Let the zeroes of q(x)=x³+2x²+a are also the zeroes of the polynomial p(x)=x⁵-x⁴-4x³+3x²+3x+b, i.e q(x) is a factor of p(x). Now, let us divide p(x) by q(x).
Then, the division process is
Since, (x³+2x²+a) is a factor
(x⁵-x⁴-4x³+3x²+3x+b), so remainder should be zero.
i.e -(1+a)x²+(3+3a)x+(b-2a)=0 or (1+a)x²+(3+3a)x+(b-2a)=0=0.x²+0.x+0
On comparing the coefficient of x² and constant term, we get
a+1=0 and b-2a=0
⇒ a=-1 and b=2a
⇒ a=-1 and b=2(-1)=-2           [∵ a=-1]
Hence, for a=-1 and b=-2, the zeroes of q(x) are also the the zeroes of the polynomial p(x).

Sum of zeroes = \(\frac{Coefficient \ of \ x}{Coefficient \ of \ x^{2}}\)
  \(-1=\frac{-(2a+3)}{a +1}\) 
= -2        

\(y^{ 2 }-\frac { 2y }{ 21 } -\frac { 8 }{ 21 } \)

Quotient=x²+x-3, remainder=8

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

a=5, b=-2

-3 and 2

a=0, b=-6

Let given polynomial be p(x)=x²+kx+k.
On comparing with ax²+bx+c, we get|a=1, b=k and c=k
Let the quadratic polynomial have equal zeroes say α and α.
Then, \(\alpha +\alpha =-\frac { b }{ \alpha } =\frac { -k }{ 1 } \)
\(\Rightarrow \quad 2\alpha =\frac { -k }{ 1 } \) 
\(\\ \Rightarrow \quad a=\frac { -k }{ 2 } \)
and \(\alpha .\alpha =\frac { c }{ a } =\frac { k }{ 1 } \Rightarrow a^{ 2 }=k\)
\(\therefore \ \left( \frac { -k }{ 2 } \right) ^{ 2 }=k \ \left[ \because \ =\frac { -k }{ 2 } \right] \) 
\(\\ \Rightarrow \ k^{ 2 }-4k=0\) 
\(\\ \Rightarrow \ k(k-4)=0\) 
\(\\ \Rightarrow \ k=0,4\)
Thus, equal roots are possible only when k is even.
Hence, the given statement is wrong.

Given, a and β are the zeroes of f(x)=ax²+bx+c.
\(\alpha +\beta =-\frac { Coefficient \ of \ x }{ Coefficient \ of\ x^{ 2 } } =-\frac { b }{ a } \)    ....(i)
and \(\alpha \beta =\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } =\frac { c }{ a } \) .....(ii)
We know that,
\(\left( \alpha -\beta \right) ^{ 2 }=\left( \alpha +\beta \right) ^{ 2 }-4\alpha \beta \)
\(\left( \alpha -\beta \right) ^{ 2 }=\left( \frac { -b }{ a } \right) ^{ 2 }-4\left( \frac { c }{ a } \right) \)   [from Eqs. (i) and (ii)]
\(=\frac { b^{ 2 } }{ a^{ 2 } } -\frac { 4c }{ a } =\frac { b^{ 2 }-4c }{ a^{ 2 } } \)
On taking square root both sides, we get
\(\alpha -\beta =\pm \frac { 1 }{ a } \sqrt { b^{ 2 }-4c } \)
which is the required difference between their zeroes.

Given, a and β are the zeroes of the polynomial
p(x)=ax²+bx+c.
Sum of zeroes, a+β=\(-b\over a\) and product of zeroes, aβ=\(c\over a\)
Now, a²β+aβ²=aβ(a+β)
\(=\frac { c }{ a } \times \frac { (b) }{ a } =\frac { -bc }{ a^{ 2 } } \)

Let other zeroes be a.
Then, according to the given condition,
a+7=-18
⇒ a=-25
Required quadratic polynomial will be (x-7)(x+25)
or x²+18x-175

Let f(x)=x²+7x+10.
By splitting the middle term, we get
f(x)=x²+5(5+2)x+10     [∵ 7=5+2 and 5x²=10]
⇒ f(x)=x²+5x+2x+10=x(x+5)+2(x+5)
⇒ f(x)=(x+5)(x+2)
On putting f(x)=0, we get
(x+5)(x+2)=0
⇒ x+5=0 or x+2=0
⇒ x=-2
Thus, the zeroes of given polynomial are a=-5 and b=-2
Verification
Here, sum zeros, a+β=-7=-\(\frac { Coefficient \ of \ x }{ Coefficient \ of \ x^{ 2 } } \)
and product of zeroes, aβ=10=\(\frac { Constant \ term }{ Coefficient \ of \ x^{ 2 } } \)
So, the relationship between the zeros and the coefficients is verified.

Given, 2 and -3 are the zeros of x²+(a+1)x+b.
On putting x=2, we get 
(2)²+(a+1)2+b=0
⇒ 4+2a+2+b=0
⇒ 2a+b=6         ..(i)
Again putting x=-3, we get
(-3)²+(a+1)(-3)+b=0
⇒ 9-3a-3+b=0
⇒ -3a+b=-6  ..(ii)
On subtracting Eq. (ii) from Eq. (i), we get
5a=0⇒a=0
On putting a=0 in Eq. (ii), we get
-3(0)+b=-6⇒0+b=-6⇒b=6
Now, a+b=0-6=-6
Hence, the value of (a+b) is -6.

Let p(x)=2x²+2ax+5x+10
Since, x+a is a factor of p(x).
Therefore, p(-a)=0
⇒ 2(-a)²+2a(-a)+5(-a)+10=0
⇒ 2a²-2a²-5a+10=0⇒5a=10a=2