Question

algebra 2
5-1 operations with polynomials
divide.

1. ((x^3 + 3x^2 - 5x + 1) div (x - 1))

Explanation:

Step1: Use Polynomial Long Division

Divide the leading term of the dividend \(x^3 + 3x^2 - 5x + 1\) by the leading term of the divisor \(x - 1\). The leading term of the dividend is \(x^3\) and of the divisor is \(x\), so \(\frac{x^3}{x}=x^2\). Multiply the divisor \(x - 1\) by \(x^2\) to get \(x^3 - x^2\). Subtract this from the dividend:
\[

$$\begin{align*} &(x^3 + 3x^2 - 5x + 1)-(x^3 - x^2)\\ =&x^3 + 3x^2 - 5x + 1 - x^3 + x^2\\ =&4x^2 - 5x + 1 \end{align*}$$

\]

Step2: Divide the New Leading Term

Now, divide the leading term of \(4x^2 - 5x + 1\) (which is \(4x^2\)) by the leading term of the divisor \(x\), so \(\frac{4x^2}{x}=4x\). Multiply the divisor \(x - 1\) by \(4x\) to get \(4x^2 - 4x\). Subtract this from \(4x^2 - 5x + 1\):
\[

$$\begin{align*} &(4x^2 - 5x + 1)-(4x^2 - 4x)\\ =&4x^2 - 5x + 1 - 4x^2 + 4x\\ =&-x + 1 \end{align*}$$

\]

Step3: Divide the Next Leading Term

Divide the leading term of \(-x + 1\) (which is \(-x\)) by the leading term of the divisor \(x\), so \(\frac{-x}{x}=-1\). Multiply the divisor \(x - 1\) by \(-1\) to get \(-x + 1\). Subtract this from \(-x + 1\):
\[

$$\begin{align*} &(-x + 1)-(-x + 1)\\ =&-x + 1 + x - 1\\ =&0 \end{align*}$$

\]

Step4: Combine the Quotient Terms

The quotient terms we found were \(x^2\), \(4x\), and \(-1\), and the remainder is \(0\). So the result of the division is \(x^2 + 4x - 1\).

Answer:

\(x^2 + 4x - 1\)