Answer:
The remainder is zero
Step-by-step explanation:
To find the remainder we will use the long division
[tex]\frac{x^{3}-6x^{2}-5x-14}{x-7}=x^{2}\frac{x^{2}-5x-14 }{x-7}[/tex]⇒(1)
[tex]\frac{x^{2}-5x-14}{x-7}=x\frac{2x-14}{x-7}[/tex]⇒(2)
[tex]\frac{2x-14}{x-7}=2[/tex]⇒(3)
From (1) , (2) and (3)
The quotient of the long division is [tex]x^{2}+x+2[/tex] and no remainder
So the remainder is zero
* If you want to check your answer Multiply the quotient by the divisor
[tex](x^{2}+x+2)(x-7)=x^{3}-6x^{2}-5x-14[/tex]
