Question

11 if $f(x) = x^4 - x^3 + x^2$ and $g(x) = -x^2$, where $x \
eq 0$, \
options: \
$x^2 - x + 1$ \
$x^2 + x + 1$ \
$-x^2 + x - 1$ \
$-x^2 - x - 1$

Explanation:

Step1: Divide \( f(x) \) by \( g(x) \)

We have \( f(x) = x^4 - x^3 + x^2 \) and \( g(x) = -x^2 \). To find \( \frac{f(x)}{g(x)} \) (assuming the problem is to divide \( f(x) \) by \( g(x) \) as the options are linear/quadratic and \( f(x) \) is quartic), we divide each term of \( f(x) \) by \( g(x) \).

Step2: Divide each term

• For the first term: \( \frac{x^4}{-x^2} = -x^{4 - 2} = -x^2 \)? Wait, no, wait. Wait, maybe I misread. Wait, \( f(x) = x^4 - x^3 + x^2 \), factor out \( x^2 \): \( f(x) = x^2(x^2 - x + 1) \). Then \( g(x) = -x^2 \). So \( \frac{f(x)}{g(x)} = \frac{x^2(x^2 - x + 1)}{-x^2} \). The \( x^2 \) terms cancel (since \( x

eq 0 \)), so we get \( - (x^2 - x + 1) = -x^2 + x - 1 \)? Wait, no, wait, maybe the problem is \( \frac{f(x)}{g(x)} \) but let's check the options. Wait, the options include \( -x^2 + x - 1 \)? Wait, no, wait, let's do the division properly.

Wait, \( f(x) = x^4 - x^3 + x^2 \), \( g(x) = -x^2 \). So divide each term:

\( \frac{x^4}{-x^2} = -x^{2} \)

\( \frac{-x^3}{-x^2} = x \)

\( \frac{x^2}{-x^2} = -1 \)

Wait, that gives \( -x^2 + x - 1 \)? But wait, the first term: \( x^4 / (-x^2) = -x^{4-2} = -x^2 \), second term: \( -x^3 / (-x^2) = x \), third term: \( x^2 / (-x^2) = -1 \). So combining, \( -x^2 + x - 1 \). Wait, but let's check the options. The third option is \( -x^2 + x - 1 \). Wait, but let's re-express \( f(x) \): \( f(x) = x^2(x^2 - x + 1) \), \( g(x) = -x^2 \), so \( f(x)/g(x) = - (x^2 - x + 1) = -x^2 + x - 1 \). Yes, that matches the third option.

Wait, but let's check again. Maybe the problem is \( \frac{f(x)}{g(x)} \). So:

\( \frac{x^4 - x^3 + x^2}{-x^2} = \frac{x^4}{-x^2} - \frac{x^3}{-x^2} + \frac{x^2}{-x^2} = -x^2 + x - 1 \).

Answer:

\( \boldsymbol{-x^2 + x - 1} \) (which is the third option: -x² + x - 1)