Question

algebra 2
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dividing polynomials
divide.

1. $(n^3 - 10n^2 + 26n - 8) \div (n - 4)$

Explanation:

Step1: Divide the leading terms

Divide the leading term of the dividend \(n^3\) by the leading term of the divisor \(n\), we get \(n^2\). Multiply the divisor \((n - 4)\) by \(n^2\) to get \(n^3 - 4n^2\).
\[

$$\begin{align*} (n^3 - 10n^2 + 26n - 8)-(n^3 - 4n^2)&=n^3 - 10n^2 + 26n - 8 - n^3 + 4n^2\\ &=-6n^2 + 26n - 8 \end{align*}$$

\]

Step2: Divide the new leading terms

Divide the leading term of the new dividend \(-6n^2\) by the leading term of the divisor \(n\), we get \(-6n\). Multiply the divisor \((n - 4)\) by \(-6n\) to get \(-6n^2 + 24n\).
\[

$$\begin{align*} (-6n^2 + 26n - 8)-(-6n^2 + 24n)&=-6n^2 + 26n - 8 + 6n^2 - 24n\\ &=2n - 8 \end{align*}$$

\]

Step3: Divide the new leading terms

Divide the leading term of the new dividend \(2n\) by the leading term of the divisor \(n\), we get \(2\). Multiply the divisor \((n - 4)\) by \(2\) to get \(2n - 8\).
\[

$$\begin{align*} (2n - 8)-(2n - 8)&=2n - 8 - 2n + 8\\ &=0 \end{align*}$$

\]

Step4: Combine the results

The quotient is the sum of the terms we got in each step: \(n^2 - 6n + 2\), and the remainder is \(0\).

Answer:

\(n^2 - 6n + 2\)