(2x³+4x²-x+8) / (x² + 3x + 2)

1 answer:

3 0

$\dfrac{2x^3+4x^2 -x+8}{x^2 +3x+2}\\\\\\=\dfrac{2x^3+6x^2+4x-2x^2-6x-4+x+12}{}\\\\=\dfrac{2x(x^2 +3x+2)-2(x^2 +3x+2)+x+12}{x^2 +3x+2}\\\\\\=\dfrac{2x(x^2+3x+2)}{x^2 +3x +2 }- \dfrac{2(x^2 +3x +2)}{x^2 +3x +2 } + \dfrac{x+12}{x^2 +3x+2}\\\\\\=2x-2 + \dfrac{x+2}{x^2 +3x +2}\\\\\\=2x-2 + \dfrac{x+12}{x^2 +2x +x +2}\\\\\\=2x-2 + \dfrac{x+12}{x(x+2) + x+2}\\\\\\=2x-2 + \dfrac{x+12}{(x+1)(x+2)}\\$

You might be interested in

The population od a small country is 5,740,000. Express this number using scientific notation. Say steps to.

bija089 [108]

5,740,000( you have to make the number less than 10

5.740000 ( you put a decimal to make it less )

5.74e6 (you count how many times you skipped over to get at the decimal point , and that's your awnser .

5 0

3 years ago

The University of Montana ski team has thirteen entrants in a men's downhill ski event. The coach would like the first, second,

sleet_krkn [62]

Answer:

1716 ways

Step-by-step explanation:

Given that :

Number of entrants = 13

The number of ways of attaining first, second and third position :

The number of ways of attaining first ; only 1 person can be first ;

Using permutation :

nPr = n! ÷(n-r)!

13P1 = 13! ÷ 12! = 13

Second position :

We have 12 entrants left :

nPr = n! ÷(n-r)!

12P1 = 12! ÷ 11! = 12

Third position :

We have 11 entrants left :

nPr = n! ÷(n-r)!

11P1 = 11! ÷ 10! = 11

Hence, Number of ways = (13 * 12 * 11) = 1716 ways

8 0

3 years ago

HELP! Find the value of sin 0 if tan 0 = 4; 180 < 0< 270

BabaBlast [244]

Hi there! Use the following identities below to help with your problem.

$\large \boxed{sin \theta = tan \theta cos \theta} \\ \large \boxed{tan^{2} \theta + 1 = {sec}^{2} \theta}$

What we know is our tangent value. We are going to use the tan²θ+1 = sec²θ to find the value of cosθ. Substitute tanθ = 4 in the second identity.

$\large{ {4}^{2} + 1 = {sec}^{2} \theta } \\ \large{16 + 1 = {sec}^{2} \theta } \\ \large{ {sec}^{2} \theta = 17}$

As we know, sec²θ = 1/cos²θ.

$\large \boxed{sec \theta = \frac{1}{cos \theta} } \\ \large \boxed{ {sec}^{2} \theta = \frac{1}{ {cos}^{2} \theta} }$

And thus,

$\large{ {cos}^{2} \theta = \frac{1}{17}} \\ \large{cos \theta = \frac{ \sqrt{1} }{ \sqrt{17} } } \\ \large{cos \theta = \frac{1}{ \sqrt{17} } \longrightarrow \frac{ \sqrt{17} }{17} }$