Find b if h(x) = 2x−1 is a factor of f(x) = 6x2 + bx + 2.

If $h(x) = 2x-1$ is a factor, then $h(x)=0$ for some value of $x$.

$$2x - 1 = 0$$ $$2x = 1$$ $$x = \frac{1}{2}$$

According to the Factor Theorem, if $(x-k)$ is a factor of a polynomial $f(x)$, then $f(k)=0$. In this case, $k = \frac{1}{2}$, so $f\left(\frac{1}{2}\right)$ must be equal to 0.
Substitute $x = \frac{1}{2}$ into $f(x) = 6x^2 + bx + 2$:

$$f\left(\frac{1}{2}\right) = 6\left(\frac{1}{2}\right)^2 + b\left(\frac{1}{2}\right) + 2 = 0$$

$$6\left(\frac{1}{4}\right) + \frac{b}{2} + 2 = 0$$ $$\frac{6}{4} + \frac{b}{2} + 2 = 0$$ $$\frac{3}{2} + \frac{b}{2} + 2 = 0$$

Multiply the entire equation by 2 to eliminate the denominators:

$$2\left(\frac{3}{2}\right) + 2\left(\frac{b}{2}\right) + 2(2) = 2(0)$$ $$3 + b + 4 = 0$$ $$b + 7 = 0$$ $$b = -7$$

The value of $b$ is $\boxed{-7}$.