Given that y = (x−2)(2x2−3x−5), find dy/dx and find the gradient of the tangents to the curve where it cuts the y-axis.

$$y = x(2x^2-3x-5) - 2(2x^2-3x-5)$$ $$y = 2x^3 - 3x^2 - 5x - 4x^2 + 6x + 10$$ $$y = 2x^3 - 7x^2 + x + 10$$

Using the power rule for differentiation, $\frac{d}{dx}(ax^n) = nax^{n-1}$:

$$\frac{dy}{dx} = \frac{d}{dx}(2x^3) - \frac{d}{dx}(7x^2) + \frac{d}{dx}(x) + \frac{d}{dx}(10)$$ $$\frac{dy}{dx} = 2(3)x^{3-1} - 7(2)x^{2-1} + 1x^{1-1} + 0$$ $$\frac{dy}{dx} = 6x^2 - 14x + 1$$

The curve cuts the y-axis when $x=0$.
Substitute $x=0$ into the original equation for $y$:

$$y = (0-2)(2(0)^2-3(0)-5)$$ $$y = (-2)(-5)$$ $$y = 10$$

So, the curve cuts the y-axis at the point $(0, 10)$.

Substitute $x=0$ into the derivative $\frac{dy}{dx}$ to find the gradient:

$$\frac{dy}{dx}\Big|_{x=0} = 6(0)^2 - 14(0) + 1$$ $$\frac{dy}{dx}\Big|_{x=0} = 0 - 0 + 1$$ $$\frac{dy}{dx}\Big|_{x=0} = 1$$

The gradient of the tangent to the curve where it cuts the y-axis is $\boxed{1}$.