Sol. Since x = –1 is the root of the given equation ⇒ x + 1 is a factor of the elements of at least one row or one column of Δ . In particular, elements of R 1 are

x 2 – 1 = (x +1) (x – 1)

x 2 + 2x + 1 = (x + 1) 2 2x

2 + 3x + 1 = (x + 1) (2x + 1)

Similarly, 2x 2 + x –1 = (x + 1) (2x –1)

2x 2 + 5x –3 = (2x –1) (x + 3)

4x 2 + 4x –3 = (2x –1) (2x + 3)

6x 2 –x –2 = (3x –2) (2x + 1)

6x 2 –7x + 2 = (3x –2) (2x + 1)

12x 2 –5x –2 = (3x –2) (4x + 1)

∴ Δ (x) = (x + 1) (2x –1) (3x –2)

Now Operate R 2 → R 2

– R 1 ; R 3 –R 3 –2 R 1

=

= 2

Operate C 2 → C 2 –C 1 ; C 3 → C 3 –C 1

= 2

= –2 (6x + 4)

Hence other roots of Δ (x) = 0 are – , ,