Let p(x) = xⁿ + aⁿ
The zero of x + a is - a. | x + a = 0 ⇒ x = - a
Now,
p(- a) = (- a)ⁿ + aⁿ = (- 1)ⁿaⁿ + aⁿ
         = (- 1)aⁿ + aⁿ 
    
 n is an odd positive interger
   (-1)ⁿ = - 1
= - aⁿ + aⁿ = 0
∴ By Factor Theorem, x + a is a factor of xⁿ + aⁿ for any odd positive integer n.

x² + 2x - 3
= x² + 3x - x - 3
= x(x + 3) - 1 (x + 3)
= (x + 3) (x - 1)
Let p(x) = x⁴ + 2x³ - 2x² + 2x - 3
We see that
p(-3) = (-3)⁴ + 2(-3)³ - 2(-3)² + 2(-3) - 3
= 81 - 54 - 18 - 6 - 3
= 0
Hence by converse of factor theorem, (x + 3) is a factor of p(x).
Also, we see that
p(1) = (1)⁴ + 2(1)³ - 2(1)² + 2(1) - 3
= 0
Hence by converse of factor theorem, (x - 1) is a factor of p(x).
From above, we see that
(x + 3) (x - 1), i.e., x² + 2x - 3 is a factor of p(x)
⇒ p(x) is exactly divisible by (x² + 2x - 3).

Let p(x) = 5x³ - x² + 4x + b
1 - 5x = 0



  The zero of 1 - 5x is 
If p(x) is divisivle by 1 - 5x, then 
                                         | By Factor Theorem

Let p(x) = 2x³ + ax² + 11x + a + 3
If p(x) is exactly divisible by 2x - 1, then by factor theorem,

Hence, the required polynomial is
2x³ - 7x² + 11x - 7 + 3
or 2x³ - 7x² + 11x - 4

Let p (x) = x⁴ + ax³ - 3x² + 2x + b
If (x + 1) and (x - 1) are factors of p(x), then by factor theorem,
p(-1) = 0 ...(1) | x + 1 = 0 ⇒ x = -1
and    p(1) = 0 ...(2) | x - 1 = 0 ⇒ x = 1
Now,
P(-1) = 0
⇒ (-1)⁴ + a(-1)³ - 3(-1)² + 2 (-1) + b = 0
⇒ 1 - a - 3 - 2 + b = 0
⇒    -a + b = 4    ...(3)
and    p(1) = 0
⇒ (1)⁴ + a(1)³ - 3(1)² + 2 (1) + b = 0
⇒ 1 + a - 3 + 2 + b = 0
⇒    a + b = 0    ...(4)
Solving (3) and (4), we get
a = -2,b = 2