Answer:
trying by Polynom division:
(2x^n+ax^2–6) / (x-1) = 2x^(n-1)+2x^(n-2)+…+2x^2+(a+2)x+(a+2)
-(2x^n-2x^(n-1))
———————
2x^(n-1)+ax^2–6
-(2x^(n-1)-2x^(n-2))
—————————
2x^(n-2)+ax^2–6
…..
….
…
———————
2x^(3)+ax^2–6
-(2x^3–2x^2)
——————-
(a+2)x^2–6
-((a+2)x^2-(a+2)x)
————————-
(a+2)x-6
-((a+2)x-(a+2))
———————
-6+(a+2)
So some explanation:
Upon trying polynomial division I realized 2 things:
If 2x^i is the highest expression on the current step, I will add a +2x^(i-1) on the right side of the equation sign.
And especially, after subtracting the correponding expression, we came from
the initial
2x^i+ax^2–6
to
2x^(i-1)+ax^2–6
So only the exponent decreased by one, nothing else changed!
So it’s obvious that this is a pattern that will continue.
So we repeat this pattern until we arrive at 2x^3+ax^2–6
Now we the pattern is over and we do normal polynomial division.
Anyways, for the pattern induced part, we would actually have to prove via induction or something similar to it that there evn is such a pattern.
Now, since you understand how the polynomial division worked there, let’s think about the results we got:
The remainder is obviously
-6+(a+2)=-6+2+a=-4+a
But as stated the remainder should equal to -7
So a has to be -3 since
-4+a=-4–3=-7
And what about n?
Actually, with the conditions you gave , all positive n equal or greater than 3 fit there. (the value of a above is actually based on the fact that n>=3)
For n=2 we get
2x^2+ax^2–6=(2+a)x^2–6
and therefor
((2+a)x^2–6) / (x-1)= (2+a)x+(2+a)
-((2+a)x^2-(2+a)x)
————————-
(2+a)x-6
-((2+a)x-(2+a))
———————
-6+(2+a)
Since again remainder has to be equal to -7, we get
-7=-6+2+a=-4+a
therefor a=-3
Son the answer a=-3 actually works for all n>=2.
Now n=1:
f(x)=2x^1+ax^2–6=ax^2+2x-6
so we get
(ax^2+2x-6) / (x-1) = ax+(2+a)
-(ax^2-ax)
—————
(2+a)x-6
-((2+a)x-(2+a))
———————-
-6+(2+a)
Oh miracle, we get a=-3 again.
So a=-3 for all n>=1
finally n=0:
f(x)=2x^0+ax^2–6=2*1+ax^2–6=-4+ax^2=ax^2–4
so we get
(ax^2–4) / (x-1) = ax+a
-(ax^2-ax)
————-
ax-4
-(ax-a)
———
-4+a
So….
-4+a=-7
-> a=-3
So a=-3 for all n>=0!!!
Why don’t we look at negative n?
Cause for example
2x^(-1)+ax^2–6
=2/x+ax^2–6
Is no polynom anymore and therefor polynom division is no longer possible!
So to sum it up:
a=-3 when n>=0
(or respectively a,n=(-3,r) with r>=0 satisfies your conditions)
how to right the polynomial function:
f(x)=2x^n+ax^2–6 =righthandside*(x-1)-7
righthandside means the polynom right from the = in the above polynom divisions.
And if n<0, there exists no solution.
(Okay, there may actually a solution. But I both don’t know how to find it if it exists and I’m too lazy to try it too.)
