using the remainder theorem to find p(2) for p(x)=2x^3-2x^2-4x+7 Give the quotient and the remainder for the associated division and the value of p(2)

Answer:In the given polynomial p(x) [tex] Q(x)  = (2x^2 + 2x)[/tex]Remainder = p(2) = 7Step-by-step explanation:Here, the given polynomial is [tex]p(x)=2x^3-2x^2-4x+7[/tex]Also, p (2) is to be determined,⇒ x = 2 is the zero of the given p(x)⇒ (x- 2) is he Root of the polynomial. Now, by REMAINDER THEOREM:The remainder theorem states that when a polynomial, f(x), is divided by a linear polynomial , x - a, the remainder of that division will be equivalent to f(a).So, here when p(x) is divided by ( x- 2) the remainder = p(2)[tex]\frac{P(x)}{(x-2)}   = Q(x)  + p(2)[/tex]Now, [tex]p(2) =2(2)^3-2(2)^2-4(2)+7  =16 -8-8+7 = 7[/tex]⇒ p(2) = 7Now, the given equation becomes:Now, [tex]\frac{P(x)}{(x-2)}   = Q(x)  + 7\\\implies Q(x) =  \frac{P(x)}{(x-2)}  - 7 =  \frac{2x^3-2x^2-4x+7}{x-2}  -7\\= (2x^2 + 2x)  - 7[/tex][tex]\implies Q(x)  = (2x^2 + 2x)[/tex]