algebra

[verbal cypress]

How to approach this?

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[warm crow]

Obviously, not all values are of the same sign.
Looking at the graph of, say, f(x) = x^3 - x, we can assume the following for a > 0:
P(1) = -a
P(2) = a
P(3) = a
P(5) = -a
P(6) = -a
P(7) = a
The polynomial should be odd around x = 4, so we have it in the form:
P(x) = A(x - 4)^3 + B(x - 4)
Let's find P(1), P(2) and P(3).
P(1) = -27A - 3B = -a
P(2) = -8A - 2B = a
P(3) = -A - B = a
We can solve the system and find the coefficients. But that doesn't determine P(0) uniquely.

Maybe they meant P(4)?

Because that can be determined even without the system.

[verbal cypress]

They did something like this

I am so confused

[verbal cypress]

Why did we assume the function x³-x out of the blue

[warm crow]

Just by looking at it. Can't honestly say anything else 😄
Though, their logic isn't correct. We can multiply the resulting polynomial by any constant and the property will still work, and it will change the value of P(0). The only values that never change are the zeroes of P(x), which are at x = 4 and x = 4 ± √(7).

[verbal cypress]

How did we get p(4) = 0

[oak echo]

Hmm. Okay, let's start by assuming P(x) is a general cubic, ax^3 + bx^2 + cx + d.

Then we can just plug in x = 1, 2, 3, 5, 6, 7. LordDarpinger and the illustration are both right that we would get a pattern that looks like P(1) = -P(2) = -P(3) = P(5) = P(6) = -P(7).

[verbal cypress]

Yes yes, but why did we make two lines

For x axis

That is