Thread: (r-s)(x)

Suppose r(x) = 3+4/x^2+1, s(x)= x^+6/6x-1. Write the expression (r-s)(x) as a simplified ratio with the numerator and denominator each written as a sum of terms of the form cx^m and c>0 for the term with the highest power in the numerator. I ended up with (-x^4+11x^2+21x-10)/-6x^3+x^2-6x+1. Where did I go wrong here?

Without knowing the supposition, it is difficult to say where you went wrong.

This is the right answer but I don't know how to get it.

The links to images in your first post are broken for me.

Edited.

Originally Posted by Eraser147

Suppose r(x) = 3+4/x^2+1, s(x)= x^+6/6x-1.

Why not add 3 + 1 and simplify r(x) to 4 + (4/x^2)? And why do you write + in front of 6 in x^+6?

The only way I got the answer right was when I subtracted s with r instead of r - s. It looked like this with the polynomial subtraction (x^4 + 7x^2 + 6)-(18x^2+21x-4). And it gives me x^4-11x^2+21x+10 for the numerator.

Despite three request to write the question correctly, I am still doubting this has been done.

Here. it has to be correct. Just click on the image to enlarge.

Why are the signs flipped again at the end?

Changing the sign of both the numerator and the denominator (i.e., multiplying both by -1) does not change the fraction. The problem asks for a fraction with a positive leading coefficient in the numerator.

Originally Posted by Eraser147

Suppose r(x) = 3+4/x^2+1,

Do you mean 3+ 4/(x^2+ 1)?

s(x)= x^+6/6x-1.

What should be after the "^"? Do you mean x^?+ 6/(6x- )? Or (x^?+ 6)(6x- 1)?

Write the expression (r-s)(x) as a simplified ratio with the numerator and denominator each written as a sum of terms of the form cx^m and c>0 for the term with the highest power in the numerator. I ended up with (-x^4+11x^2+21x-10)/-6x^3+x^2-6x+1. Where did I go wrong here?

Oh, I already figured it out. Emakarov showed it. My mistake was not flipping the signs by multiplying -1 from numerator and denominator since the problem asked for the leading term to be a positive number and the constant to be more than zero. Read the post above to find the image of the actual equation if you are interested.