Question

$\text{Given f (x)=}\left|\begin{array}{ccc}1& x& x^2\\ 1& a& a^2\\ 1& b& b^2\end{array}\right|\text{Then, is (x - a) a factor to this determinant}$

• A. True
• B. False

Answer 1

The correct option is A

True

If we put x = a, f(a) = 0 . It’s important to see that the determinant f(x) is polynomial in $x^2$. So when we put a value x = k for which the function goes to 0 then it’s a root of f(x) or (x-k) is factor for f(x). Therefore (x-a) is a factor for the determinant.