Question

use synthetic division to divide the following. dont forget to add placeholders of 0!$(x^3 - 125)\div(x - 5)$$(5x^4 + 2x^2 - 15x + 10)\div(x + 2)$

Explanation:

Problem 1: $(x^3 - 125) \div (x - 5)$

Step1: Identify root & coefficients

Root of divisor $x-5$ is $5$. Coefficients of dividend: $1, 0, 0, -125$ (add 0 placeholders for $x^2$ and $x$ terms)

Step2: Set up synthetic division

$$\begin{array}{r|rrrr} 5 & 1 & 0 & 0 & -125 \\ \hline & & 5 & 25 & 125 \\ \hline & 1 & 5 & 25 & 0 \end{array}$$

Step3: Write quotient

The final row (excluding last 0) gives coefficients of quotient: $x^2 + 5x + 25$, remainder = 0

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Problem 2: $(5x^4 + 2x^2 - 15x + 10) \div (x + 2)$

Step1: Identify root & coefficients

Root of divisor $x+2$ is $-2$. Coefficients of dividend: $5, 0, 2, -15, 10$ (add 0 placeholder for $x^3$ term)

Step2: Set up synthetic division

$$\begin{array}{r|rrrrr} -2 & 5 & 0 & 2 & -15 & 10 \\ \hline & & -10 & 20 & -44 & 118 \\ \hline & 5 & -10 & 22 & -59 & 128 \end{array}$$

Step3: Write quotient & remainder

The final row gives quotient coefficients $5x^3 -10x^2 +22x -59$, remainder = 128

Answer:

1. $(x^3 - 125) \div (x - 5) = x^2 + 5x + 25$
2. $(5x^4 + 2x^2 - 15x + 10) \div (x + 2) = 5x^3 - 10x^2 + 22x - 59 + \frac{128}{x+2}$