Factorise x2 + x5

(i) x² + x⁵

Take out common in all terms we get,

x²(1 + x³)

x²(1³ + x³)

We know that, a³ + b³ = (a + b) (a² – ab + b²)

Where, a = 1, b = x

= x² [(1 + x) (1² – (1 × x) + x²)]

= x² (1 + x) (1 – x + x²)

(ii)32x⁴ – 500x

Take out common in all terms we get,

4x(8x³ – 125)

Above terms can be written as,

4x((2x)³ – 5³)

We know that, a³ – b³ = (a – b) (a² + ab + b²)

Where, a = 2x, b = 5

= 4x(2x – 5) ((2x)² + (2x × 5) + 5²)

= 4x(2x – 5) (4x² + 10x + 25)

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