Question

. Find the two points on the curve that have a common tangent line

Answer 1

If y=mx+n is an equation of the tangent to the graph of the polynomial f

then there are real a and polynomial q for which f(x)−mx−n=(x−a)2q(x).

Here a it's an abscissa of the touch point.

Indeed, f(a)=ma+n and f′(a)=m.

Thus, f(x)−mx−n=0 for x=a,

which by Polynomial remainder theorem givesf(x)−mx−n=(x−a)h(x)for some polynomial h.

Now,(f(x)−mx−n)′=((x−a)h(x))′orf′(x)−m=h(x)+(x−a)h′(x).Hence,0=f′(a)−m=h(a),which by Polynomial remainder theorem again

says that there is polynomial q,

for which h(x)=(x−a)q(x).

Id est,f(x)−mx−n=(x−a)2q(x).

Since a degree of x4−2x2−x is four,

we needx4−2x2−x−(mx+n)=(x2+px+q)2,which gives the followingx4−2x2−x=x4−2x2+1−x−1=(x2−1)2−(x+1).Thus, y=−x−1 is an equation of the tangent.

Now, we got touching points: (1,−2) and (−1,0).