# The polynomial 4x^3+ ax^2+ 9x + 9, where a is a constant, is denoted by p(x). It is given that whenp(x) is divided by (2x −1) the remainder is 10. (i) Find the value of a and hence verify that (x −3) is a factor of p(x). (ii) When a has this value, solve the equation p(x) = 0. Answer part (ii) only. part (i) is answered in http://www.enotes.com/homework-help/polynomial-4x-3-ax-2-9x-9-where-constant-denoted-456218

From the previous part we have obtained that a = -16 and (x-3) is a factor of p(x)

`p(x)=0`

`4x^3-16x^2+9x+9=0`

We know that (x-3) is a factor of p(x).Then we can rewrite p(x) as;

`4x^3-16x^2+9x+9 =(x-3) (px^2+qx+r) =px^3+qx^2+rx-3px^2-3qx-3r`

By comparing coefficients we will get;

`x^3 -> 4=p`

`x^2-> -16=q-3p`

`x-> 9=r-3q`

`const->9=-3r`

Solving above yields;

`p = 4`

`r = - 3`

`q = - 4`

`4x^3-16x^2+9x+9 =(x-3) (4x^2- 4x- 3)`

`4x^2-4x-3`

`= 4x^2-6x+2x-3`

`= 2x(2x-3)+1(2x-3)`

`= (2x-3)(2x+1)`

`4x^3-16x^2+9x+9 =(x-3)(2x-3)(2x+1)`

`p(x) = 4x^3-16x^2+9x+9 =(x-3)(2x-3)(2x+1) = 0`

So the roots of p(x) = 0 are x = 3,3/2,-1/2