Question

name:date: wednesday, february 11, 2026perioda2: 6, 13, 19, 20adifferentiated task #1problem: determine if $(x - 3)$ is a factor of $f(x)=x^{3}-2x^{2}-5x + 6$.step a: the setup• what number will go outside of the box? (inverse of -3): ____• list the coefficients: __, __, __, __step b: the tablestep c: the conclusion (factor theorem)1. what is the remainder? __2. is the remainder 0? (yes / no)3. therefore, is $(x - 3)$ a factor? (yes / no)4. if yes, where does the graph cross the x-axis? at $x=$ ____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:

Step1: Setup for first problem

For \(f(x) = x^3 - 2x^2 - 5x + 6\) and divisor \((x-3)\):

• Number outside box: \(3\) (inverse of \(-3\))
• Coefficients: \([1, -2, -5, 6]\)

Step2: Synthetic division table

$$\begin{array}{r|rrrr} 3 & 1 & -2 & -5 & 6 \\ & & 3 & 3 & -6 \\ \hline & 1 & 1 & -2 & 0 \\ \end{array}$$

Step3: Answer first problem questions

1. Remainder is the final value: \(0\)
2. Check if remainder = 0: Yes
3. Apply Factor Theorem: Yes
4. x-intercept is \(3\) (from \(x-3=0\))

Step4: Solve \((x^3 - 125)\div(x-5)\)

Rewrite dividend as \(x^3 + 0x^2 + 0x - 125\), coefficients \([1,0,0,-125]\), divisor uses \(5\):

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

Result: \(x^2 + 5x + 25\)

Step5: Solve \((5x^4 + 2x^3 -15x +10)\div(x+2)\)

Rewrite dividend as \(5x^4 + 2x^3 + 0x^2 -15x +10\), coefficients \([5,2,0,-15,10]\), divisor uses \(-2\):

$$\begin{array}{r|rrrrr} -2 & 5 & 2 & 0 & -15 & 10 \\ & & -10 & 16 & -32 & 94 \\ \hline & 5 & -8 & 16 & -47 & 104 \\ \end{array}$$

Result: \(5x^3 -8x^2 +16x -47 + \frac{104}{x+2}\)

Answer:

For \(f(x) = x^3 - 2x^2 -5x +6\) and \((x-3)\):

1. Remainder: \(0\)
2. Is the remainder 0? Yes
3. Is \((x-3)\) a factor? Yes
4. Graph crosses x-axis at \(x=3\)

For \((x^3 - 125)\div(x-5)\):

\(x^2 + 5x + 25\)

For \((5x^4 + 2x^3 -15x +10)\div(x+2)\):

\(5x^3 -8x^2 +16x -47 + \frac{104}{x+2}\)