Given, polynomial, f(x) = ax² - 5x + c
Let the zeroes of f(x) are \(\alpha\) and \(\beta\), then according to the question
Sum of zeroes, \((\alpha+\beta)\) = Product of zeroes, \((\alpha\beta)\) = 10
Now \(\alpha +\beta =-\frac { Coeff.of \ x }{ Coeff \ of \ { x }^{ 2 } } =\frac { -5 }{ a } \)
\(\Rightarrow \quad 10=\frac{+5}{a}\)
\(\therefore \quad a=\frac{1}{2}\)
and \(\alpha \beta =\frac { Constant \ term }{ Coeff \ of \ { x }^{ 2 } } \)
\(\Rightarrow \quad 10 = 2c\)
\(\therefore \quad c=5\)
Hence \(a=\frac{1}{2}\) and c = 5

Here, p(x)=x⁴+x³+8x²+ax+b and g(x)=x²+1, both are in standard form.
Now, on dividing p(x) by g(x), we get the following division process.
Thus quotient=x²+x+7 and remainder=x(a-1)+(b-7)
Since, x⁴+x³+8x²+ax+b is exactly divisible by x²+1, therefore the remainder should be zero.
So, put x(a-1)+(b-7)=0
⇒ x(a-1)+(b-7)=0.x+0
On comparing the coefficients of x and constant terms, we get
a-1=0 and b-7=0
⇒ a=1 and b=7

\(-\frac { 2 }{ \sqrt { 3 } } ,\quad \frac { \sqrt { 3 } }{ 4 } \)

Let the zeroes be α, β and \(\gamma \).
Then, we have α+β+\(\gamma \)=2
αβ+β\(\gamma \)+\(\gamma \)α=-7 and αβ\(\gamma \)=-14
Now required poynomial is given by
\(x^{ 3 }-(\alpha +\beta +\gamma )x^{ 2 }+(\alpha \beta +\beta \gamma +\gamma \alpha )x-\alpha \beta \gamma =x^{ 3 }-2x^{ 2 }-7x+14\)

Since, a and β are the zeroes of polynomial x² -px-p+c.
So, sum of zeroes, a+β =p      ...(i)
and product of zeroes, aβ =c-p|   ...(ii)
Also, (a+1)(β+1)=0  [given]
aβ+(a+β)+1=0
⇒ c-p+p+1=0 [from Eqs. (i) and (ii)]
⇒ c-1

Given polynomial is p(x)=4x²+2x-5a.
Since, 2 is a zero of polynomial.
p(2)=0
⇒ 4(2)²+2(2)-5a=0     [putting x=2]
⇒ 16+4-5a=0 ⇒ 5a=20
\(\therefore\) \(a=\frac { 20 }{ 5 } =4\)

Let a=\(2\sqrt { 7 } \) and β=\(-5\sqrt { 7 } \) . Then,
α+β=\(2\sqrt { 7 } \)\(-5\sqrt { 7 } \)=\(-3\sqrt { 7 } \)
and αβ=\((2\sqrt { 7 } )\)\((-5\sqrt { 7 } )\)=-70
Now, required quadratic polynomial is given by x²-(sum of zeroes)x+(product of zeroes)=x²+\(3\sqrt { 7x } -70\)

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

(i) 5
(ii) 3

Sum of zeroes = 6, Product of zeroes = 9
\(\therefore\)  Quadratic polynomial is x² - 6x + 9
Also x² - 6x + 9 = 0
\(\Rightarrow\)  (x - 3)(x - 3) = 0
\(\Rightarrow\)  x = 3, 3
Hence zeroes are 3, 3

Let f(x)=x³+(k+8)x+k.
Now, divide f(x) by x-2 and x+1, respectively.
Then, the division process is
Thus, on dividing f(x) by (x-2), we get remainder =3k+24 and on dividing f(x) by (x+1), we get remainder=-9
Now, according to the given condition,
Sum of remainders=0
∴ 24+3k+(-9)=0
⇒ 3k+15=0
⇒ k=-5

k=7

α³+β³=(α³+β³)-3αβ(α+β) \(\frac { 215 }{ 27 } \)

f(2)=0 and f(3)=0⇒ 2k-m=6k-2m=27
On solving these equations, we get m=9 and k=\(\frac { 15 }{ 2 } \)

2(3)²+3k=0⇒k=-21