1. ## 3 questions

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 .

Find the quotient and remainder using long division for

Simplify the expression

2. 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 .

Find the quotient and remainder using long division for

Simplify the expression

For Q.3 note that

$\displaystyle 7x^3 - 8x^2 - 12x = x(x-2)(7x+3)$

and

$\displaystyle 6x^2 - 16x + 8 = 2(x - 2)(3x - 2)$.

So $\displaystyle \frac{7x^3 - 8x^2 - 12x}{6x^2 - 16x + 8} = \frac{x(x-2)(7x+3)}{2(x - 2)(3x - 2)}$

$\displaystyle = \frac{x(7x + 3)}{2(3x - 2)}$.

Therefore $\displaystyle f(x) = x(7x + 3)$ and $\displaystyle g(x) = 2(3x - 2)$.

3. 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:

• $\displaystyle x<4$

• $\displaystyle 4<x<5$

• $\displaystyle 5<x$

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 $\displaystyle -5$.