Q. 2 C4.4( 16 Votes )

# Factorize the fol

(x2 – 6x) 2 – 8(x2 – 6x + 8) – 64

(x2 – 6x) 2 – 8(x2 -6x) - 64 – 64

(x2 – 6x) 2 – 8(x2 -6x) – 128

Put (x2 -6x) = a

(a) 2 – 8(a) – 128

a2 – 8a – 128

a2 – 16a + 8a - 128

a (a-16) + 8(a – 16)

(a + 8) × (a-16)

But a = (x2 -6x)

((x2 -6x) + 8) × ((x2 -6x) – 16)

(x2 -6x + 8) × (x2 -6x – 16)

(x2 -4x – 2x + 8) × (x2 -8x + 2x – 16)

(x(x-4) – 2(x-4)) × (x(x-8) + 2(x-8)

(x-2)(x-4)(x-8)(x+2)

Therefore, the factorized form = (x-2) (x-4) (x-8) (x+2)

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