A polyomial problem to show that Q (x)/$x^k $ is strictly positive 
I am extremely sorry I couldnot type as it requires much time to do MathJax.
This problem Question 8 was asked in an entrance examination at the 10+2 level.
My approqch for part (a) was  to use differentiation and find Q'(x) but it is not so ....helping me to arrive at the result.
Any hints or help is appreciable.
Thank you
 A: (Fill in the gaps as needed. If you're stuck, show what you've considered and explain why you're stuck.)
Hint: Show that $ |x^k | > |\sum a_i x^{k+i}|$.
A: I don't know if it could be the correct method, but I show you what I was thinking of.
So, imagine to plug arbitrary numbers inside the inequality. You'd notice that the sum (with sigma symbol) is always positive (because it starts with 1 and approaches to positive infinity). Additionally, it has an absolute value, so it is surely always positive.
the fraction is always positive too, because it has "1+[always positive stuff]". It approaches to 0 from the right (to prove that, imagine to plug every positive or negative number and see what would happen). Therefore, it's always positive. zero isn't included because of the sign of the inequality "i.e <, which means "is less than and not equal to [something]".
Moreover, x is inside the absolute value too, so you could surely say Q(X) is always positive.
EDIT (because I forgot the main part of this exercise): imagine to plug arbitrary values inside "k" variable (because k is a variable), you'd notice that it doesn't matter what you plug into it, because you'd get a positive value.
secondly, if x is negative, it's not a problem because in the inequality you have the absolute values.
