How to find $P(x)$ here?

Question

For such polynomial $P(x)$,

$$P(x^4) = ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2$$

Find the remainder of the division of $P(x^3-3)$ with $x+1$.

We might have to find $P(x)$ first. My attempt is shown below

$$P(x^4) = a(x^4)^2 + (b+1)(x^4)x+(ab)x^4 + (a-1)\dfrac{x^4}{x^2}+2b-2$$

So

$$P(x) = ax^2 +(b+1)x^2 + (ab)x + \color{red}{(a-1)x^{-1}}+2b-2$$

However, the degree of a polynomial should be a natural number. Correspondingly I've gone wrong somewhere.

Regards

Answer 1

From $$P(t) = at^2+(b+1)t^{5/4}+(ab)t+(a-1)t^{1/2} + 2b-2$$ you deduce that $b=-1$ and $a=1$ for $P$ to indeed be a polynomial.

Then the requested remainder is

$$P((-1)^3-3)=(-4)^2-(-4)-4=16.$$

Answer 2

Just should be $b=-1$ and $a=1$.

The rest is smooth.

Let $p(x)=a_0x^{n}+a_1x^{n-1}+...+a_n,$ where $a_0\neq0.$

Thus, by the given $$a_0x^{4n}+a_1x^{4n-4}+...+a_n= ax^8+(b+1)x^5+(ab)x^4+(a-1)x^2 + 2b-2.$$ We know that two polynomials are equal iff all coefficients before powers are equal.

Thus, $b+1=0$ and $a-1=0$, which gives $$p(x)=x^2-x-4.$$

Now, the remainder it's $p(x^3-3),$ where $x=-1$ and since $$p(-4)=(-4)^2-(-4)-4=16,$$ we are done!

Answer 3

Obviously $P$ must be a quadratic polynomial:

$$ P(x) = cx^2+dx+e \implies P(x^4) = cx^8+dx^4+e$$

So $b+1=0$ and $a-1=0$.