Question

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.

Answer 1

a = - 2, remainder = 21

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 =
`\sqrt[3]{-8}` = - 2

Then

f(x) = 2x³ - 7x² - 14x + 16

Evaluate f(-
`(1)/(2)` ) for remainder on division by (2x + 1)

f(-
`(1)/(2)` ) = 2(-
`(1)/(2)` )³ - 7(-
`(1)/(2)` )² - 14(-
`(1)/(2)` ) + 16

= 2(-
`(1)/(8)` ) - 7(
`(1)/(4)` ) + 7 + 16

= -
`(1)/(4)` -
`(7)/(4)` + 23

= -
`(8)/(4)` + 23

= - 2 + 23

= 21