Question

if the quotient obtained on dividing (3x³+11x²+34x+106) by (x-3) is (3x²+ax+b) , then find a and b and also the remainder​

Answer 1

$\dfrac{\left(3x^3+11x^2+34x+106\right)}{\left(x-3\right)}$

$=3x^2+\dfrac{20x^2+34x+106}{x-3}$

$=3x^2+20x+\dfrac{94x+106}{x-3}$

$=3x^2+20x+94+\dfrac{388}{x-3}$

Thus a = 20, b = 94 + (388)/(x-3)