Question

question
use the long division method to find the result when
$6x^4 + 4x^3 + 23x^2 - 6x + 6$ is
divided by $2x^2 + 1$. if there is a
remainder, express the result in the
form $q(x) + \frac{r(x)}{b(x)}$.

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(6x^4\) by the leading term of the divisor \(2x^2\). So, \(\frac{6x^4}{2x^2} = 3x^2\). This is the first term of the quotient.

Step2: Multiply and subtract

Multiply the divisor \(2x^2 + 1\) by \(3x^2\) to get \(6x^4 + 3x^2\). Subtract this from the dividend:

$$
LATEXBLOCK0
$$

Step3: Divide the new leading term

Divide the leading term of the new dividend \(4x^3\) by the leading term of the divisor \(2x^2\). So, \(\frac{4x^3}{2x^2} = 2x\). This is the next term of the quotient.

Step4: Multiply and subtract again

Multiply the divisor \(2x^2 + 1\) by \(2x\) to get \(4x^3 + 2x\). Subtract this from the new dividend:

$$
LATEXBLOCK1
$$

Step5: Divide the next leading term

Divide the leading term of the new dividend \(20x^2\) by the leading term of the divisor \(2x^2\). So, \(\frac{20x^2}{2x^2} = 10\). This is the next term of the quotient.

Step6: Multiply and subtract the last time

Multiply the divisor \(2x^2 + 1\) by \(10\) to get \(20x^2 + 10\). Subtract this from the new dividend:

$$
LATEXBLOCK2
$$

Now, the quotient \(q(x)\) is \(3x^2 + 2x + 10\) and the remainder \(r(x)\) is \(-8x - 4\), and the divisor \(b(x)\) is \(2x^2 + 1\).

Answer:

\(3x^2 + 2x + 10 + \frac{-8x - 4}{2x^2 + 1}\) (or simplified as \(3x^2 + 2x + 10 - \frac{8x + 4}{2x^2 + 1}\))