We have, \(q(x)=\sqrt { 3x^{ 2 } } +10x+7\sqrt { 3 } \)
On splitting the middle term, i.e.10x into two parts, we get
\(q(x)=\sqrt { 3x^{ 2 } } +3x+7+7\sqrt { 3 } \)
\(=\sqrt { 3x } (x+\sqrt { 3 } )+7(x+\sqrt { 3 } )\)
\(=(x+\sqrt { 3 } )\sqrt { 3x } +7)\)
For the zeroes of q(x), put q(x)=0
\(\therefore \quad (x+\sqrt { 3 } )\sqrt { 3x } +7)=0\)
\(\Rightarrow \quad x+\sqrt { 3 } =0\) and \(\sqrt { 3x } +7=0\)
\(\Rightarrow \quad x=-\sqrt { 3 } \) and \(x=\frac { -7 }{ \sqrt { 3 } } \)
Hence, the zeroes of q(x) are \(\alpha =-\sqrt { 3 } \) and \(\beta =\frac { -7 }{ \sqrt { 3 } } \)
Verification
Here, \(\alpha +\beta =-\sqrt { 3 } +\left( \frac { -7 }{ \sqrt { 3 } } \right) =\frac { -3-7 }{ \sqrt { 3 } } =-\frac { 10 }{ \sqrt { 3 } } \)
\(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad x^{ 2 } } \)
and \(\alpha \beta =-\sqrt { 3 } \times \left( \frac { -7 }{ \sqrt { 3 } } \right) =\frac { 7\sqrt { 3 } }{ \sqrt { 3 } } =-\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \)
Hence, the relations between the zeroes and the coefficients of the polynomial is verified.

Given, f(x)=ax+b
For zero of f(x), put p(x) = 0⇒ ax+b=0
⇒ x=b/a
So, the zero of f(x) is -b/a.

Given, two zeroes are \(\sqrt { \frac { 5 }{ 3 } } \) and \(-\sqrt { \frac { 5 }{ 3 } } \)
.So, \(\left( x-\sqrt { \frac { 5 }{ 3 } } \right) \) and \(\left( x+\sqrt { \frac { 5 }{ 3 } } \right) \).
polynomial.
⇒ \(\left( x-\sqrt { \frac { 5 }{ 3 } } \right) \)\(\left( x+\sqrt { \frac { 5 }{ 3 } } \right) \)\(=x^{ 2 }-\frac { 5 }{ 3 } =\frac { 3x^{ 2 }-5 }{ 3 } \)
given polynomial
Consequently, 3x²-5 is a factor of the given polynomial.
Now, let us divide 3x⁴+6x³-2x²-10x-5.

Let p(x)=2x⁴-3x³-3x²+6x-2.
Since, its two zeroes are \(\sqrt { 2 } \) and \(-\sqrt { 2 } \).
So, (x-\(\sqrt { 2 } \)) and (x+\(-\sqrt { 2 } \)) are the factors of p(x).
⇒  (x-\(\sqrt { 2 } \)) (x+\(-\sqrt { 2 } \)) is a factor of p(x).
⇒ x²-2 also a factor of p(x).
Now, let us divide the given polynomial p(x) by g(x)=x²-2. Then division process is
Here, quotient=2x²-3x+1 and remainder=0
Now, factorise the quotient by splitting the middle term, i.e. write
2x²-3x+1=2x²-2x-x+1
=2x(x-1)-1(x-1)=(2x-1)(x-1)
So, other zeroes of f(x) are given by
(2x-1)=0 and (x-1)=0
⇒ \(x=\frac { 1 }{ 2 } \) and x=1
Hence, all the zeroes of  2x⁴-3x³-3x²+6x-2 are \(\sqrt { 2 } \), \(-\sqrt { 2 } \), \(\frac{1}{2}\) and 1.

Recall the identity a² - b² = (a - b)(a + b). Using it, we can write:
x² - 3 = (x - \(\sqrt{3}\))(x + \(\sqrt{3}\))
So, the value of x² - 3 is zero when x = \(\sqrt{3}\) or x = -\(\sqrt{3}\)
Therefore, the zeroes of x² - 3 are \(\sqrt{3}\) and -\(\sqrt{3}\)
Now,
sum of zeroes = \(\sqrt{3}\) - \(\sqrt{3}\) = 0 = \(\frac{-(Coefficient \quad of \quad x)}{Coefficient \quad of \quad x^{2}}\)
product of zeroes = (\(\sqrt{3}\))(-\(\sqrt{3}\)) = -3 = \(\frac{-3}{1}=\frac{Constant \quad term}{Coefficient \quad of \quad x^{2}}\)

Quotient=x-7 and remainder=17x+1

We have, α+β=-3and α-β=-10 (assuming α<β)
\(\alpha =-\frac { 13 }{ 2 } \) and \(\beta =\frac { 7 }{ 2 } \)
Now, \(\alpha ^{ 2 }\beta ^{ 2 }=30\)

Let the quadratic polynomial be ax² + bx + c, and its zeroes be \(\alpha \text { and } \beta \text { . }\) 
We have
\(\alpha+\beta=-3=\frac{-b}{a}\) 
and \(\alpha \beta=2=\frac{c}{a}\) 
If a = 1, then b = 3 and c = 2.
So, one quadratic polynomial which fits the given conditions is x² + 3x + 2.
You can check that any other quadratic polynomial that fits these conditions will be of the form k(x² + 3x + 2), where k is real.
Let us now look at cubic polynomials. Do you think a similar relation holds between the zeroes of a cubic polynomial and its coefficients
Let us consider p(x) = 2x³ – 5x² – 14x + 8.
You can check that p(x) = 0 for x = 4,\(-2, \frac{1}{2}\) .Since p(x) can have atmost three zeroes, these are the zeores of 2x³ – 5x² – 14x + 8. Now,
sum of the zeroes = \(4+(-2)+\frac{1}{2}=\frac{5}{2}=\frac{-(-5)}{2}=\frac{-\left(\text { Coefficient of } x^{2}\right)}{\text { Coefficient of } x^{3}}\) 
product of the zeroes = \(4 \times(-2) \times \frac{1}{2}=-4=\frac{-8}{2}=\frac{-\text { Constant term }}{\text { Coefficient of } x^{3}}\) 
However, there is one more relationship here. Consider the sum of the products of the zeroes taken two at a time. We have
\(\{4 \times(-2)\}+\left\{(-2) \times \frac{1}{2}\right\}+\left\{\frac{1}{2} \times 4\right\}\) 
\(=-8-1+2=-7=\frac{-14}{2}=\frac{\text { Coefficient of } x}{\text { Coefficient of } x^{3}}\) 
In general, it can be proved that if \(\alpha, \beta, \gamma\)  are the zeroes of the cubic polynomial ax³ + bx² + cx + d, then
\(\alpha+\beta+\gamma=\frac{-b}{a}\) 
\(\alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}\) 
\(\alpha \beta \gamma=\frac{-d}{a}\) 

If x + 7 is a factor then (-7) is a root.
So, f(-7) = (-7)⁴ + 10(-7)³ + 25(-7)² + 15(-7) + k = 0
(when it is a root the polynomial should be equal to zero when value is substituted)
2401 - 3430 + 1225 - 105 + k = 0
\(\Rightarrow\)  3626 - 3535 + k = 0
\(\Rightarrow\)  91 + k = 0
\(\therefore\)  k = - 91

Let f(x) = 2x² - 5x - 3
Let the zeroes of polynomial be \(\alpha\) and \(\beta\), then
Sum of zeroes = \(\alpha+\beta = \frac{5}{2}\)
Product of zeroes = \(\alpha\beta=-\frac{3}{2}\)
According to the question, zeroes of x² + px + q are \(2\alpha\) and \(2\beta\)
Sum of zeroes \(-\frac { Coeff.of \ x }{ Coeff.of \ { x }^{ 2 } } =\frac { -p }{ 1 } \)
\(\Rightarrow \quad -p= 2\alpha+2\beta=2(\alpha+\beta)\)
\(\Rightarrow \quad -p=2\times\frac{5}{2}=5\Rightarrow p=-5\)
Product of zeroes \(=\frac { Constant \ term }{ Coeff \ of \ { x }^{ 2 } } =\frac { q }{ 1 } \)
\(\Rightarrow \quad q = 2\alpha\times2\beta=4\alpha\beta\)
\(\Rightarrow \quad q=4(-\frac{3}{2})=-6\)
\(\therefore \quad p=-5\) and \(q=-6\)

Since \(\alpha\) and \(\beta\) are the zeroes of the polynomial, then
\(\alpha +\beta =-\frac { Coefficient \ of\ x }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad \quad \alpha +\beta =-\left( \frac { -1 }{ 1 } \right) =1\)   .... (i)
Given \(\alpha-\beta=9\)  ... (ii)
From (i) and (ii), \(\alpha=5,\beta=-4\)
\(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ { x }^{ 2 } } \)
\(\alpha \beta =-k\)
\(\Rightarrow \quad (5)(-4) = -k\)
\(\Rightarrow \quad k = 20\)

Let  p(x) = 2x² + 3x + \(\lambda \) 
One of the zero is \(\frac{1}{2}\),
so  \(p\left( \frac { 1 }{ 2 } \right) =0\)
\(\Rightarrow \quad 2\left( \frac { 1 }{ 2 } \right) ^{ 2 }+3\left( \frac { 1 }{ 2 } \right) +\lambda =0\)
\(\Rightarrow \quad \frac { 1 }{ 2 } +\frac { 3 }{ 2 } +\lambda =0\)
\(\Rightarrow \quad \lambda +2=0\)
\(\therefore \quad \lambda =-2\)
Now, Product of zeroes = \(\alpha \beta =\frac { constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha \left( \frac { 1 }{ 2 } \right) =\frac { \lambda }{ 2 } =-\frac { 2 }{ 2 } =-1\)
\(\Rightarrow \quad \alpha=-2\)
Hence, other zero = - 2

f(x) = 2x² - 7x + 3
Sum of roots = p + q \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(=-\left( \frac { -7 }{ 2 } \right) =\frac { 7 }{ 2 } \)
Product of roots = pq \(=-\frac { Constant\quad term }{ Coefficient\quad of\quad { x }^{ 2 } } =\frac{3}{2}\)
We know that 
(p + q)² = p² + q² + 2pq
\(\Rightarrow\)  p² + q² = (p + q)² - 2pq
\(=\left( \frac { 7 }{ 2 } \right) ^{ 2 }-3=\frac { 49 }{ 4 } -\frac { 3 }{ 1 } =\frac { 37 }{ 4 } \)

f(x) = x² - 7x - 8
Let other zero be k, then
Sum of zeroes \(-1+k=-\left( \frac { -7 }{ 1 } \right) =7\)
\(\Rightarrow k = 8\)

p(x) = 3x² - kx + 6
Sum of the zeroes = 3 \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad 3=-\frac { \left( -k \right) }{ 3 } \)
k = 9

Sum of the roots \(=-\frac { Coefficient\quad of\quad x }{ Coefficient\quad of\quad { x }^{ 2 } } \)
\(\Rightarrow \quad \alpha +\beta =-\left( \frac { b }{ a } \right) \)
\(=\frac{b}{a}\)

x²-x-56

α+β=7  ...(i)
α-β=1 ....(ii)
On solving Eqs. (i) and (ii), we get α=4 and β=3
Now, k=αβ=12

2(3)²+3k=0⇒k=-21