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Solve the cubic equation $z^3+6z^2+12z+16=0$

and show the three solutions on an Argand diagram

HINT: $(a+b)^3$

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    Is that a question? What are your thoughts?2011-12-30
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    @David, the OP's own hint, namely recognizing the binomial expansion of $(z+2)^3+8=0$, seems to be easier than dividing out one of the solutions.2011-12-30
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    @Henning Makholm Indeed... Deleting my previous comments.2011-12-30

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Okay, lets proceed with your hint then.

$$z^3+6z^2+12z+16=0$$ $$(z+2)^3+8=0$$ $$(z+4)(z^2+2z+4)=0$$ $$(z+4)[(z+1)^2+3]=0$$

So you obtain $z=-4, -1+i\sqrt{3},-1-i\sqrt{3}$

It should be easy to represent these on an argand diagram.

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    But you might miss an important feature of the diagram if you do it this way. I'd go $x+2=-2Q$, where $Q$ is any of the three cube roots of 1. Now if you know the geometry of the cube roots of 1, you can plot them (without having to know anything about how big $\sqrt3$ is), then multiply by $-2$, then shift left by 2.2011-12-30
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    @GerryMyerson, you are absolutely right. Thanks.2011-12-30
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    why not just take cube roots of minus eight and shift by minus 2?2011-12-30