Given the binomial raising to the 3rd power forumula: $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3,$ how can we reverse the action? I mean from the result produce the original binomial. For example how to translate this expression : $2x^3 +18x^2 -3x -27$ to the original $(a+b)^3$ form? I know how to find the $b$ variable but I'm not sure about the $a$.
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1$\begingroup$ $2x^3 +18x^2 -3x -27$ is not the third power of any polynomial. It factors as $(x+9) (2 x^2-3)$. $\endgroup$– hmakholm left over MonicaCommented Nov 9, 2015 at 15:57
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$\begingroup$ you should look at $$(ax+b)^3$$ to make it easier for you to determine the conditions on $a,b$ if they indeed exist. $\endgroup$– Chinny84Commented Nov 9, 2015 at 15:58
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$\begingroup$ @HenningMakholm can you show me the steps how you factored? $\endgroup$– BsDCommented Nov 9, 2015 at 16:06
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1$\begingroup$ $2x^3+18x^2-3x-27=2x^2(x+9)-3(x+9)=(x+9)(2x^2-3).$ $\endgroup$– Jack FrostCommented Nov 9, 2015 at 16:07
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$\begingroup$ Only polyinomials with triple roots can be factored thusly. The vast majority of cubics do not have triple roots and can't be so factored. $\endgroup$– fleabloodCommented Nov 9, 2015 at 17:25
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1 Answer
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As chinny84 suggested:
$(ax +b)^3 = a^3x + 3a^2bx^2 + 3ab^2x + b^3$
So to reduce $2x^3 +18x^2 -3x -27$ to $(ax + b)^3$ or $c(ax + b)^3$ we must solve the following four equations
$a^3 = 2 $
$3a^2b = 18 $
$3ab^2 = -3 $
$b^3 = -27 $
(or multiples of 1/c of these equations.) These clearly have no solutions (rational or otherwise) so it can't be factored as $(ax + b)^3$. This shouldn't surprise us as only cubics with a triple root can be so factored.
