Let's say that I have a function $f(x)=6x^3-5x^2-12x+k$. What should I put in Mathematica to solve the value of $k$ such that $3x+2$ is a factor of $f(x)$?
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2 Answers
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One liner:
Solve[PolynomialRemainder[6 x^3 - 5 x^2 - 12 x + k, 3 x + 2, x] == 0, k]
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$\begingroup$ Whoops, didn't notice the flip. Thanks @Runny! $\endgroup$ Commented Dec 22, 2016 at 10:11
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$\begingroup$ @RunnyKine I observe that you have largely stopped posting here. Is everything OK? I hope that you find the time and interest to participate more actively as you once did. $\endgroup$ Commented Dec 24, 2016 at 18:55
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$\begingroup$ @Mr.Wizard. Thanks for the concern. I hope to start participating actively again soon. Just not in the right state of mind currently. Thanks again. $\endgroup$ Commented Dec 24, 2016 at 20:27
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$\begingroup$ @RunnyKine I am glad you are well and I completely understand needing to be in the right mood for it. Should you move on to other things let me say thanks for all your contributions; I have some sense of the time involved in that. $\endgroup$ Commented Dec 24, 2016 at 20:37
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f[x_] := 6 x^3 - 5 x^2 - 12 x
poly = f[x] - f[-2/3]
so
PolynomialRemainder[poly, 3 x + 2, x]
yields 0
Table[{k, Factor[6 x^3 - 5 x^2 - 12 x + k]}, {k, -6, 6}] // Columnshowsk==-4. More finely:f[x_, k_] := 6 x^3 - 5 x^2 - 12 x + k; Solve[f[-(2/3), k] == 0, k]$\endgroup$