Explore topic-wise InterviewSolutions in .

We know that The zero polynomial is the additive identity of the additive group of polynomials.

Zero of zero polynomial is not defined as the zero polynomial cannot have a variable

i.e 0 does not have a variable and equating it to zero will give no result.

1. c) ‘a’ is a non zero real number and b and c are any real numbers.

2. d) D = 0

3. b)

4. c) Neither touches nor intersects x‐axis.

5. c) k (x² + px - \(\cfrac1p\))

Yes, because every quadratic polynomial has at the most two zeros.

The shape of a quadratic polynomial is either upward or downward U - shaped curve i.e., an upward or downward parabola. Also, the graph of the quadratic equation cuts the X - axis at the most at two points, but in fig(D) it cuts the X - axis at three points.

∴ Fig (D) is not the graph of a quadratic polynomial.

If the graph of a polynomial intersects the x - axis at only one point it need not be a quadratic polynomial: True.

The quadratic polynomial cuts the x - axis at most at two points i.e. it can either touch x - axis at two points or one point or doesn’t touch the axis at all.

Hence the polynomial intersecting the x - axis at only one point need not be a quadratic polynomial.

A. has no linear term and the constant term is negative

Let p(x) = x² + ax + b

And let α be one of the zeroes

∴ - α is the other zero of the polynomial p(x)

Product of the zeroes = constant term ÷ coefficient of x²

Product of the zeroes = b

α(- α) = b

- α² = b

i.e. b is negative.

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

α - α = - a

0 = - a

⇒ a = 0

∴ The polynomials can be written as x² - α² (No linear term and negative constant term)

If the zeroes of a quadratic polynomial ax² + bx + c are both positive, then a, b and c all have the same sign: False.

Let α, β be the zeroes of the polynomial p(x) = ax² + bx + c

Sum of the zeroes = - (coefficient of x) ÷ coefficient of x²

α + β = - b/a > 0 (∵ α > 0, β > 0 ⇒ α + β > 0)

∴ for – b/a > 0, b and a must have opposite signs.

Product of the zeroes = constant term ÷ coefficient of x²

αβ = c/a > 0 (∵ α,β > 0⇒ αβ > 0)

∴ for c/a > 0, c and a must have same signs.

Case 1: when a > 0

⇒ - b > 0 and c > 0

= b < 0 and c > 0

Case 2: when a < 0

⇒ - b < 0 and c < 0

= b > 0 and c < 0

Hence, the coefficients have different signs.

Given: 3 is one of zeroes of the polynomial 2x² + x + k , which means x = 3 will satisfy it.

2x² + x + k = 0

2(3)² + 3 + k = 0

18 + 3+ k = 0

k = -21

The value of k is -21

Given: 2 is one of zeroes of the polynomial kx² + 3x + k , which means x = 2 will satisfy it.

k(2)² + 3( 2) + k = 0

4k + 6 + k= 0

5k + 6 = 0

k = -6 / 5

The value of k is -6 / 5

Given: 1 is one of zeroes of the polynomial ax² ‒ 3 (a ‒ 1) x ‒ 1 , which means x = 1 will satisfy it.

a(1)² – 3(a – 1) x 1 – 1 = 0

a – 3a + 3 – 1 = 0

-2a + 2 = 0

a = 1

The value of a is 1.

By using the relationship between the zeroes of the quadratic polynomial. 

We have 

Sum of zeroes = \(\frac{-(coefficient\,of\,x)}{(coefficient\,of\,x^2)}\)

⇒ 1 = \(\frac{-(-3)}k\)

⇒ k = 3

Let f(x) = x² – x – 6

= x² – 3x + 2x – 6

= x(x – 3) + 2(x – 3)

= (x – 3)(x + 2)

To find the zeros of f(x), let sat f(x) = 0, we get

(x – 3)(x + 2) = 0

Either x – 3 = 0 or x + 2 = 0

x = 3 or x = -2

Therefore, 3, -2 are zeros.

Given: -2 is one of zeroes of the polynomial 3x² + 4x + 2k , which means x = -2 will satisfy it.

3(-2)² + 4(-2) + 2k = 0

12 – 8 + 2k = 0

4 + 2k=0

k = -2

The value of k is -2.

Given: 

x = 1 is one zero of the polynomial ax² – 3(a – 1) x – 1 

Therefore, 

it will satisfy the above polynomial. 

Now, we have 

a(1)² – (a – 1)1 – 1 = 0 

⇒ a – 3a + 3 – 1 = 0 

⇒ –2a = – 2 

⇒ a = 1

f(x) = x² – x – 6 

= x² – 3x + 2x – 6 

= x(x – 3) + 2(x – 3) 

= (x – 3) (x + 2) 

f(x) = 0 

⇒ (x – 3) (x + 2) = 0

⇒ (x – 3) = 0 or (x + 2) = 0 

⇒ x = 3 or x = –2 

So, 

the zeroes of f(x) are 3 and –2.

Given: 

x = –4 is one zero of the polynomial x² – x – (2k + 2) 

Therefore, 

it will satisfy the above polynomial. 

Now, we have 

(–4)² – (–4) – (2k + 2) = 0 

⇒ 16 + 4 – 2k – 2 = 0 

⇒ 2k = – 18 

⇒ k = 9

Given, Polynomial

3x² + 4x + 2k

Zero of the polynomial = – 2

As we have zero,

f(x) = 3x² + 4x + 2k = 0

By putting x = – 2

f(– 2) = 3(– 2)² + 4 (– 2) + 2k = 0

f(– 2) = 12 – 8 + 2k = 0

= 4 + 2k = 0

= 2k = – 4

k = – 4/2 = – 2

So we have the value of k = – 2

Given, A polynomial

x² – x – (2k + 2)

– 4 is a zero of the given polynomial.

As – 4 is the zero,

so at x = -4,

the value of the polynomial

x² – x – (2k + 2) will be 0.

So we get,

⇒x ² – x – (2k + 2) = 0

⇒(– 4)² – (– 4) – (2k + 2) = 0

⇒16 + 4 – (2k + 2) = 0

⇒20 – (2k + 2) = 0

⇒ – (2k + 2) = – 20

⇒2k + 2 = 20

⇒ 2k = 20 - 2

⇒2k = 18

⇒ k = 18/2 = 9

Given: 

x = –2 is one zero of the polynomial 3x² + 4x + 2k 

Therefore, 

it will satisfy the above polynomial. 

Now, we have 

3(–2)² + 4(–2)1 + 2k = 0 

⇒ 12 – 8 + 2k = 0 

⇒ k = – 2

Given: 

x = 3 is one zero of the polynomial 2x² + x + k 

Therefore, 

it will satisfy the above polynomial. 

Now, 

we have 2(3)² + 3 + k = 0 

⇒ 21 + k = 0 

⇒ k = – 21

(a) k = 3 

Let α and \(\frac{1}α\) be the zeroes of 3x² – 8x + k. 

Then the product of zeroes = \(\frac{k}3\)

⇒ α × \(\frac{1}α\) = \(\frac{k}3\)

⇒ 1 = \(\frac{k}3\)

⇒ k = 3

Given: 

x = 2 is one zero of the quadratic polynomial kx² + 3x + k 

Therefore, 

it will satisfy the above polynomial. 

Now, we have 

k(2)² + 3(2) + k = 0 

⇒ 4k + 6 + k = 0 

⇒ 5k + 6 = 0

⇒ k = - \(\frac{6}5\)

(c) a = –2, b = –6 

Given: 

–2 and 3 are the zeroes of x² + (a + 1) x + b. 

Now,

 (–2)² + (a + 1) × (–2) + b = 0 

⇒ 4 – 2a – 2 + b = 0 

⇒ b – 2a = –2 ….(1) 

Also, 3² + (a + 1) × 3 + b = 0 

⇒ 9 + 3a + 3 + b = 0 

⇒ b + 3a = –12 ….(2)

On subtracting (1) from (2), 

we get a = –2 

∴ b = –2 – 4 = –6 [From (1)]

Correct option is (B) x² – 3

\((x-\sqrt3)\,(x+\sqrt3)\) \(=x^2-(\sqrt3)^2\)

\(=x^2-3\)

Correct option is B) x² – 3

Correct option is (A) 13/36

\(\because\) \(\alpha+\beta\) \(=\frac{-(-5)}1\) = 5 and \(\alpha\beta\) \(=\frac61\) = 6

\(\therefore\) \(\frac{\alpha+\beta}{\alpha\beta}=\frac56\)

\(\Rightarrow\) \(\frac{1}{\alpha}+\frac{1}{\beta}\) \(=\frac56\)

\(\therefore\) \((\frac{1}{\alpha}+\frac{1}{\beta})^2\) \(=(\frac56)^2\)

\(\Rightarrow\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}+\frac2{\alpha\beta}\) \(=\frac{25}{36}\)

\(\Rightarrow\) \(\frac{1}{\alpha^2}+\frac{1}{\beta^2}\) \(=\frac{25}{36}-\frac2{\alpha\beta}\)

\(=\frac{25}{36}-\frac26\) \(=\frac{25-12}{36}=\frac{13}{36}\)

Correct option is A) 13/36

Since –4 is a zero of (k – 1) x² + kx + 1, 

we have: 

(k – 1) × (-4)² + k × (-4) + 1 = 0 

⇒ 16k – 16 – 4k + 1 = 0 

⇒ 12k – 15 = 0

⇒ k = \(\frac{15}{12}\)

⇒ k = \(\frac{5}{4}\)

(i) x² – 4 = (x + 2)(x – 2)

x² + 4x + 4 = (x + 2)²

H.C.F. of coefficient = 1

H.C.F. of other factors = (x + 2)¹ = x + 2

H.C.F. = x + 2

(ii) 4x⁴ – 16x³ + 12x² = 4x²(x² – 4x + 3) 

= 4x²(x – 1)(x – 3)

6x³ + 6x² – 72x = 6x(x² + x – 12) 

= 6x(x + 4)(x – 3)

= 2x(x – 3) [because H.C.F. of coefficients is 2]

= 2x² – 6x

Answer is (A) 2a²

\(x = {-b \pm \sqrt{b^2-4ac} \over 2a}\)

Answer is (C) q² = 4pr

Answer is (A) b² < 4ac

Discriminant (D) = b² – 4ac

Correct option is (B) -1

Given that \(\alpha\;and\;\beta\) are zeros of \(x^2+x+1.\)

\(\therefore\) Sum of roots \(=\frac{\text{-coefficient of x}}{\text{coefficient of }x^2}=\frac{-1}1\) = -1

\(\Rightarrow\) \(\alpha+\beta\) = -1

And product of zeros \(=\frac{\text{constant term}}{\text{coefficient of }x^2}=\frac11\) = 1

\(\Rightarrow\) \(\alpha\beta=1\)

Now, \(\frac{\alpha+\beta}{\alpha\beta}=\frac{-1}1\) = -1

\(\Rightarrow\) \(\frac1\alpha+\frac1\beta\) = -1

Correct option is B) -1

(i) If (b² – 4ac) > 0, then roots will be real and distinct.

(ii) If (b² – 4ac) = 0, then roots will be equal and real.

(iii) If (b² – 4ac) < 0, then roots will be imaginary.

Answer: (D) 4, -2

We have,
f(x) = x²-2x -8
Now, put f(x) = 0
x²-2x -8 = 0
x²- 4x + 2x -8 = 0
x(x - 4) + 2(x - 4) = 0

(x - 4) (x + 2) = 0
x - 4 = 0 or x + 2 = 0
x = 4 or x = -2
So, the zeros of given polynomial are 4 and -2.

Correct option is (B) -6

Given that roots of \(x^2+6x+5=0\) are \(\alpha\;and\;\beta.\)

\(\therefore\) Sum of roots \(=\frac{\text{-coefficient of x}}{\text{coefficient of }x^2}=\frac{-6}1\) = -6

\(\Rightarrow\) \(\alpha+\beta\) = -6

Correct option is B) -6

Answer:(A) 2, -5

We have,
f(x) = x² + 3x - 10
Now, put f(x) = 0
x² + 3x - 10 = 0
x² + 5x - 2x - 10 = 0
x(x + 5) - 2(x + 5) = 0
(x - 2) (x + 5) = 0
x - 2 = 0 or x + 5 = 0
x = 2 or x = -5
So, the zeros of given polynomial are 2 and -5.

Let the first natural number be x.

Sum of two natural numbers is 8 then other natural numbers will be 8 – x.

According to question.

Product of both natural numbers = 15

⇒ x (8 – x) = 15

⇒ 8x – x² = 15

⇒ x² – 8x + 15 = 0

⇒ x² – (5 + 3)x + 15 = 0

⇒ x² – 5x – 3x + 15 = 0

⇒ (x² – 5x) – (3x – 15) = 0

⇒ x (x – 5) – 3 (x – 5) = 0

⇒ (x – 5) (x – 3) = 0

⇒ x – 5 = 0 or x – 3 = 0

⇒ x = 5 or x = 3

Thus, if First natural no. = 5

then Second natural no. = 8

if First natural no. = 3

or Second natural no. = 8

Correct option is (D) -2

p(4) = 0

\(\Rightarrow\) \(4^2+4k-8=0\)

\(\Rightarrow4k=8-4^2\) = 8 - 16 = -8

\(\Rightarrow\) \(k=\frac{-8}4=-2\)

Correct option is D) -2

Answer:(C) 3/2, -1/2

We have,
f(x) = 4x² - 4x - 3
Now, put f(x) = 0
4x² - 4x - 3 = 0
4x² - 6x + 2x - 3 = 0
2x(2x - 3) + 1(2x - 3) = 0
(2x - 3) (2x + 1) = 0
2x - 3 = 0 or 2x + 1 = 0
2x = 3 or 2x = -1
x = 3/2 or x = -1/2 
So, the zeros of given polynomial are 3/2 and -1/2.