Question

3x² - 7x - 10 divided by x - 4
enter only the numerator of the remainder!!!
question 3
2 pts
divide
(x³ - 6x² + 1)/(x + 2)
include a 0x term
options:
○ x² - 9x + 5 - 12/(x+2)
○ x² - 8x + 16 - 31/(x+2)
○ x² - 3x + 1 + 9/(x+2)
○ x² - 6x + 12 + 17/(x+2)

Explanation:

First Problem (Finding the numerator of the remainder when \(3x^2 - 7x - 10\) is divided by \(x - 4\))

Step 1: Recall the Remainder Theorem

The Remainder Theorem states that if a polynomial \(f(x)\) is divided by \(x - a\), the remainder is \(f(a)\). Here, \(a = 4\) (since the divisor is \(x - 4\)), so we need to find \(f(4)\) where \(f(x)=3x^2 - 7x - 10\).

Step 2: Substitute \(x = 4\) into \(f(x)\)

\[

$$\begin{align*} f(4)&=3(4)^2 - 7(4) - 10\\ &=3(16) - 28 - 10\\ &=48 - 28 - 10\\ &=10 \end{align*}$$

\]

First, rewrite the dividend as \(x^3 - 6x^2 + 0x + 1\) (to include the \(0x\) term). We can use polynomial long division or synthetic division. Let's use synthetic division. The root for the divisor \(x + 2\) is \(x=-2\).

Set up synthetic division:

• Coefficients of the dividend: \(1\) (for \(x^3\)), \(-6\) (for \(x^2\)), \(0\) (for \(x\)), \(1\) (constant term)
• Root: \(-2\)

Bring down the first coefficient: \(1\)

Multiply by \(-2\): \(1\times(-2)=-2\). Add to the next coefficient: \(-6 + (-2)=-8\)

Multiply \(-8\) by \(-2\): \(16\). Add to the next coefficient: \(0 + 16 = 16\)

Multiply \(16\) by \(-2\): \(-32\). Add to the last coefficient: \(1+(-32)=-31\)

So the quotient is \(x^2 - 8x + 16\) and the remainder is \(-31\). Thus, \(\frac{x^3 - 6x^2 + 1}{x + 2}=x^2 - 8x + 16-\frac{31}{x + 2}\)

Answer:

10

Second Problem (Dividing \((x^3 - 6x^2 + 1)\) by \((x + 2)\) with a \(0x\) term)