Hello wannous Originally Posted by

**wannous** Solve the following inequality. Write the answer in interval notation.

** Note: ** If the answer includes more than one interval write the intervals separated by the "union" symbol, U. If needed enter

as

* infinity * and

as

* -infinity * .

To solve this type of inequality, find the values of $\displaystyle x$ that make each factor zero, and then look at their *signs *as $\displaystyle x$ moves along the number line from left to right, through each of these 'zero values' in turn.

Now the 'zeros' occurs when $\displaystyle x =4$ and $\displaystyle x=5$. So we look at what happens when:

First, if $\displaystyle x<4,\, (x-5)$ is negative and $\displaystyle (x-4)$ is also negative; and $\displaystyle -^{ve}\times -^{ve} = +^{ve}$. So $\displaystyle (x-5)(x-4)>0$ in this range.

Next, if $\displaystyle 4<x<5,\, (x-5)$ is $\displaystyle -^{ve}$ and $\displaystyle (x-4)$ is $\displaystyle +^{ve}$; $\displaystyle -^{ve}\times+^{ve} = -^{ve}$. So $\displaystyle (x-5)(x-4) < 0$ in this range.

Finally, if $\displaystyle x>5$, we get $\displaystyle +^{ve}\times+^{ve}=+^{ve}$. So $\displaystyle (x-5)(x-4)>0$ in this range.

So the values we want are $\displaystyle 4\le x \le 5$, or in interval notation $\displaystyle [4,5]$. Find the quotient and remainder using long division for

See the attached image.

So the quotient is $\displaystyle x^2-4x+1$ and the remainder is $\displaystyle -5$.

Grandad