How do you factor completely $3 x^{3} - 36 x^{2} + 96 x$ $3x^3-36x^2+96x$ ?

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Answer 1 · 2016-08-03T15:20:22

$3 x (x - 8) (x - 4)$ $3x(x-8)(x-4)$ .

Observe that $3 x$ $3x$ is common in every term. We take it out, & get,

The Expression $= 3 x (x^{2} - 12 x + 32)$ $=3x(x^2-12x+32)$

$=3x{x^2-8x-4x+32}...............[8xx4=32, 8+4=12]$

$=3x{x(x-8)-4(x-8)}$

$=3x(x-8)(x-4)$ .

How do you factor completely 3x^3-36x^2+96x3x3−36x2+96x 3x^3-36x^2+96x?