(i) x2 + 1/x2 – 11
(ii) x4 + 5x2 + 9
Answer:
(i) x2 + 1/x2 – 11
Above terms can be written as,
x2 + (1/x2) – 2 – 9
Then, (x2 + (1/x2) – 2) – 32
We know that, (a – b)2 = a2 – 2ab + b2,
(x2 – (2 × x2 × (1/x2)) + (1/x)2)
(x – 1/x)2 – 32
We know that, a2 – b2 = (a + b) (a – b)
(x – 1/x + 3) (x – 1/x – 3)
(ii) x4 + 5x2 + 9
x4 + 5x2 + 9
x4 + 6x2 – x2 + 9
(x4 + 6x2 + 9) – x2
((x2)2 + (2 × x2 × 3) + 32)
We know that, (a + b)2 = a2 + 2ab + b2,
((x2)2 + (2 × x2 × 3) + 32)
So, (x2 + 3)2 – x2
We know that, a2 – b2 = (a + b) (a – b)
(x2 + 3 + x) (x2 + 3 – x)
More Solutions: