Question

use the long division method to find the result when $12x^3 - 5x^2 + 12x + 5$ is divided by $3x + 1$. 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 \(12x^{3}\) by the leading term of the divisor \(3x\). So, \(\frac{12x^{3}}{3x} = 4x^{2}\). Multiply the divisor \(3x + 1\) by \(4x^{2}\) to get \(12x^{3}+4x^{2}\). Subtract this from the dividend: \((12x^{3}-5x^{2}+12x + 5)-(12x^{3}+4x^{2})=-9x^{2}+12x + 5\).

Step2: Divide the new leading term

Divide the leading term of \(-9x^{2}\) by \(3x\), which gives \(\frac{-9x^{2}}{3x}=-3x\). Multiply the divisor \(3x + 1\) by \(-3x\) to get \(-9x^{2}-3x\). Subtract this from \(-9x^{2}+12x + 5\): \((-9x^{2}+12x + 5)-(-9x^{2}-3x)=15x + 5\).

Step3: Divide the new leading term

Divide the leading term of \(15x\) by \(3x\), which gives \(\frac{15x}{3x}=5\). Multiply the divisor \(3x + 1\) by \(5\) to get \(15x + 5\). Subtract this from \(15x + 5\): \((15x + 5)-(15x + 5)=0\). Since the remainder is \(0\), the quotient \(q(x)\) is \(4x^{2}-3x + 5\) and the remainder \(r(x)\) is \(0\).

Answer:

\(4x^{2}-3x + 5+\frac{0}{3x + 1}\) (or simply \(4x^{2}-3x + 5\) as the remainder is \(0\))