Factorize using identities (i) 1-64x³+48x²-12x

Mathematics

1 answer:

enot [183]3 years ago

6 0

Answer:

(1 - 4x)³

Step-by-step explanation:

The first 2 terms are a difference of cubes and factor in general as

a³ - b³ = (a - b)(a² + ab + b²), thus

1 - 64x³

= 1³ - (4x)³

= (1 - 4x)(1 + 4x + 16x²)

Thus

1 - 64x³ + 48x² - 12x ← factor out 12x from each of the 2 terms

= (1 - 4x)(1 + 4x + 16x²) + 12x(4x - 1) ← factor out - 1 from (4x - 1)

= (1 - 4x)(1 + 4x + 16x²) - 12x(1 - 4x) ← factor out (1 - 4x) from the terms

= (1 - 4x)(1 + 4x + 16x² - 12x)

= (1 - 4x)(1 - 8x + 16x²) ← perfect square

= (1 - 4x)(1 - 4x)²

= (1 - 4x)³ ← in factored form

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