Q:

Show that $\det(xA+(1-x)B)=0$ for only finitely many values of $x$

Show that $$\det(xA+(1-x)B) = 0$$ for only finitely many values of $x$ where $A,B$  are $n \times n$ real invertible matrices. 

Consider $$f(x)=\det(xA+(1-x)B)$$
Obviously, $f(1),f(0)\neq 0$. Hence, $f \equiv 0$ is false.  But, how to show that it is zero for finitely many $x$? 

A:

Hint The quantity $\det[x A + (1 - x) B]$ is a polynomial in $x$, and, as you've observed, it is not identically zero. 

A:

Hint $:$ $\det (xA + (1-x)B)$ is a polynomial in $x$ of degree at most $n.$

