Question

what is the rational expression \\(\frac{2x^4 + 4x^3 - 6x - 12}{x + 1}\\) in the form of \\(q(x) + \frac{r(x)}{b(x)}\\)? use the keyboard to type your answer in the box. you may use geogebra cas calculator or any cas calculator. the rational expression \\(\frac{2x^4 + 4x^3 - 6x - 12}{x + 1}\\) in the form of \\(q(x) + \frac{r(x)}{b(x)}\\) is \\(\square\\).

Explanation:

Step1: Perform Polynomial Long Division

Divide the numerator \(2x^4 + 4x^3 - 6x - 12\) by the denominator \(x + 1\).

First, divide \(2x^4\) by \(x\) to get \(2x^3\). Multiply \(x + 1\) by \(2x^3\) to get \(2x^4 + 2x^3\). Subtract this from the numerator:

$$
LATEXBLOCK0
$$

Step2: Continue the Division

Divide \(2x^3\) by \(x\) to get \(2x^2\). Multiply \(x + 1\) by \(2x^2\) to get \(2x^3 + 2x^2\). Subtract this from the previous remainder:

$$
LATEXBLOCK1
$$

Step3: Continue the Division

Divide \(-2x^2\) by \(x\) to get \(-2x\). Multiply \(x + 1\) by \(-2x\) to get \(-2x^2 - 2x\). Subtract this from the previous remainder:

$$
LATEXBLOCK2
$$

Step4: Continue the Division

Divide \(-4x\) by \(x\) to get \(-4\). Multiply \(x + 1\) by \(-4\) to get \(-4x - 4\). Subtract this from the previous remainder:

$$
LATEXBLOCK3
$$

Step5: Write the Result

Putting it all together, the division gives \(2x^3 + 2x^2 - 2x - 4\) with a remainder of \(-8\). So the rational expression can be written as:

$$ 2x^3 + 2x^2 - 2x - 4 + \frac{-8}{x + 1} $$

Answer:

\(2x^3 + 2x^2 - 2x - 4 - \frac{8}{x + 1}\)