Question

question
use the long division method to find the result when
2x⁴ + 3x³ − 8x² + 24x − 27 is
divided by 2x² + 5x − 9. 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 \(x^2\).
Multiply the divisor \(2x^2 + 5x - 9\) by \(x^2\): \(x^2(2x^2 + 5x - 9)=2x^4 + 5x^3 - 9x^2\).
Subtract this from the dividend: \((2x^4 + 3x^3 - 8x^2 + 24x - 27)-(2x^4 + 5x^3 - 9x^2)= - 2x^3 + x^2 + 24x - 27\).

Step2: Divide leading terms again

Divide the leading term of the new dividend \(-2x^3\) by the leading term of the divisor \(2x^2\), we get \(-x\).
Multiply the divisor \(2x^2 + 5x - 9\) by \(-x\): \(-x(2x^2 + 5x - 9)=-2x^3 - 5x^2 + 9x\).
Subtract this from the new dividend: \((-2x^3 + x^2 + 24x - 27)-(-2x^3 - 5x^2 + 9x)=6x^2 + 15x - 27\).

Step3: Divide leading terms once more

Divide the leading term of the new dividend \(6x^2\) by the leading term of the divisor \(2x^2\), we get \(3\).
Multiply the divisor \(2x^2 + 5x - 9\) by \(3\): \(3(2x^2 + 5x - 9)=6x^2 + 15x - 27\).
Subtract this from the new dividend: \((6x^2 + 15x - 27)-(6x^2 + 15x - 27)=0\).

Answer:

The quotient \(q(x)=x^2 - x + 3\) and the remainder \(r(x) = 0\), so the result is \(x^2 - x + 3+\frac{0}{2x^2 + 5x - 9}\) (or simply \(x^2 - x + 3\) since the remainder is 0).