Question

What must be added to x 3 - 3x 2 - 12x + 19 so that the result is exactly divisible by x 2 + x -6 ?

Answer 1

x²+x-6
x²+3x-2x-6
x(x+3)-2(x+3)
(x-2)(x+3)
now take (x-2)
x-2=0
x=2
p(2)=x³-3x²-12x+19
(2)³-3(2)²-12(2)+19
8-12-24+19
27-36
-9
therefore -9 must be added to be exactly divisible.

Answer 2

Answer:

$x^2+x+2x-1$                

Step-by-step explanation:

Dividend = $x^3 - 3x^2 - 12x + 19$

Divisor = $x^2+x-6$

We know that

$Dividend= (Divisor \times quotient)+Remainder$

$x^3 - 3x^2 - 12x + 19= (x^2+x-6 \times x)+(-4x^2-6x+19)$

$x^3 - 3x^2 - 12x + 19= (x^2+x-6 \times x-4)+(-2x-5)$

So, quotient = x-4

Remainder = -2x-5

Polynomial should be added to Dividend = Divisor - Remainder

                                                                   = $x^2+x-6-(-2x-5)$    

                                                                   = $x^2+x-6+2x+5$                                                                                                                                                                                                      

                                                                   = $x^2+x+2x-1$                

So, $x^2+x+2x-1$ must be added to  $x^3 - 3x^2 - 12x + 19$ so that the result is exactly divisible by $x^2+x-6$