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
From the previous part we have obtained that a = -16 and (x-3) is a factor of p(x)
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`
Solving above yields;
`p = 4`
`r = - 3`
`q = - 4`
`4x^3-16x^2+9x+9 =(x-3) (4x^2- 4x- 3)`
`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