How do you factor completely: $14 b x^{2} - 7 x^{3} - 4 b + 2 x$ $14bx^2 − 7x^3 − 4b + 2x$ ?

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Answer 1 · 2015-08-19T12:56:54

$(2 b - x) (7 x^{2} - 2)$ $color(green)((2b-x)(7x^2-2)$

$14 b x^{2} - 7 x^{3} - 4 b + 2 x$ $14bx^2-7x^3-4b+2x$

We can factor this expression by making groups of two terms:
$(14 b x^{2} - 7 x^{3}) + (- 4 b + 2 x)$ $(14bx^2-7x^3) +(-4b+2x)$

$7 x^{2}$ $7x^2$ is common to both the terms in the first group, and $- 2$ $-2$ is common to both the terms in the second group

$= 7 x^{2} (2 b - x) - 2 (2 b - x)$ $= 7x^2(2b-x) -2(2b-x)$
$2 b - x$ $2b-x$ is common to both the terms now:

$(2 b - x) (7 x^{2} - 2)$ $color(green)((2b-x)(7x^2-2)$ is the factorised form

How do you factor completely: 14bx^2 − 7x^3 − 4b + 2x14bx2−7x3−4b+2x14bx^2 − 7x^3 − 4b + 2x?