Question

question
use the long division method to find the result when $3x^3 + 4x^2 - 9x - 17$ is divided by $3x + 4$. 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 \(3x^{3}\) by the leading term of the divisor \(3x\). So, \(\frac{3x^{3}}{3x}=x^{2}\).

Step2: Multiply and subtract

Multiply the divisor \(3x + 4\) by \(x^{2}\) to get \(3x^{3}+4x^{2}\). Subtract this from the dividend: \((3x^{3}+4x^{2}-9x - 17)-(3x^{3}+4x^{2})=-9x-17\).

Step3: Divide the new leading terms

Now, divide the leading term of the new dividend \(-9x\) by the leading term of the divisor \(3x\). So, \(\frac{-9x}{3x}=-3\).

Step4: Multiply and subtract again

Multiply the divisor \(3x + 4\) by \(-3\) to get \(-9x-12\). Subtract this from the new dividend: \((-9x - 17)-(-9x - 12)=-5\).

Answer:

The quotient \(q(x)=x^{2}-3\), the remainder \(r(x)=-5\) and the divisor \(b(x)=3x + 4\). So the result is \(x^{2}-3+\frac{-5}{3x + 4}\) or \(x^{2}-3-\frac{5}{3x + 4}\).