Question

If f(x) = x^4 − x^3 + x^2 and g(x) = −x^2, where x ≠ 0, what is (f ⁄g)(x)?

Answer 1

The correct answer is

$(\frac{f}{g})(x)=-x^2+x-1$

To solve this, first let's write the division:

$f(x)=x^4-x^3+x^2,g(x)=-x^2\Rightarrow(\frac{f}{g})(x)=\frac{x^4-x^3+x^2}{-x^2}$

Now we can factor out a x^2 on the top and the bottom of the expression:

$(\frac{f}{g})(x)=\frac{x^4-x^3+x^2}{-x^2}\Rightarrow(\frac{f}{g})(x)=\frac{x^2(x^2-x+1)}{x^2(-1)}$

Now we can cancel out and divide by (-1), or the same thing, multiply by (-1):

$(\frac{f}{g})(x)=\frac{x^2(x^2-x+1)}{x^2(-1)}=-(x^2-x+1)=-x^2+x-1$

Then the answer is (f/g)(x) = -x^2 + x - 1. That's the third option