Question

use the long division method to find the result when $2x^3 + 3x + 10$ is divided by $x + 1$. 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 \(2x^{3}\) by the leading term of the divisor \(x\), we get \(2x^{2}\). Multiply the divisor \(x + 1\) by \(2x^{2}\), we have \(2x^{2}(x + 1)=2x^{3}+2x^{2}\). Subtract this from the dividend: \((2x^{3}+3x + 10)-(2x^{3}+2x^{2})=- 2x^{2}+3x + 10\).

Step2: Divide the new leading terms

Divide the leading term of \(-2x^{2}+3x + 10\) (which is \(-2x^{2}\)) by the leading term of the divisor \(x\), we get \(-2x\). Multiply the divisor \(x + 1\) by \(-2x\), we have \(-2x(x + 1)=-2x^{2}-2x\). Subtract this from \(-2x^{2}+3x + 10\): \((-2x^{2}+3x + 10)-(-2x^{2}-2x)=5x + 10\).

Step3: Divide the new leading terms

Divide the leading term of \(5x + 10\) (which is \(5x\)) by the leading term of the divisor \(x\), we get \(5\). Multiply the divisor \(x + 1\) by \(5\), we have \(5(x + 1)=5x + 5\). Subtract this from \(5x + 10\): \((5x + 10)-(5x + 5)=5\).

Answer:

The quotient \(q(x)=2x^{2}-2x + 5\) and the remainder \(r(x) = 5\), the divisor \(b(x)=x + 1\). So the result is \(2x^{2}-2x + 5+\frac{5}{x + 1}\)