_Catra_
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Sorry for the whole load of questions lately but how would you factorise x^3+x+10 I know the answer is (x+2)(x^2-2x+5) butwhat would the steps be to get the answer. Thank you in advance
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lhabgabdgbfa
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you do factor theorem, basically the number -2 can be substituted into x^3+x+10 to give 0. then you can do algebraic long division and factorise the quotient
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RDKGames
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(Original post by _Catra_)
Sorry for the whole load of questions lately but how would you factorise x^3+x+10 I know the answer is (x+2)(x^2-2x+5) butwhat would the steps be to get the answer. Thank you in advance
As above. Once you know x+2 is a factor, its basic division from there.

A shortcut would be to generate the term x+2 in the cubic as below:

\begin{aligned} x^3+x+10 & = x^3+(2x^2-2x^2)+(-4x+4x)+x + 10 \\ & = (x^3+2x^2)+(-2x^2-4x)+(5x+10) \\ & = x^2(x+2)-2x(x+2)+5(x+2) \\ & = (x+2)(x^2-2x+5)\end{aligned}.
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lhabgabdgbfa
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(Original post by RDKGames)
As above. Once you know x+2 is a factor, its basic division from there.

A shortcut would be to generate the term x+2 in the cubic as below:

\begin{aligned} x^3+x+10 & = x^3+(2x^2-2x^2)+(-4x+4x)+x + 10 \\ & = (x^3+2x^2)+(-2x^2-4x)+(5x+10) \\ & = x^2(x+2)-2x(x+2)+5(x+2) \\ & = (x+2)(x^2-2x+5)\end{aligned}.
wow, never seen that method before
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_Catra_
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(Original post by lhabgabdgbfa)
you do factor theorem, basically the number -2 can be substituted into x^3+x+10 to give 0. then you can do algebraic long division and factorise the quotient
(Original post by RDKGames)
As above. Once you know x+2 is a factor, its basic division from there.

A shortcut would be to generate the term x+2 in the cubic as below:

\begin{aligned} x^3+x+10 & = x^3+(2x^2-2x^2)+(-4x+4x)+x + 10 \\ & = (x^3+2x^2)+(-2x^2-4x)+(5x+10) \\ & = x^2(x+2)-2x(x+2)+5(x+2) \\ & = (x+2)(x^2-2x+5)\end{aligned}.
Thank you to both of you
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