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3x^3-12x^2-17x+10= 0
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i assume you want the zeros of this using factor theorem one root is x=5
u can find the others by dividing x-5 into your expression to get a quadratic - solve this to get the other 2 roots
\[3x^3-12x^2-17x+10 = 0\]\[(x-5)(3x^2+3x-2) = 0\] so \[x = 5\space or \space \frac{-3 \pm \sqrt{3^2-4(3)(-2}}{6}\] \[x = 5 \space or \space \frac{1}{6}(-3 \pm \sqrt{33})\]
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