Is it possible to factor $y = x^{3} - 2 x^{2} + 3 x - 6$ $y=x^3 - 2x^2 + 3x -6$ ? If so, what are the factors?

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Is it possible to factor $y = x^{3} - 2 x^{2} + 3 x - 6$ $y=x^3 - 2x^2 + 3x -6$ ? If so, what are the factors?

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Answer 1 · 2016-05-28T08:49:26

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

Explanation:

Group the terms in 'pairs' thus

$[x^{3} - 2 x^{2}] + [3 x - 6]$ $[x^3-2x^2]+[3x-6]$

now factorise each 'pair'

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

A common factor of (x -2) can now be removed

$\Rightarrow (x - 2) (x^{2} + 3)$ $rArr(x-2)(x^2+3)$

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Is it possible to factor $y = x^{3} - 2 x^{2} + 3 x - 6$ $y=x^3 - 2x^2 + 3x -6$ ? If so, what are the factors?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

... and beyond

Is it possible to factor y=x3−2x2+3x−6y=x^3 - 2x^2 + 3x -6 ? If so, what are the factors?

1 Answer

Explanation:

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Impact of this question

Is it possible to factor $y = x^{3} - 2 x^{2} + 3 x - 6$ $y=x^3 - 2x^2 + 3x -6$ ? If so, what are the factors?