If $\alpha $ is a root of equation $4x^2+2x-1=0$ and $f(x) =4x^3 -3x+1$ then $2[f(\alpha)+1]= $

Question

Since $\alpha$ is a root of equation therefore $4x^2 +2x-1 =0 $ can be written as $4\alpha^2 +2\alpha -1 =0$

and $f(\alpha ) = 4\alpha^3 -3\alpha +1 = \alpha ( 4\alpha^2) -3\alpha +1 = 0 $ $\Rightarrow \alpha ( 1-2\alpha ) -3\alpha +1 =0 $

$\Rightarrow \alpha -2\alpha^2 -3\alpha +1 = 0 $

$\Rightarrow -2\alpha^2 -2\alpha +1 = 0$

But answer is 1 , can you please help on this.

Answer 1

Hint

$$f(x){=4x^3-3x+1=4x^3+2x^2-x+x-2x^2-x-2x+0.5+0.5\\=x(4x^2+2x-1)-{1\over 2}(4x^2+2x-1)-x+0.5}$$

Answer 2

Using Euclidean division of polynomials, we find $$ 2 (f(x)+ 1)= (2 x - 1)(4 x^2 + 2 x - 1)+(3 - 2 x) $$