (a - 1)(0)² + b(0) + c = c
Since 0 is a zero, p(0) = 0, so c = 0

Step 2: Use the zero x = 2
p(2) = (2)³ + (a - 1)(2)² + b(2) + 0 = 8 + 4(a - 1) + 2b = 0
8 + 4a - 4 + 2b = 0
4a + 2b + 4 = 0
2a + b + 2 = 0 ...(1)

Step 3: Use the zero x = -3
p(-3) = (-3)³ + (a - 1)(-3)² + b(-3) + 0 = -27 + 9(a - 1) - 3b = 0
-27 + 9a - 9 - 3b = 0
9a - 3b - 36 = 0
3a - b - 12 = 0 ...(2)

Step 4: Solve equations (1) and (2)
From (1): b = -2a - 2
Substitute in (2): 3a - (-2a - 2) - 12 = 0
3a + 2a + 2 - 12 = 0
5a - 10 = 0
a = 2

∴ a = 2, c = 0

Write any two linear polynomials having one term under three terms. (J'19)

Step 1: Understand linear polynomials
A linear polynomial has degree 1 and can have 1, 2, or 3 terms.

Step 2: Examples with one term
1. 5x (degree 1, one term)
2. -3x (degree 1, one term)

Step 3: Examples with three terms
1. 2x + 3y + 4 (degree 1 in x and y, three terms)
2. x + y - 2 (degree 1 in x and y, three terms)

∴ One-term examples: 5x, -3x
Three-term examples: 2x + 3y + 4, x + y - 2

If p(x) = x² + 3x + 4, then find the values of p(0) and p(1). (May 2022)

Step 1: Calculate p(0)
p(0) = (0)² + 3(0) + 4 = 0 + 0 + 4 = 4

Step 2: Calculate p(1)
p(1) = (1)² + 3(1) + 4 = 1 + 3 + 4 = 8

∴ p(0) = 4, p(1) = 8

If p(x) = 2x² + 5x – 7, then find the value of p(0) and p(1). (Aug 2022)

Step 1: Calculate p(0)
p(0) = 2(0)² + 5(0) - 7 = 0 + 0 - 7 = -7

Step 2: Calculate p(1)
p(1) = 2(1)² + 5(1) - 7 = 2 + 5 - 7 = 0

∴ p(0) = -7, p(1) = 0

Polynomial Problems - 2 Marks

Polynomial-2 Marks

For what value of k, –4 is a zero of the polynomial x² – x – (2k + 2). (J'15)

Step 1: Substitute x = -4 in the polynomial
p(x) = x² - x - (2k + 2)
p(-4) = (-4)² - (-4) - (2k + 2)

Step 2: Simplify the expression
p(-4) = 16 + 4 - 2k - 2 = 18 - 2k

Step 3: Set p(-4) = 0 (since -4 is a zero)
18 - 2k = 0

Step 4: Solve for k
2k = 18
k = 9

∴ k = 9

Use the table given below to draw the graph. Use the graph drawn to find the values of a and b. (J'15)

Given table:

x	-2	0	2	1	b
y	-3	1	a	3	-7

Step 1: Assume the relationship is linear (y = mx + c)
Using points (-2, -3) and (0, 1):
When x = 0, y = 1 ⇒ c = 1
Using (-2, -3): -3 = m(-2) + 1 ⇒ -3 = -2m + 1 ⇒ -2m = -4 ⇒ m = 2
∴ Equation: y = 2x + 1

Step 2: Find value of a
When x = 2: y = 2(2) + 1 = 4 + 1 = 5
∴ a = 5

Step 3: Find value of b
When y = -7: -7 = 2b + 1 ⇒ 2b = -8 ⇒ b = -4

∴ a = 5, b = -4

Length of a rectangle is 5 units more than its breadth. Express its perimeter in polynomial form.

Step 1: Define variables
Let breadth = x units
Then length = (x + 5) units

Step 2: Write perimeter formula
Perimeter = 2 × (length + breadth)
= 2 × [(x + 5) + x]
= 2 × (2x + 5)

Step 3: Simplify to polynomial form
= 4x + 10

∴ Perimeter = 4x + 10

Show that 2 and −1/3 are zeros of the polynomial 3x² – 5x – 2. (J'16)

Given polynomial: p(x) = 3x² - 5x - 2

Step 1: Check if x = 2 is a zero
p(2) = 3(2)² - 5(2) - 2 = 3(4) - 10 - 2 = 12 - 10 - 2 = 0
∴ 2 is a zero.

Step 2: Check if x = -1/3 is a zero
p(-1/3) = 3(-1/3)² - 5(-1/3) - 2 = 3(1/9) + 5/3 - 2 = 1/3 + 5/3 - 2 = 6/3 - 2 = 2 - 2 = 0
∴ -1/3 is a zero.

∴ Both 2 and -1/3 are zeros of the polynomial 3x² - 5x - 2.

Which of √2 and 2 is a zero of the polynomial p(x) = x³ – 2x? Why? (M'17)

Given polynomial: p(x) = x³ - 2x

Step 1: Check if x = √2 is a zero
p(√2) = (√2)³ - 2(√2) = 2√2 - 2√2 = 0
∴ √2 is a zero.

Step 2: Check if x = 2 is a zero
p(2) = (2)³ - 2(2) = 8 - 4 = 4 ≠ 0
∴ 2 is not a zero.

∴ √2 is a zero of the polynomial, but 2 is not.

Divide x³ – 3x² + 5x – 3 by x² – 2. And verify the division lemma. (J'17)

Division:

         x - 3
        -----------
x² - 2 | x³ - 3x² + 5x - 3
          x³ + 0x² - 2x
        -----------
             -3x² + 7x - 3
             -3x² + 0x + 6
             -----------
                  7x - 9

Step 1: Result of division
Quotient = x - 3
Remainder = 7x - 9

Step 2: Verify division lemma
Division lemma: Dividend = Divisor × Quotient + Remainder
(x² - 2)(x - 3) + (7x - 9) = x³ - 3x² - 2x + 6 + 7x - 9 = x³ - 3x² + 5x - 3
This matches the original dividend.

∴ Quotient = x - 3, Remainder = 7x - 9
The division lemma is verified.

Complete the following table for the polynomial y = p(x) = x³ – 2x + 3. (M'18)

Given polynomial: y = x³ - 2x + 3

x	-1	0	1	2
x³	(-1)³ = -1	0³ = 0	1³ = 1	2³ = 8
-2x	-2(-1) = 2	-2(0) = 0	-2(1) = -2	-2(2) = -4
3	3	3	3	3
y	-1 + 2 + 3 = 4	0 + 0 + 3 = 3	1 - 2 + 3 = 2	8 - 4 + 3 = 7
(x, y)	(-1, 4)	(0, 3)	(1, 2)	(2, 7)

∴ The completed table is shown above.

If one of the zeros of the cubic polynomial p(x) = ax³ + bx² + cx + d is zero, then find the product of other two zeros of p(x). (a ≠ 0) (J'18)

Step 1: Let the zeros be α, β, γ with γ = 0
So p(x) = a(x - α)(x - β)(x - 0) = a(x - α)(x - β)x

Step 2: Expand the polynomial
p(x) = a[x³ - (α + β)x² + αβx] = ax³ - a(α + β)x² + aαβx

Step 3: Compare with given form
p(x) = ax³ + bx² + cx + d
Comparing coefficients:
-a(α + β) = b ⇒ α + β = -b/a
aαβ = c ⇒ αβ = c/a
d = 0 (since constant term is 0 when one zero is 0)

Step 4: Product of other two zeros
The product of the other two zeros = αβ = c/a

∴ Product of other two zeros = c/a

Divide x³ – 4x² + 5x – 2 by x – 2. (M'19)

Division using synthetic division:

         x² - 2x + 1
        -----------
x - 2 | x³ - 4x² + 5x - 2
          x³ - 2x²
        -----------
             -2x² + 5x
             -2x² + 4x
             -----------
                   x - 2
                   x - 2
                   -----
                     0

Step 1: Result of division
Quotient = x² - 2x + 1
Remainder = 0

Step 2: Verify
(x - 2)(x² - 2x + 1) = x³ - 2x² - 2x² + 4x + x - 2 = x³ - 4x² + 5x - 2
This matches the original dividend.

∴ Quotient = x² - 2x + 1, Remainder = 0

Polynomial Problems - 4 Mark Questions

Lakshmi does not want to disclose the length, breadth and height of a cuboid of her project. She has constructed a polynomial x³ – 6x² + 11x – 6 by taking the values of length, breadth and height as its zeros. Can you open the secret [i.e., find the measures of length, breadth and height]? (M'15)

Step 1: Find the zeros of the polynomial
Given polynomial: p(x) = x³ - 6x² + 11x - 6

Step 2: Check for possible rational zeros
Possible zeros: ±1, ±2, ±3, ±6

Step 3: Check x = 1
p(1) = 1 - 6 + 11 - 6 = 0
∴ x = 1 is a zero.

Step 4: Factor out (x - 1)
Using synthetic division:
Coefficients: 1, -6, 11, -6
Dividing by (x - 1):
Bring down 1, multiply by 1 → 1, add to -6 → -5, multiply by 1 → -5, add to 11 → 6, multiply by 1 → 6, add to -6 → 0
Quotient: x² - 5x + 6

Step 5: Factor the quadratic
x² - 5x + 6 = (x - 2)(x - 3)

Step 6: Complete factorization
p(x) = (x - 1)(x - 2)(x - 3)

∴ The zeros are 1, 2, and 3
Therefore, the dimensions of the cuboid are:
Length = 3 units, Breadth = 2 units, Height = 1 unit (or any permutation of these values).

Draw the graph for the polynomial p(x) = x² + 3x - 4 and find its zeroes from the graph. (M'15, J'19)

Step 1: Create a table of values

x	-5	-4	-3	-2	-1	0	1	2
p(x)	6	0	-4	-6	-6	-4	0	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1.5, -6.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -4 and x = 1.

Step 4: Verify algebraically
x² + 3x - 4 = 0
(x + 4)(x - 1) = 0
x = -4 or x = 1

∴ The zeros are x = -4 and x = 1

Draw the graph of the polynomial p(x) = 3x² + 2x - 1 on the graph paper. Find its zeros from the graph. (J'15)

Step 1: Create a table of values

x	-2	-1	-0.5	0	0.5	1	2
p(x)	7	0	-1.25	-1	0.75	4	15

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1/3, -4/3).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 1/3.

Step 4: Verify algebraically
3x² + 2x - 1 = 0
(3x - 1)(x + 1) = 0
x = 1/3 or x = -1

∴ The zeros are x = -1 and x = 1/3

Draw the graph for the polynomial p(x) = x² - 3x + 2 and find the zeroes from the graph. (M'16)

Step 1: Create a table of values

x	-1	0	1	1.5	2	3	4
p(x)	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 2.

Step 4: Verify algebraically
x² - 3x + 2 = 0
(x - 1)(x - 2) = 0
x = 1 or x = 2

∴ The zeros are x = 1 and x = 2

Draw the graph of the polynomial p(x) = x² - 5x + 4 on the graph paper. Find its zeros from the graph. (J'16)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	10	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 4.

Step 4: Verify algebraically
x² - 5x + 4 = 0
(x - 1)(x - 4) = 0
x = 1 or x = 4

∴ The zeros are x = 1 and x = 4

On dividing x³ - 3x² + 5x - 7 by x² - 2x + 4, if the remainder is in the form of Ax + B, find the values of A and B. (J'16)

Step 1: Perform polynomial division

         x - 1
        ---------------
x² - 2x + 4 | x³ - 3x² + 5x - 7
          x³ - 2x² + 4x
        ---------------
             -x² + x - 7
             -x² + 2x - 4
             -----------
                  -x - 3

Step 2: Identify remainder
Remainder = -x - 3

Step 3: Compare with Ax + B
-x - 3 = Ax + B
A = -1, B = -3

∴ A = -1, B = -3

Divide 3x⁴ - 5x³ + 4x² + 3x - 5 by x² - 3 and verify the division algorithm. (M'17)

Step 1: Perform polynomial division

         3x² - 5x + 13
        ---------------
x² - 3 | 3x⁴ - 5x³ + 4x² + 3x - 5
         3x⁴ + 0x³ - 9x²
        ---------------
             -5x³ + 13x² + 3x
             -5x³ + 0x² + 15x
             ---------------
                  13x² - 12x - 5
                  13x² + 0x - 39
                  ---------------
                       -12x + 34

Step 2: Identify quotient and remainder
Quotient = 3x² - 5x + 13
Remainder = -12x + 34

Step 3: Verify division algorithm
Division algorithm: Dividend = Divisor × Quotient + Remainder
(x² - 3)(3x² - 5x + 13) + (-12x + 34)
= 3x⁴ - 5x³ + 13x² - 9x² + 15x - 39 - 12x + 34
= 3x⁴ - 5x³ + 4x² + 3x - 5
This matches the original dividend.

∴ Quotient = 3x² - 5x + 13, Remainder = -12x + 34
The division algorithm is verified.

The perimeter of a right-angle triangle is 60 cm and its hypotenuse is 25 cm. Then find the remaining two sides. (M'17)

Step 1: Set up equations
Let the sides be a, b, and c where c = 25 cm (hypotenuse)
Perimeter: a + b + c = 60 ⇒ a + b = 35
Pythagorean theorem: a² + b² = c² = 625

Step 2: Solve the system
From a + b = 35, we get b = 35 - a
Substitute in a² + b² = 625:
a² + (35 - a)² = 625
a² + 1225 - 70a + a² = 625
2a² - 70a + 1225 - 625 = 0
2a² - 70a + 600 = 0
Divide by 2: a² - 35a + 300 = 0

Step 3: Solve the quadratic
a² - 35a + 300 = 0
(a - 15)(a - 20) = 0
a = 15 or a = 20

Step 4: Find the corresponding sides
If a = 15, then b = 35 - 15 = 20
If a = 20, then b = 35 - 20 = 15

∴ The remaining two sides are 15 cm and 20 cm

Draw the graph of the polynomial p(x) = x² - 5x + 6 and find the zeros from the graph. (M'17)

Step 1: Create a table of values

x	-1	0	1	2	2.5	3	4	5
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 2 and x = 3.

Step 4: Verify algebraically
x² - 5x + 6 = 0
(x - 2)(x - 3) = 0
x = 2 or x = 3

∴ The zeros are x = 2 and x = 3

Draw the graph of p(x) = x² - 2x - 8 and find the zeros of the polynomial from it. (J'17)

Step 1: Create a table of values

x	-3	-2	-1	0	1	2	3	4	5
p(x)	7	0	-5	-8	-9	-8	-5	0	7

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (1, -9).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 4.

Step 4: Verify algebraically
x² - 2x - 8 = 0
(x - 4)(x + 2) = 0
x = 4 or x = -2

∴ The zeros are x = -2 and x = 4

Total number of pencils required are given by 4x⁴ + 2x³ - 2x² + 62x - 66. If each box contains x² + 2x - 3 pencils, then find the number of boxes to be purchased. (M'18)

Step 1: Perform polynomial division
We need to divide 4x⁴ + 2x³ - 2x² + 62x - 66 by x² + 2x - 3

Step 2: Division process

         4x² - 6x + 22
        ---------------
x² + 2x - 3 | 4x⁴ + 2x³ - 2x² + 62x - 66
         4x⁴ + 8x³ - 12x²
        ---------------
             -6x³ + 10x² + 62x
             -6x³ - 12x² + 18x
             ---------------
                  22x² + 44x - 66
                  22x² + 44x - 66
                  ---------------
                            0

Step 3: Interpret the result
The quotient is 4x² - 6x + 22 and the remainder is 0.

∴ Number of boxes to be purchased = 4x² - 6x + 22

Draw the graph of the polynomial p(x) = x² + x - 2 on the graph paper. Find its zeroes from the graph. (J'18)

Step 1: Create a table of values

x	-3	-2	-1	-0.5	0	1	2
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -2 and x = 1.

Step 4: Verify algebraically
x² + x - 2 = 0
(x + 2)(x - 1) = 0
x = -2 or x = 1

∴ The zeros are x = -2 and x = 1

Draw the graph of the polynomial p(x) = x² - 7x + 12, then find its zeroes from the graph. (M'19)

Step 1: Create a table of values

x	0	1	2	3	3.5	4	5	6
p(x)	12	6	2	0	-0.25	0	2	6

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (3.5, -0.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 3 and x = 4.

Step 4: Verify algebraically
x² - 7x + 12 = 0
(x - 3)(x - 4) = 0
x = 3 or x = 4

∴ The zeros are x = 3 and x = 4

Draw the graph of the polynomial p(x) = x² + 2x - 3 and find the zeroes of the polynomial from the graph. (May 2022, Jun'23)

Step 1: Create a table of values

x	-4	-3	-2	-1	0	1	2
p(x)	5	0	-3	-4	-3	0	5

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (-1, -4).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -3 and x = 1.

Step 4: Verify algebraically
x² + 2x - 3 = 0
(x + 3)(x - 1) = 0
x = -3 or x = 1

∴ The zeros are x = -3 and x = 1

Draw the graph of the quadratic polynomial p(x) = x² - 4x + 3 and find the zeroes of the polynomial from the graph. (Apr'23)

Step 1: Create a table of values

x	-1	0	1	2	3	4	5
p(x)	8	3	0	-1	0	3	8

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (2, -1).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = 1 and x = 3.

Step 4: Verify algebraically
x² - 4x + 3 = 0
(x - 1)(x - 3) = 0
x = 1 or x = 3

∴ The zeros are x = 1 and x = 3

Draw the graph of the polynomial p(x) = x² - x - 2 and find the zeros of the polynomial from the graph. (Aug 2022)

Step 1: Create a table of values

x	-2	-1	0	0.5	1	2	3
p(x)	4	0	-2	-2.25	-2	0	4

Step 2: Graph description
The graph is a parabola opening upwards with vertex at (0.5, -2.25).

Step 3: Find zeros from the graph
From the table, we see p(x) = 0 when x = -1 and x = 2.

Step 4: Verify algebraically
x² - x - 2 = 0
(x - 2)(x + 1) = 0
x = 2 or x = -1

∴ The zeros are x = -1 and x = 2

Author

zaheerpashamd

M.Sc,B.Ed: SDVR ZPHS B.Gangaram Sathupally(md) Khammam(dt)
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