Quartic function sharing three common roots with another function

Question

So the question is

"The quartic function f(x) = (x^2+x-20)(x^2+x-2) has three roots in common with the function g(x) = f(x-k), where k is a constant. Find the two possible values of k."

So my understanding so far is that function of g is translated k units. Function of g would look something like

g(x) = ((x-k)^2+(x-k)-20)*((x-k)^2+(x-k)-2)

Since these two functions share the same coordinates, therefore to find x:

f(x) = g(x)

(x^2+x-20)(x^2+x-2) = ((x-k)^2+(x-k)-20)((x-k)^2+(x-k)-2)

When expanded, this produces 45 terms and when like terms are added/subtracted, I get:

-4kx^3 + 6k^2x^2 - 6kx^2 - 4k^3x + k^4 + 7k^2x - 2k^3 - 21k^2 + 22k = 0

Maybe I think I have not taken the right direction so any help will be much appreciated. And my apologies in advance for any confusion caused.

Thank You

Answer 1

The roots of $f(x)=(x^2+x-20)(x^2+x-2)$ are $-5$, $-2$, $1$, and $4$. So, the roots of $f(x-k)$ are $k-5$, $k-2$, $k+1$, and $k+4$. So, those two possible values of $k$ are $k=3$ (in which case the roots of $f(x-k)$ are $-2$, $1$, $4$, and $7$) and $k=-3$ (in which case the roots of $f(x-k)$ are $-8$, $-5$, $-2$, and $1$).

Answer 2

The first thing you could have noticed is f(X) is being given as product of two quadratic polynomials. And more astonishing is that those two quadratic polynomials are themselves factorised. So you easily get the roots of f(X). Now comes the play of g(X). Put X =x-k everywhere in f(X) such that you get the function g(X). And now you can easily get roots of g(X) in terms of k. After that it's just manually checking values of k. And yes, don't directly jump into expansion because it is feasible upto cubic but after that it's much more complicated. Thank You!