Question

P( x) =x^3-6x^2-5x-14 What is the remainder R when the polynomial p(x) is divided by (x-7)

Answer 1

Answer:

The remainder is zero

Step-by-step explanation:

To find the remainder we will use the long division

$\frac{x^{3}-6x^{2}-5x-14}{x-7}=x^{2}\frac{x^{2}-5x-14 }{x-7}$⇒(1)

$\frac{x^{2}-5x-14}{x-7}=x\frac{2x-14}{x-7}$⇒(2)

$\frac{2x-14}{x-7}=2$⇒(3)

From (1) , (2) and (3)

The quotient of the long division is $x^{2}+x+2$ and no remainder

So the remainder is zero

* If you want to check your answer Multiply the quotient by the divisor

$(x^{2}+x+2)(x-7)=x^{3}-6x^{2}-5x-14$