Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$

Question

Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$

my attempt: $$(x+1)=(x-1)\frac{(x+1)}{(x-1)}$$ but this seems useless? I want to use synthetic division but I don't know how

You could simply substitute $x=y-1$ and expand the LHS. – Martin R Jul 24 '19 at 11:46 — Martin R, Jul 24 '19 at 11:46
The LHS simplifies to $x^3 - 5x$. – ViktorStein Jul 24 '19 at 11:48 — ViktorStein, Jul 24 '19 at 11:48

Michael Rozenberg · Accepted Answer · 2019-07-24T11:55:34.703

4

It's $$(x+1-2)^3+3(x+1-2)^2-2(x+1-2)-4=(x+1)^3-3(x+1)^2-2(x+1)+4.$$ Can you end it now?

edited Jul 24 '19 at 11:55

answered Jul 24 '19 at 11:47

Michael Rozenberg

194,933

score 0 · Answer 2 · answered Jul 24 '19 at 11:48

0

HINT:

If $F(t)=c_0+c_1t+c_2t^2+\cdots$ then $$ c_n=\frac1{n!}\left.\frac{d^nF}{dt^n}\right|_{t=0}. $$

answered Jul 24 '19 at 11:48

Andrea Mori

26,969

score 0 · Answer 3 · answered Jul 24 '19 at 12:17

0

In case you don't know how to solve this most elegantly, there is still the straightforward possibility by expanding both sides. The LHS is given by $$ x^3-5x, $$ whereas the RSH is $$ ax^3 + (3a + b)x^2 + (3a + 2b + c)x + a + b + c + d. $$

answered Jul 24 '19 at 12:17

Dietrich Burde

130,978

Given $(x-1)^3+3(x-1)^2-2(x-1)-4=a(x+1)^3+b(x+1)^2+c(x+1)+d$, find$(a,b,c,d)$

3 Answers3