Polynomial tangents.

I found something interesting regarding the tangents of a polynomial curves, whenever we solve the curve with the equation of a tangent, we get a quadratic perfect square factor of the resulting polynomial which gives us the value of abcissa of the point of tangency. Here's a quick proof of it. Let f(x) be any polynomial(degree>1). Now the equation of tangent at an arbitrary point say (x1,f(x1)) will be y-f(x1)=f'(x1)(x-x1) [using elementary calculus]. On solving it with the curve y=f(x), we get the polynomial, f(x)-f(x1)-f'(x1)(x-x1)=g(x) (suppose). Now all we need to do is show that g(x) has (x-x1)² as a factor. Clearly g(x1)=0, and g'(x)=f'(x)-f'(x1), hence,g'(x1)=0. Thus g(x) has a quadratic factor (x-x1)². Hence proved.

#Algebra

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Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
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