Question

what should be added to the polynomial x^3-6x^2+11x+8 so that it is completely divisible by x^2-3x+2

Answer 1

Answer:

$x^2-3x-12$

Step-by-step explanation:

Divisor : $x^2-3x+2$

Dividend : $x^3-6x^2+11x+8$

Since we now that :

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

$x^3-6x^2+11x+8 = (x^2-3x+2 \times x)+(-3x^2+9x+8)$

$x^3-6x^2+11x+8 = (x^2-3x+2 \times x-3)+(14)$

Now the number need to be added to polynomial  $x^3-6x^2+11x+8$ is Divisor -Remainder

= $x^2-3x+2-14$

= $x^2-3x-12$

So, $x^2-3x-12$ need to be added to polynomial  $x^3-6x^2+11x+8$

Hence  $x^2-3x-12$ need to be added to polynomial  $x^3-6x^2+11x+8$  so that it is completely divisible by $x^2-3x+2$

Answer 2

Step-by-step explanation:

Hope you mark it as the best answer