Question

polynomial long division (level 3)
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use the long division method to find the result when $4x^3 + 9x^2 + 12x + 27$ is divided by $x + 2$. if there is a remainder, express the result in the form $q(x) + \frac{r(x)}{b(x)}$.
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Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(4x^{3}+9x^{2}+12x + 27\) (which is \(4x^{3}\)) by the leading term of the divisor \(x + 2\) (which is \(x\)). We get \(4x^{2}\). Multiply the divisor \(x + 2\) by \(4x^{2}\) to get \(4x^{3}+8x^{2}\). Subtract this from the dividend: \((4x^{3}+9x^{2}+12x + 27)-(4x^{3}+8x^{2})=x^{2}+12x + 27\).

Step2: Divide the new leading term

Divide the leading term of the new dividend \(x^{2}+12x + 27\) (which is \(x^{2}\)) by the leading term of the divisor \(x\). We get \(x\). Multiply the divisor \(x + 2\) by \(x\) to get \(x^{2}+2x\). Subtract this from the new dividend: \((x^{2}+12x + 27)-(x^{2}+2x)=10x + 27\).

Step3: Divide the new leading term again

Divide the leading term of the new dividend \(10x + 27\) (which is \(10x\)) by the leading term of the divisor \(x\). We get \(10\). Multiply the divisor \(x + 2\) by \(10\) to get \(10x+20\). Subtract this from the new dividend: \((10x + 27)-(10x + 20)=7\).

Step4: Write the result

The quotient \(q(x)\) is \(4x^{2}+x + 10\), the remainder \(r(x)\) is \(7\), and the divisor \(d(x)\) is \(x + 2\). So the result is \(4x^{2}+x + 10+\frac{7}{x + 2}\).

Answer:

\(4x^{2}+x + 10+\frac{7}{x + 2}\)