Given that f(x) = 2x³-7x²+7ax+ 16 is divisible by x-a, find
(i) the value of the constant a
(ii) the remainder when f(x) is divided by 2x+1.

Respuesta :

Answer:

a = - 2, remainder = 21

Step-by-step explanation:

The Remainder theorem states that if f(x) is divided by (x - a) the remainder is f(a)

Since f(x) is divisible by (x - a) then remainder is zero , then

f(a) = 2a³ - 7a² + 7a² + 16 = 0 , that is

2a³ + 16 = 0 ( subtract 16 from both sides )

2a³ = - 16 ( divide both sides by 2 )

a³ = - 8 ( take the cube root of both sides )

a = [tex]\sqrt[3]{-8}[/tex] = - 2

Then

f(x) = 2x³ - 7x² - 14x + 16

Evaluate f(- [tex]\frac{1}{2}[/tex] ) for remainder on division by (2x + 1)

f(- [tex]\frac{1}{2}[/tex] ) = 2(- [tex]\frac{1}{2}[/tex] )³ - 7(- [tex]\frac{1}{2}[/tex] )² - 14(- [tex]\frac{1}{2}[/tex] ) + 16

        = 2(- [tex]\frac{1}{8}[/tex] ) - 7([tex]\frac{1}{4}[/tex] ) + 7 + 16

        = - [tex]\frac{1}{4}[/tex] - [tex]\frac{7}{4}[/tex] + 23

        = - [tex]\frac{8}{4}[/tex] + 23

       = - 2 + 23

       = 21