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Correct option is (A) 0

Let the third zero be \(\alpha.\)

Sum of zeros \(=\frac{-b}a=\frac{-(-5)}1\) = 5

\(\therefore\) \(\alpha+2+3=5\)

\(\Rightarrow\) \(\alpha=5-5=0\)

Thus, the third zero be 0.

Correct option is A) 0

To find the zeroes of the quadratic polynomial we will equate f(x) to 0 

∴f(x) = 0 

⇒ 6x² – 3 = 0 

⇒ 3(2x² – 1) = 0 

⇒ 2x² – 1 = 0 

⇒ 2x² = 1 

⇒ x² = \(\frac{1}2\)

⇒ x = ± \(\frac{1}{\sqrt2}\) 

Hence, 

the zeroes of the quadratic polynomial f(x) = 6x² – 3 are \(\frac{1}{\sqrt2}\), - \(\frac{1}{\sqrt2}\)

Correct option is (D) \(\sqrt2x^{2} + \frac{1}{2}\)

Answer:(A) 1/2, 1/2

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

The polynomial long division is an algorithm for dividing a polynomial by another polynomial of the same or lower degree; it is a generalized version of the familiar arithmetic technique called long division.

It can be done manually because it separates a complex division problem into smaller ones.

Let’s take the Example:

f(x) and g(x) are two polynomials with,

g(x)≠0,

Now we can find the polynomials p(x) and q(x) such that,

f(x) = p(x) × g(x) × q(x)

Where q(x) = 0 or degree of q(x) < is degree of g(x).

The result is;

Dividend = Quotient × Divisor + Remainder

This is known as the Division Algorithm for polynomials.

Answer:(B) 3√2, -2√2

We have,
f(x) = x²-√2x -12
Now, put f(x) = 0
x²-√2x –12 = 0
x² - 3√2x + 2√2x – 12 = 0
x(x - 3√2) + 2√2(x - 3√2) = 0
(x - 3√2)(x + 2√2) = 0
x - 3√2 = 0 or x + 2√2 = 0
x = 3√2 or x = -2√2
So, the zeros of given polynomial are 3√2 and -2√2.

“If f(x) and g(x) are two polynomials such that degree of f(x) is greater than degree of g(x) where g(x) ≠ 0, there exists unique polynomials q(x) and r(x) such that 

f(x) = g(x) × q(x) + r(x),

where r(x) = 0 or degree of r(x) ˂ degree of g(x)

We can find the quadratic polynomial if we know the sum of the roots and product of the roots by using the formula 

x² – (sum of the zeroes)x + product of zeroes 

⇒ x² – (− 1/2 )x + (–3) 

⇒ x² + 1/2x – 3 

Hence, the required polynomial is x² + 1/2x – 3.

Given : Sum of zeros = 5 and product of zeros = 17

So, quadratic polynomial is given by

=> f(x) = k {x² - x(sum of zeros) + product of zeros}

=> f(x) = k{x² - 5x + 17}, where, k is any non-zero real number,

 (d) \(\frac{7}{4}, 0\)

Given, (9x² + 3px + 6q), when divided by (3x + 1) leaves a remainder \(-\frac{3}{4}\)

∴ f(x) = 9x² + 3px + 6q – \(\big(-\frac{3}{4}\big)\) = \(\big(9x^2+3px+6q+\frac{3}{4}\big)\)

is exactly divisible by (3x + 1) 

∴ f\(\big(-\frac{3}{4}\big)\) = 0 ⇒ 9\(\big(-\frac{3}{4}\big)\)² + 3 p . \(\big(-\frac{3}{4}\big)\) + 6q + \(\frac{3}{4}\) = 0

⇒ 6q - p + \(\frac{7}{4}\) 0 

⇒ 24q - 4p + 7 = 0                      ......(i)

Now, the expression g(x) = qx² + 4px + 7 is exactly divisible by x + 1 

⇒ g(–1) = 0 ⇒ q – 4p + 7 = 0                     ...(ii) 

Solving equations (i) and (ii), we get q = 0, p = \(\frac{7}{4}\).

Given, 

p(x) = x² + 2√2x – 6 

We put p(x) = 0 

⇒ x² + 2√2x – 6 = 0 

⇒  x² + 3√2x – √2x – 6 = 0 

⇒ x(x + 3√2) – √2 (x + 3√2) = 0 

⇒ (x – √2)(x + 3√2) = 0 

This gives us 2 zeros, for

x = √2 and x = -3√2 

Hence, the zeros of the quadratic equation are √2 and -3√2. 

Now, for verification 

Sum of zeros = \(\frac{– coefficient\, of\, x}{coefficient\, of\, x^2}\) 

√2 + (-3√2) = – \(\frac{(2\sqrt2)}{1}\) 

-2√2 = -2√2 

Product of roots = \(\frac{constant}{coefficient\, of\, x^2}\)

√2 x (-3√2) = \(\frac{(-6)}{ 2\sqrt2}\) 

-3 x 2 = -6/1

-6 = -6 

Therefore, the relationship between zeros and their coefficients is verified.

(B) 1

Explanation:

According to the question,

p(x) = x² – 2√2x + 1

To get p(2√2),

We substitute x = 2√2,

p(2√2) = (2√2)² – (2√2 × (2√2)) + 1

= (4 × 2) – (4 × 2) + 1

= 8 – 8 + 1

= 1

Hence, option B is the correct answer

B) x³ – 23x² + 142x – 120 

The given zeroes of polynomials in ‘x’ are : 1, 2 and – 1. 

The factors of the polynomial are : x – 1, x – 2 and x + 1 

The required polynomial is : (x – 1) (x – 2) (x + 1) 

= (x² – 1) (x – 2) = x³ – 2x² – x + 2.

(d) 2 

Since α and β are the zeroes of 2x² + 5x + k, 

we have:

α + β = \(\frac{-5}2\) and αβ = \(\frac{k}2\)

Also, it is given that α² + β² + αβ = \(\frac{21}4\)

⇒ (α + β)² – αβ = \(\frac{21}4\)

⇒ \((\frac{-5}2)^2\) - \(\frac{k}2\) = \(\frac{21}4\)

⇒ \(\frac{25}4\)  - \(\frac{k}2\) = \(\frac{21}4\)

⇒ \(\frac{k}2\) = \(\frac{25}4\) - \(\frac{21}4\) = \(\frac{4}4\) = 1

⇒ k = 2

(a) -1 

It is given that α, β and γ are the zeroes of x³ – 6x² – x + 30.

∴ (αβ + βγ + γα) = \(\frac{coefficient\,of\,x}{coefficient\,of\,x^3}\) = \(\frac{-1}1\) = - 1

Given: 

p(x) = 6x³ + 3x² – 5x + 1 

= 6x³ – (–3) x² + (–5) x – 1 

Comparing the polynomial with x³ – x² (α + β + γ) + x (αβ + βγ + γα) – αβγ, 

we get: 

αβ + βγ + γα = –5 

and αβγ = – 1

∴ \((\frac{1}{\alpha}+\frac1{\beta}+\frac1{\gamma})\)

= \((\frac{\beta\gamma+\alpha\gamma+\alpha\beta}{\alpha\beta\gamma})\)

= \((\frac{-5}{-1})\)

= 5

(a) –3 

Since, α, β and γ are the zeroes of 2x³ + x² – 13x + 6, 

we have:

αβγ = \(\frac{-(constant\,term)}{coefficient\,of\,x^3}\) = \(\frac{-6}2\) = - 3

(a) -1 

Here, 

p(x) = x³ – 6x² – x + 3 

Comparing the given polynomial with x³ – (α + β + γ) x² + (αβ + βγ + γα) x – αβγ, 

we get:

(αβ + βγ + γα) = -1

Correct option is (A) \(4x^2+9y^2+16z^2+12xy+24yz+16xz\)

\((2x+3y+4z)^2\) \(=(2x)^2+(3y)^2+(4z)^2+2\times2x\times3y\) \(+2\times3y\times4z+2\times2x\times4z\)

\(=4x^2+9y^2+16z^2+12xy+24yz+16xz\)

A) 4x² + 9x² + 16z² + 12xy + 24yz + 16xz 

Correct option is (B) t² – 2t + 1

(t - 1)(t - 1) \(=t^2-t-t+1\)

= \(t^2-2t+1\)

B) t² – 2t + 1 

Correct option is (A) (y – 1)(3y + 5)

\(3y^2+2y-5\) \(=3y^2+5y-3y-5\)

= y(3y+5) - 1(3y+5)

= \((y-1)(3y+5)\)

A) (y – 1)(3y + 5)

Correct option is (C) 1 + x² + 2x

\((1+x)\) \((1+x)\) \(=1+x+x+x^2\)

= \(1+2x+x^2\)

C) 1 + x² + 2x 

B) i – C, ii – A, iii – B

Correct option is (B) 10

\(\because\) \(p(x)=4x^3+3x^2+2x+1\)

\(\therefore\) \(p(1)=4.1^3+3.1^2+2.1+1\)

= 4+3+2+1 = 10

Correct option is  B) 10

Correct option is (C) trinomial

Polynomial \(2x^3+7x^2-8\) has 3 terms.

\(\therefore\) It is a trinomial.

Correct option is  C) trinomial

Correct option is (B) binomial

A polynomial with two term is called binomial.

Correct option is  B) binomial

B) A is correct, R is incorrect.

Correct option is (B) 28

\(\because\) \(x=2\)

\(\therefore\) \((x+5)\) \((x+2)\) = (2+5) (2+2)

\(=7\times4=28\)

Correct option is  B) 28