If using properties of determinants find the value of f(2x) – f(x).

We have,

Taking a common from C₁

Applying C₂→ C₁+ C₂

Expanding along R₁ , we get

f(x) = a[1{a²+ax-(-1)(ax+x²)}]

= a(a²+ax+ax+ x²) = a(a²+2ax + x²)

Now, f(2x) = a(a²+2a× 2x + (2x)²)= a(a²+4ax + 4x²)

∴ f(2x)- f(x) = a(a²+4ax + 4x²) - a(a²+2ax + x²)

= a[a²+4ax + 4x²- a²-2ax - x²]

=a[2ax+3x²] = ax[2a+3x]