Rewrite Cubic equation f(x) = x^3 + bx^2 + cx + d, using x = z - (b/3)

S

susanf

New member
Joined
Apr 3, 2019
Messages
2
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.
 
Hello, and welcome to FMH! :)

Okay, I am assuming we are given:

[MATH]f(x)=x^3+bx^2+cx+d[/MATH]
If we let:

[MATH]x=z-\frac{b}{3}[/MATH]
Then:

[MATH]x^2=\left(z-\frac{b}{3}\right)^2=z^2-2\frac{b}{3}z+\frac{b^2}{9}[/MATH]
[MATH]x^3=\left(z-\frac{b}{3}\right)^3=z^3-3z^2\frac{b}{3}+3z\frac{b^2}{9}-\frac{b^3}{27}[/MATH]
Can you proceed?
 
susanf said:
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.
Click to expand...
SEE HERE.
 
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
 
susanf said:
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
Click to expand...

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?
 
susanf said:
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
Click to expand...
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
 
Jomo 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
Click to expand...
OP 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
 
MarkFL said:
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?
Click to expand...

To follow up, we next get:

[MATH]f(z)=z^3-bz^2+\frac{1}{3}b^2z-\frac{1}{27}b^3+bz^2-\frac{2}{3}b^2z+\frac{1}{9}b^3+cz-\frac{1}{3}bc+d[/MATH]
Combine like terms:

[MATH]f(x)=z^3+\left(c-\frac{1}{3}b^2\right)z+\left(\frac{2}{27}b^3-\frac{1}{3}bc+d\right)[/MATH]
 
You must log in or register to reply here.
Top