Question

How can I find the factor of

`243 {w}^(4) z - 48z`

Answer 1

3z(3w - 2)(3w + 2)(9w² + 4)

Explanation:

take out a common factor 3z from both terms

= 3z(81
`w^(4)` - 16)

81
`w^(4)` - 16 ← is a difference of squares

• a² - b² = (a - b)(a + b)

81
`w^(4)` = (9w²)² → a = 9w² and 16 = 4² → b = 4

= 3z(9w² - 4)(9w² + 4)

9w² - 4 ← is also a difference of squares with a = 3w and b = 2

= 3z(3w - 2)(3w + 2)(9
`w^(4)` + 4)