My textbooks lists the set of values b may take as $b>-4 $ and $b≠4$
One of the roots is given as x=2 in the first part of the question, which asks for the value of a such that, for all values of b, one root of the equation $2x^3+ax+4=b(x-2)$ I calculated this as a=-10
If the RHS of the equation was just b I would have taken the derivative to find the stationary points, substituted those values into the original function and calculated the product of $y(x1)y(x)2$. If the product of that is <0 then the cubic has three distinct real roots. when I do this I get the following
The trouble I am having is that when I take the derivative of the function as follows $f'(x) = 6x^2-10-b=0$, the unknown constant remains and I'm not sure what to do next. If one of the roots of the equation is 2, then isn't 14 the only value that b can take to satisfy f'(x) ?
Cancel out the $x-2$ factors on the both sides to end up with:
$$2x^2 + 4x -2 = b$$
This is a quadratic equation and will have two distinct solutions if the discriminant is positive. The discriminant in our case is $D = 16 + 8(b+2)$. And thus we obtain one condition which says that $D>0 \iff b> -4$.
Finally we have to be sure that $2$ isn't a zero of the quadratic equation. To get when it is a zero of the equation just plug 2 in it to get:
$$b = 8 + 8 - 2 = 14$$
Thus the condition is $b > -4$ and $b \not = 14$