Given the cubic polynomial $P(x) = ax^3 + bx^2 + cx + d; a ≠ 0$,
Let A(x, P(x)) be a point on the graph of the polynomial P. The tangent of the function P at the point A intersects the function once more in some other point B(y, P(y)). The tangent line of the function P at the point B intersects the function P once more in some other point C(z, P(z)). Show that the area surrounded by the function P and the tangent line at C is 16 times the area surrounded by the function P and the tangent line at B (originally posted as A).
I drew some graphs that would meet the requirements. I tried to find the area surrounded by the tangent line at point A and the graph of P like this
$$\int_{b}^{a}P(x)dx - \int_{b}^{a}P'(a)(x-a)+P(a)dx$$
but that's as far as I've gotten. What confuses me the most I think is firstly that I don't know the order the points appear on the graph (whether $a>b>c$ or $b>c>a$, or something else).
If I knew that, I think would try to find coordinates of B by using the tangent line at A, the coordinates of C by using the tangent line at B and then the coordinates of the point that the tangent line at C intersects. Then I'd try to somehow compare the areas so that we can see that one is 16 times more than the other.
I think the first thing to do is to shift the origin to the first point you are considering, which makes $d=0$. This doesn't change the areas. Then re-scale so that $a=1$ which doesn't change the relative areas.
Without loss of generality you then have $p(x)=x^3+bx^2+cx$. The tangent at the origin has gradient $p'(0)=c$ and therefore has the equation $y=cx$. When $cx=x^3+bx^2+cx$ you have the expected double root $x=0$ and find the additional point $B$ for which $x=-b, y=bc$.
Now the tangent at $x=-b$ has gradient $p'(b)=3b^2-2b^2+c=b^2-+c$ and is therefore $$y-bc=(b^2+c)(x+b)$$
This meets the curve at $$bc+(b^2+c)(x+b)=x^3+bx^2+cx$$Here the sum of the roots is $-b$ and we have a double-root $x=-b$ and hence a single root $x=b$, so the tangent meets the curve again at point $C$, $x=b, p(x)=2b^3+bc$
The tangent at $C$ will meet the curve at a third point. We can do the sum of the roots trick again - we still have sum of roots equal to $-b$ and we know a double root at $x=b$, so the third root is $x=-3b$.
So the first area involves an integral from $0$ to $-b$ and the second from $b$ to $-3b$ of the difference between the value on the curve and the value on the chord joining the points. We know that the chord does not cross the curve because of the way it was constructed.
Note, the sum of the roots trick can be done without the initial simplifications. But I think the integrals might then be messier. It might be easier to shift so that $b=0$ so that the sum of the roots is zero (the tangent is linear so the equation would always have zero coefficient of $x^2$). If the first point is then $x=r$ then the next ones would be at $-2r, 4r, -8r$.