SEE HERE.Consider the monic (leading coefficient is 1) cubic equation f (x) = x3 + bx2 + cx + d, where b, c, d ∈ R. Letting x = z-(b/3) , rewrite f (x) as another monic cubic equation f (z) with real coefficients but which does not have a squared term.
yes. and I have sub those into the original f(x) equation. I am unsure of what to do next and what my final answer should look like
This is exactly why the forum rules ask that you submit your work, even if you know that it is wrong, so we can see exactly where you need help. So please obey the forum rules and present us with your work so that you can get the best help from this forum.yes. and I have sub those into the original f(x) equation. I am unsure of what to do next and what my final answer should look like
OP said:This is exactly why the forum rules ask that you submit your work, even if you know that it is wrong, so we can see exactly where you need help. So please obey the forum rules and present us with your work so that you can get the best help from this forum.
Your final answer should be in the form Az3+Bz2+ Cz + D
Okay, so you should have something like:
[MATH]f(z)=\left(z^3-3z^2\frac{b}{3}+3z\frac{b^2}{9}-\frac{b^3}{27}\right)+b\left(z^2-2\frac{b}{3}z+\frac{b^2}{9}\right)+c\left(z-\frac{b}{3}\right)+d[/MATH]
I would next distribute \(b\) and \(c\) to remove all parentheses, and then combine like terms. What do you get?
Yeah, I knew that. So B will end up as 0OP said:
"....... f (z) with real coefficients but which does not have a squared term. " So:
Your final answer should be in the form Az3+ Cz + D