The polynomial 3x^3-13x^2+ax+b=0, a and b real numbers, has 2-i as one of its roots. Where does the graph of y=3x^3-13x^2+ax+b cross the x-axis? The solution is 1/3. Through what method(s) could i find that answer?
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Originally Posted by ceinstein1 has 2-i as one of its roots. One of the other roots is its complex conjugate 2+i, then use these with the fact that if P(a) = 0 then (x-a) is a factor. Also if you want to graph this you with have to use a complex plane with an imaginary and real axis.
If is a root of , then so is . , so we can write for some . By expanding the RHS, or otherwise, we see from the coefficient of that , and from that of we have , and as , we have . Thus it crosses the x-axis at [LaTeX ERROR: Convert failed]
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