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zmudz
 one year ago
The roots of
\(x^3 + Ax^2 + Bx + C = 0\)
are the cubes of the roots of
\(x^3 + x^2  3x 2.\)
Find \(A.\)
zmudz
 one year ago
The roots of \(x^3 + Ax^2 + Bx + C = 0\) are the cubes of the roots of \(x^3 + x^2  3x 2.\) Find \(A.\)

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If you divide \(x^3+x^23x2\) by \(x+2\) you get \(x^2x1\), so one of the roots in 2 and you can use the quadratic formula to find the other two. Cube all three of them and substitute the cubes into \(x^3+Ax^2+Bx+C=0\). The other two roots aren't rational, so it looks like you'll end up with a pretty gnarly system of equations to solve for \(A, B,\) and \(C\). \((2)^3 = 8\), so one of the equations is \(512+64A8B+C=0\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.1\[ x^3 + x^2  3x 2=0\\ (x+2)\left(x^2x1\right)=0\\ x=2,\,\frac{1}{2}\left(1+\sqrt{5}\right),\,\frac{1}{2}\left(1\sqrt{5}\right)\\ r_1=\frac{1}{2}\left(1+\sqrt{5}\right)\\ r_2=\frac{1}{2}\left(1\sqrt{5}\right) \] \[ \begin{align*} &\phantom{{}={}}\left(xr_1^3\right)\left(xr_2^3\right)\\ &=x^2\left(r_1^3+r_2^3\right)x+(r_1r_2)^3\\ &=x^2\left(\frac{1}{8}\left(1+3\sqrt{5}+15+5\sqrt{5}+13\sqrt{5}+155\sqrt{5}\right)\right)x1\\ &=x^24x1 \end{align*} \] \[ \begin{align*} &\phantom{{}={}}\left(x^24x1\right)(x+8)\\ &=x^34x^2x+8x^232x8\\ &=x^3+4x^233x8 \end{align*}\\ A=4\\ B=33\\ C=8 \]
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