Is it possible to factor $y = 2 x^{2} + 4 x + 8$ $y=2x^2+4x+8$ ? If so, what are the factors?

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Is it possible to factor $y = 2 x^{2} + 4 x + 8$ $y=2x^2+4x+8$ ? If so, what are the factors?

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Answer 1 · 2015-12-05T06:57:59

Yes it is possible to factorise $2 x^{2} + 4 x + 8$ $2x^2+4x+8$
The factors are 2, $(x + 1 - i \sqrt{3})$ $(x+1-i sqrt(3))$ , $(x + 1 + i \sqrt{3})$ $(x+1+i sqrt(3))$

Explanation:

$2 x^{2} + 4 x + 8$ $2x^2+4x+8$ =2( $x^{2} + 2 x + 4$ $x^2+2x+4$ )

$x^{2} + 2 x + 4$ $x^2+2x+4$ = $(x + 1 - i \sqrt{3})$ $(x+1-i sqrt(3))$ $(x + 1 + i \sqrt{3})$ $(x+1+i sqrt(3))$

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Is it possible to factor $y = 2 x^{2} + 4 x + 8$ $y=2x^2+4x+8$ ? If so, what are the factors?

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Explanation:

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