Câu a)Ta có P=(x+5)(ax^2 + bx + 25)
=ax^3 + bx^2 +25x + 5ax^2 + 5bx + 125
=ax^3 =(bx^2 + 5ax^2) + (25x + 5bx) + 125
=ax^3 + x^2(b+5a)+x(25+5b)+125
Câu b)
P=ax^3 + bx^2 +25x + 5ax^2 + 5bx + 125
=ax^3 +(b+5a)x^2 + (25 + 5b)x + 125
Vậy để P=Q thì
a=1
b +5a=0
25+5b=0
=> a = 1
b = -5