How do you factor the expression $3 x^{2} + 3 + x^{3} + x$ $3x^2 + 3 + x^3 + x$ ?

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How do you factor the expression $3 x^{2} + 3 + x^{3} + x$ $3x^2 + 3 + x^3 + x$ ?

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Answer 1 · 2018-03-28T18:51:04

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

Explanation:

$3 x^{2} + 3 + x^{3} + x$ $3x^2+3+x^3+x$

Grouping the $1^{s t}$ $1^(st)$ $2$ $2$ terms together and the $2^{n d}$ $2^(nd$ $2$ $2$ together, we get:

$= 3 (x^{2} + 1) + x (x^{2} + 1)$ $=3(x^2+1)+x(x^2+1)$ (taking out common factors)

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

~Hope this helps! :)

Answer 2 · 2018-03-28T19:07:23

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

Explanation:

$factorise in 'groups'$ $"factorise in 'groups'"$

$[3 x^{2} + 3] + [x^{3} + x]$ $[3x^2+3]+[x^3+x]$

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

$take out the common factor (x^{2} + 1)$ $"take out the "color(blue)"common factor "(x^2+1)$

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

$\Rightarrow 3 x^{2} + 3 + x^{3} + x = (x^{2} + 1) (3 + x)$ $rArr3x^2+3+x^3+x=(x^2+1)(3+x)$

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How do you factor the expression $3 x^{2} + 3 + x^{3} + x$ $3x^2 + 3 + x^3 + x$ ?

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