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

Question

2276 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-30T08:58:37

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

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

taking out the common terms:

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

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

making groups of two,

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

How do you factor by grouping 2x^3 - x^2 - 6x + 32x3−x2−6x+32x^3 - x^2 - 6x + 3?