Question

What must be subtracted from xcube -6 x square+13x-6 so that the resulting polynomial is exactly divisible by xsquare+×+1?

Answer 1

Check the attachment.

Answer 2

Answer:

20x+1

Step-by-step explanation:

Dividend:$x^3-6x^2+13x-6$

Divisor = $x^2+x+1$

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

So,  $x^3-6x^2+13x-6 = (x^2+x+1 \times x)+(-7x^2+13x-6)$

$x^3-6x^2+13x-6 = (x^2+x+1 \times x-7)+(20x+1))$

Since the remainder is 20x+1

So, 20x+1 must be subtracted from $x^3-6x^2+13x-6$s  o that the resulting polynomial is exactly divisible by $x^2+x+1$