Gfilui

I have a question about a polynomial being divided by another, but there are multiple unknown parameters at play making this a very difficult question to figure out, but now we are restricting ourselves to solving this question with pre-calculus techniques only.
Here is the question:

Determine $a$ and $b$ such that $ax^n+2 + bx^n + 2$ is divisible by $(x-1)^2$

Since $1$ is root of a polynomial $p(x)=ax^n+2+bx^n+2$ we have $p(1)=0$ so
$b=-2-a$. Now we have:
$$p(x)=ax^n+2-2x^n -ax^n+2 $$
$$= ax^n(x^2-1)-2(x-1)(x^n-1+...+x^2+x+1) $$
$$= (x-1)underbraceBig(ax^n(x+1)-2(x^n-1...+x^2+x+1)Big)_q(x) $$
Now since $1$ double root we have also $q(1)=0$ so $2a-2n=0$ so $a=n$ and $b=-n-2$.

Hint: $small abig((x-1)+1big)^n+2+bbig((x-1)+1big)^n+2=(x-1)^2(ldots)+abinomn+21(x-1)+a+bbinomn1(x-1)+b+2$.

Let $f(x)=ax^n+2+bx^n+2$ be divisible by $(x-1)^2$ therefore $$f(1)=0\f'(1)=0$$which means that $$a+b+2=0\(n+2)a+nb=0$$so $$a=-dfracnn+2b$$therefore $$a=n\b=-n-2$$Using pre-calculus method:

$$f(x)=(x-1)Q(x)$$therefore $$f(1)=0$$or $$b=-a-2$$ by substituting we obtain$$f(x)=ax^n+2-ax^n-2x^n+2$$here we can write $$dfracf(x)x-1=ax^n(x+1)-2(x^n-1+cdots+x+1)$$again by substituting $x=1$ we finally obtain $$a=n\b=-n-2$$

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