Question

question
use the long division method to find the
result when
$2x^4 - 6x^3 - 11x^2 + 4x + 7$ is
divided by $2x^2 - 3$. if there is a
remainder, express the result in the
form $q(x) + \frac{r(x)}{b(x)}$.

Explanation:

Step1: Divide leading terms

Divide the leading term of the dividend \(2x^4\) by the leading term of the divisor \(2x^2\). We get \(\frac{2x^4}{2x^2}=x^2\).

Step2: Multiply divisor by \(x^2\)

Multiply the divisor \(2x^2 - 3\) by \(x^2\): \(x^2(2x^2 - 3)=2x^4 - 3x^2\).

Step3: Subtract from dividend

Subtract this from the dividend: \((2x^4 - 6x^3 - 11x^2 + 4x + 7)-(2x^4 - 3x^2)= - 6x^3 - 8x^2 + 4x + 7\).

Step4: Divide leading terms again

Divide the leading term of the new dividend \(-6x^3\) by the leading term of the divisor \(2x^2\). We get \(\frac{-6x^3}{2x^2}=-3x\).

Step5: Multiply divisor by \(-3x\)

Multiply the divisor \(2x^2 - 3\) by \(-3x\): \(-3x(2x^2 - 3)=-6x^3 + 9x\).

Step6: Subtract again

Subtract this from the new dividend: \((-6x^3 - 8x^2 + 4x + 7)-(-6x^3 + 9x)= - 8x^2 - 5x + 7\).

Step7: Divide leading terms once more

Divide the leading term of the new dividend \(-8x^2\) by the leading term of the divisor \(2x^2\). We get \(\frac{-8x^2}{2x^2}=-4\).

Step8: Multiply divisor by \(-4\)

Multiply the divisor \(2x^2 - 3\) by \(-4\): \(-4(2x^2 - 3)=-8x^2 + 12\).

Step9: Subtract the last time

Subtract this from the new dividend: \((-8x^2 - 5x + 7)-(-8x^2 + 12)= - 5x - 5\).

The quotient \(q(x)=x^2 - 3x - 4\) and the remainder \(r(x)=-5x - 5\), and the divisor \(b(x)=2x^2 - 3\).

Answer:

\(x^2 - 3x - 4+\frac{-5x - 5}{2x^2 - 3}\)