Question

(2k⁴ + 10k³ - 40k² + 61k - 24) ÷ (k + 8)

Explanation:

Step1: Use Polynomial Long Division

Divide the leading term of the dividend \(2k^{4}\) by the leading term of the divisor \(k\), we get \(2k^{3}\). Multiply the divisor \(k + 8\) by \(2k^{3}\): \(2k^{3}(k + 8)=2k^{4}+16k^{3}\). Subtract this from the dividend:

$$
LATEXBLOCK0
$$

Step2: Divide the new leading term

Divide the leading term of \(-6k^{3}\) by \(k\), we get \(-6k^{2}\). Multiply the divisor \(k + 8\) by \(-6k^{2}\): \(-6k^{2}(k + 8)=-6k^{3}-48k^{2}\). Subtract this from the previous remainder:

$$
LATEXBLOCK1
$$

Step3: Divide the new leading term

Divide the leading term of \(8k^{2}\) by \(k\), we get \(8k\). Multiply the divisor \(k + 8\) by \(8k\): \(8k(k + 8)=8k^{2}+64k\). Subtract this from the previous remainder:

$$
LATEXBLOCK2
$$

Step4: Divide the new leading term

Divide the leading term of \(-3k\) by \(k\), we get \(-3\). Multiply the divisor \(k + 8\) by \(-3\): \(-3(k + 8)=-3k - 24\). Subtract this from the previous remainder:

$$
LATEXBLOCK3
$$

Answer:

The result of the division \((2k^{4}+10k^{3}-40k^{2}+61k - 24)\div(k + 8)\) is \(2k^{3}-6k^{2}+8k - 3\)