How do you factor by grouping $6 x^{3} - 2 x + 3 x^{2} - 1$ $6x^3 - 2x + 3x^2 - 1$ ?

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Answer 1 · 2015-04-30T08:49:01

we are given, 6 $x^{3}$ $x^3$ $-$ $-$ 2 $x$ $x$ $+$ $+$ 3 $x^{2}$ $x^2$ $-$ $-$ 1

on rearrangement:

$=$ $=$ 6 $x^{3}$ $x^3$ $+$ $+$ 3 $x^{2}$ $x^2$ $-$ $-$ 2 $x$ $x$ $-$ $-$ 1

The first two terms have 3 $x^{2}$ $x^2$ in common and the last two terms have $-$ $-$ 1 in common:

taking out the common terms:

$=$ $=$ 3 $x^{2}$ $x^2$ ( $2 x$ $2x$ + 1) $-$ $-$ 1 ( 2 $x$ $x$ $+$ $+$ 1)

making groups of two:
$=$ $=$ (3 $x^{2}$ $x^2$ $-$ $-$ 1) ( $2 x$ $2x$ $+$ $+$ 1 )

How do you factor by grouping 6x3−2x+3x2−16x^3 - 2x + 3x^2 - 1?