if x is a factor of P(x) , P' (x) is also a factor
P(x) = x^3 + ax^2 + bx + 12
P(1) = 0
= a+b+ 13
P'(1) = 3+2a+b
=> b = -(a+13)
substitute,
a = 10, b = -23
P(x) = x^3 + 10x^2 -23x+12 --->could also be done with product of roots etc.
When P(x) is divided by (x+1) and (x+3) -----im not sure whether this means P(-1)=P(-3)=72 or (x^2+4x+3) as a dividor will give 72 so gna leave it
long division / simultaneous / remainder theorem after to factorise