Q.9 see image........................................................

actually all these doubts that you are asking are simply based on the properties of polynomials.

let lambda= x

then det= p(x), where p represents a polyomial function.

now note that p(0)= 0 (because after taking a common from first row and then taking b common from second row, both rows become identical).

so we conclude that p(x) is divisible by x – 0 or x. however, we need to prove that it is divisible by x^2, which would be true IFF p’(0)= 0 (again a property of polynomials).

now we already know the standard formula to differentiate a determinant. apply that to find p’(x) and then put x=0. you will notice that once again it becomes zero, i.e, p’(0)= 0

hence the original determinant is divisible by x^2.

to find the last and final factor (since p is of degree 3 in lambda or x), simply expand the determinant and divide by x^2