Let’s resolve into factors

a⁴ + 4a²-5

(a²)² + 4a²-5 … (1)

To resolve it into factors we first resolve 5

5=1×5

Given, x² + (m + n)x + mn=(x + m)(x + n) … (2)

Comparing 1 and 2

m + n =4

mn =-5

∴ (1) can be written as

(a²)² + 5a²- a²-5

= (a² + 5)( a²-1)

∴ the resolved factors are (a² + 5)( a²-1)