Question

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

Explanation:

Step1: Factor the denominator

The denominator \( h^2 - 1 \) can be factored using the difference of squares formula \( a^2 - b^2=(a + b)(a - b) \). So, \( h^2 - 1=(h + 1)(h - 1) \).

Step2: Try to factor the numerator

Let's try to factor the numerator \( 2h^4 - h^3 + h^2 + h - 3 \). We can use polynomial long division or try to find roots by trial and error. Let's test \( h = 1 \):
Substitute \( h = 1 \) into the numerator: \( 2(1)^4 - (1)^3 + (1)^2 + 1 - 3=2 - 1 + 1 + 1 - 3 = 0 \). So, \( (h - 1) \) is a factor of the numerator.
Now, perform polynomial long division of \( 2h^4 - h^3 + h^2 + h - 3 \) by \( (h - 1) \).
Using long division:
\[

$$\begin{array}{r|rrrrr} h - 1 & 2h^4 & -h^3 & +h^2 & +h & -3 \\ & 2h^4 & -2h^3 & & & \\ \cline{2-6} & 0 & h^3 & +h^2 & +h & -3 \\ & & h^3 & -h^2 & & \\ \cline{3-6} & & 0 & 2h^2 & +h & -3 \\ & & & 2h^2 & -2h & \\ \cline{4-6} & & & 0 & 3h & -3 \\ & & & & 3h & -3 \\ \cline{5-6} & & & & 0 & 0 \\ \end{array}$$

\]
So, \( 2h^4 - h^3 + h^2 + h - 3=(h - 1)(2h^3 + h^2 + 2h + 3) \).
Now, let's factor \( 2h^3 + h^2 + 2h + 3 \). Test \( h=-1 \):
Substitute \( h = - 1 \) into \( 2h^3 + h^2 + 2h + 3 \): \( 2(-1)^3+(-1)^2 + 2(-1)+3=-2 + 1-2 + 3=0 \). So, \( (h + 1) \) is a factor.
Perform long division of \( 2h^3 + h^2 + 2h + 3 \) by \( (h + 1) \):
\[

$$\begin{array}{r|rrrr} h + 1 & 2h^3 & +h^2 & +2h & +3 \\ & 2h^3 & +2h^2 & & \\ \cline{2-5} & 0 & -h^2 & +2h & +3 \\ & & -h^2 & -h & \\ \cline{3-5} & & 0 & 3h & +3 \\ & & & 3h & +3 \\ \cline{4-5} & & & 0 & 0 \\ \end{array}$$

\]
So, \( 2h^3 + h^2 + 2h + 3=(h + 1)(2h^2 - h + 3) \).
Thus, the numerator \( 2h^4 - h^3 + h^2 + h - 3=(h - 1)(h + 1)(2h^2 - h + 3) \).

Step3: Simplify the fraction

Now, the fraction \( \frac{2h^4 - h^3 + h^2 + h - 3}{h^2 - 1}=\frac{(h - 1)(h + 1)(2h^2 - h + 3)}{(h - 1)(h + 1)} \).
Cancel out the common factors \( (h - 1) \) and \( (h + 1) \) (for \( h
eq\pm1 \)):
The simplified form is \( 2h^2 - h + 3 \).

Answer:

\( 2h^2 - h + 3 \) (for \( h
eq\pm1 \))