Finding value of $a$.

Question

Find $a$ for $f(x)= 6x^3-5x^2+5x-a.$ given that $3x^2-x+2$ is a factor of $f(x)$.

$3x^2-x+2$ posses no real roots hence I am unable to proceed.

Answer 1

\begin{align} f(x)&=6x^3−5x^2+5x−a\\ &=2x(3x^2-x+2)-(3x^2-x+a) \end{align} Obviously $a=2$.

Answer 2

So you have $f(x) = 6x^3 -5x^2 + 5x - a= (3x^2-x+2)g(x)$, with $g$ a polynom with degree $1$, ie: $g(x) = cx+d$.

Expanding the RHS, you can identify with the LHS in order to find the parameters $c,d$ and then you will be able to find $a$.

Answer 3

By long division we get,

$$6x^3-5x^2+5x-a=(3x^2-x+2)(2x-1)+(2-a)$$

But,remainder is $0$.

So,$2-a=0\implies a=2$