Question

1. $(20n^4 + 116n^3 + 16n^2 - 58n + 1) \div (10n - 2)$

long division work shown:
dividing $20n^4 + 116n^3 + 16n^2 - 58n + 1$ by $10n - 2$, with intermediate steps including subtraction of $20n^4 - 4n^3$ (note: likely typo, should be related to the divisor times quotient term), then continuing with $112n^3 + 16n^2$, etc.

Explanation:

Step1: Correct the first term of the quotient

The dividend is \(20n^4 + 116n^3 + 16n^2 - 58n + 1\) and the divisor is \(10n - 2\). When dividing the leading term \(20n^4\) by \(10n\), we get \(\frac{20n^4}{10n}=2n^3\), which is correct. But the next step in the given work has an error. Let's redo the polynomial long division properly.

First, multiply the divisor \(10n - 2\) by \(2n^3\): \((10n - 2)(2n^3)=20n^4 - 4n^3\). Subtract this from the dividend:

\((20n^4 + 116n^3 + 16n^2 - 58n + 1)-(20n^4 - 4n^3)=20n^4 + 116n^3 + 16n^2 - 58n + 1 - 20n^4 + 4n^3 = 120n^3 + 16n^2 - 58n + 1\)

Step2: Divide the new leading term

Now, divide the leading term \(120n^3\) by \(10n\) to get \(\frac{120n^3}{10n} = 12n^2\). Multiply the divisor \(10n - 2\) by \(12n^2\): \((10n - 2)(12n^2)=120n^3 - 24n^2\). Subtract this from the current dividend:

\((120n^3 + 16n^2 - 58n + 1)-(120n^3 - 24n^2)=120n^3 + 16n^2 - 58n + 1 - 120n^3 + 24n^2 = 40n^2 - 58n + 1\)

Step3: Divide the new leading term again

Divide the leading term \(40n^2\) by \(10n\) to get \(\frac{40n^2}{10n}=4n\). Multiply the divisor \(10n - 2\) by \(4n\): \((10n - 2)(4n)=40n^2 - 8n\). Subtract this from the current dividend:

\((40n^2 - 58n + 1)-(40n^2 - 8n)=40n^2 - 58n + 1 - 40n^2 + 8n=-50n + 1\)

Step4: Divide the new leading term

Divide the leading term \(-50n\) by \(10n\) to get \(\frac{-50n}{10n}=-5\). Multiply the divisor \(10n - 2\) by \(-5\): \((10n - 2)(-5)=-50n + 10\). Subtract this from the current dividend:

\((-50n + 1)-(-50n + 10)=-50n + 1 + 50n - 10=-9\)

So the correct quotient and remainder are found by polynomial long division:

The quotient is \(2n^3 + 12n^2 + 4n - 5\) and the remainder is \(-9\). So \((20n^4 + 116n^3 + 16n^2 - 58n + 1)\div(10n - 2)=2n^3 + 12n^2 + 4n - 5+\frac{-9}{10n - 2}\)

Answer:

The correct result of the polynomial long division is \(2n^3 + 12n^2 + 4n - 5-\frac{9}{10n - 2}\) (or in quotient - remainder form: quotient \(2n^3 + 12n^2 + 4n - 5\), remainder \(-9\))