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Factorise x² + (1/x)² + 2 - 3x - (3/x).

Solution:

Given, x² + (1/x)² + 2 - 3x - (3/x)

We have, [x + (1/x)]² = x² + (1/x)² + 2(x)(1/x) = x² + (1/x)² + 2

∴ x² + (1/x)² + 2 - 3x - (3/x)

= [x² + (1/x)² + 2] - 3[x + (1/x)]

= [x + (1/x)]² - 3[x + (1/x)]

= [x + (1/x)][x + (1/x)] - 3[x + (1/x)]

= [x + (1/x)]{[x + (1/x)] - 3}

✦ Try This: Factorise: 4x² + (1/4x²) + 6x + (3/2x) + 2

Given, 4x² + (1/4x²) + 6x + (3/2x)+ 2

We have, [2x + (1/2x)]² = (2x)² + (1/2x)² + 2(2x)(1/2x) = 4x² + 1/4x² + 2

∴4x² + (1/4x²) + 2 + 6x + (3/2x)

= [(2x)² + (1/2x)² + 2(2x)(1/2x)] + 3[2x + 1/2x]

=[2x + 1/2x]² + 3[2x + 1/2x]

= [2x + 1/2x][2x + 1/2x + 3]

☛ Also Check: NCERT Solutions for Class 8 Maths Chapter 9

NCERT Exemplar Class 8 Maths Chapter 7 Problem 118

Summary:

Factorising x² + (1/x)² + 2 - 3x - (3/x) we get [x + (1/x)]{[x + (1/x)] - 3}

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