p(x)is a polynomial of 4th degree,remainder of p(x)is 40 when divided by (x-3) or by (x+3).Find p(x)if (x-1),(x-2),(x-3) are factors of p(x)?

(x - 1), (x - 2) and (x + 2) are the factors of p(x), which is a polymomial of 4th degree.
Assume p(x) = (x - 1) (x - 2) (x + 2) (ax + b)

When p(x) is divided by (x - 3), the remainder is 40.
p(3) = 40
(3 - 1) (3 - 2) (3 + 2) (3a + b) = 40
2 * 1 * 5 * (3a + b) = 40
3a + b = 4 ……[1]

When p(x) is divided by (x + 3), the remainder is 40 :
p(-3) = 40
(-3 - 1) (-3 - 2) (-3 + 2) (-3a + b) = 40
-3a + b = -2 …… [2]

[1] + [2] :
2b = 2
b = 1

Substitute b = 1 into [1] :
3a + 1 = 4
3a = 3
a = 1

Hence, p(x) = (x - 1) (x - 2) (x + 2) (x + 1)
p(x) = x⁴ - 5x² + 4

p(x) = (x-a)(x^3 - x^2 - 4x + 4)
p(x) = -ax^3 + ax^2 + 4ax - 4a + x^4 - x^3 - 4x^2 + 4x
p(3) = 40 = -27a+9a+12a-4a+81-27-36+12
40 = 30 - 10a
a = -1
p(x) = x^3-x^2-4x+4+x^4-x^3-4x^2+4x
p(x) = -x^2+4+x^4-4x^2
p(x) = x^4-5x^2+4

remainder of p(x)is 40 when divided by (x-3) , how is (x-3) a factor of p(x)?

1.
Our polynomial P(x) as a form like
(ax+b)(x-1)(x-2)(x+2)=
=(ax+b)(x³-x²-4x+4)=
=ax⁴+(b-a)x³-(4a+b)x²+(4a-4b)x+4b=P(x)

2. first condition
Apply short division for (x+3)
P(x)=(x+3)(ax³+(b-4a)x²+(8a-4b)x+
-20a+8b)+ 60a-20b
The remainder must be 40 so
60a-20b=40 ⇒ 3a-b=2

3. second condition
Apply short division for (x-3)
P(x)=(x-3)(ax³+(b+2a)x²+(2a+2b)x+
+10a+2b)+30a+10b
The remainder must be 40 so
30a+10b=40 ⇒ 3a+b=4

3. Solve the system for a,b
{3a-b=2
{3a+b=4
a=1 & b=1
hence

4. Conclusion
P(x)=(x-1)(x+1)(x-2)(x+2)=x⁴-5x²+4