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I have obtained three points where it cuts the $x$-axis because I followed the process for quadratic equations,not cubic. Could someone explain it in detail?

Thanks to everyone who takes the time to answer.

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  • $\begingroup$ Easier method: Step one, try to find a root by inspection. Common, easy to calculate guesses are $0,1,-1$. If you get lucky enough to have correctly guessed a root, perform polynomial long division to write it in the form $(ax^2+bx+c)(x+d)$. Find the roots for the quadratic factor as normal. Harder method: apply Cardano's Method (very rarely taught). $\endgroup$ – JMoravitz Oct 17 '16 at 16:07
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By a good look ,find out that $$x=-1 \to f(-1)=0\\x=2 \to f(2)=0$$ so you can divide $x^3-5x^2+2x+8 $ to $x+1$ or $x-2$ then you will have :$$f(x)=(x+1)(x-2)(x-4) $$ so $x=-1,2,4$ cuts $x$-axis
and when you put $x=0 \to f(0)=+8$ so $y$-axis cuts in $(0,8)$

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