Question

question
use the long division method to find the result when $3x^3 + 14x^2 - 19x + 25$ is divided by $x + 6$. if there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.
answer attempt 1 out of 2

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(3x^{3}\) by the leading term of the divisor \(x\). We get \(3x^{2}\). Multiply the divisor \(x + 6\) by \(3x^{2}\): \(3x^{2}(x + 6)=3x^{3}+18x^{2}\). Subtract this from the dividend: \((3x^{3}+14x^{2}-19x + 25)-(3x^{3}+18x^{2})=-4x^{2}-19x + 25\).

Step2: Divide the new leading terms

Divide the leading term of \(-4x^{2}\) by \(x\), we get \(-4x\). Multiply the divisor \(x + 6\) by \(-4x\): \(-4x(x + 6)=-4x^{2}-24x\). Subtract this from \(-4x^{2}-19x + 25\): \((-4x^{2}-19x + 25)-(-4x^{2}-24x)=5x + 25\).

Step3: Divide the new leading terms

Divide the leading term of \(5x\) by \(x\), we get \(5\). Multiply the divisor \(x + 6\) by \(5\): \(5(x + 6)=5x+30\). Subtract this from \(5x + 25\): \((5x + 25)-(5x + 30)=-5\).

Step4: Write the result

The quotient \(q(x)=3x^{2}-4x + 5\) and the remainder \(r(x)=-5\), and the divisor \(b(x)=x + 6\). So the result is \(3x^{2}-4x + 5+\frac{-5}{x + 6}\) or \(3x^{2}-4x + 5-\frac{5}{x + 6}\).

Answer:

\(3x^{2}-4x + 5-\frac{5}{x + 6}\)