3 questions

**wannous** · September 7th 2009, 03:54 PM

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

and give your answer in the form of

Your answer for the function

is :
Your answer for the function

is :

**Prove It** · September 7th 2009, 04:39 PM

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

and give your answer in the form of

Your answer for the function

is :

Your answer for the function

is :

For Q.3 note that

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

and

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

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

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

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

**Grandad** · September 7th 2009, 10:53 PM

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 $x$ that make each factor zero, and then look at their signs as $x$ moves along the number line from left to right, through each of these 'zero values' in turn.

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

$x<4$

$4<x<5$

$5<x$

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

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

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

So the values we want are $4\le x \le 5$ , or in interval notation $[4,5]$ .

Find the quotient and remainder using long division for

See the attached image.

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

Grandad

Thread: 3 questions

LinkBack

Thread Tools

Display

3 questions

Similar Math Help Forum Discussions

More log questions

Please someone help me with just 2 questions?

Some Questions !

Plz solve these questions.(Trignometry statement questions)

Assorted algebraic questions ( Read this before asking some questions! )

Search Tags