Question

use synthetic division to find the result when $3x^4 + 14x^3 + 9x^2 - 12x + 4$ is divided by $x + 2$.

Explanation:

Step1: Identify the root

For the divisor \(x + 2\), the root \(r=-2\) (since \(x+2 = 0\) gives \(x=-2\)). The coefficients of the polynomial \(3x^{4}+14x^{3}+9x^{2}-12x + 4\) are \(3,14,9,-12,4\).

Step2: Set up synthetic division

Write the root \(r = - 2\) on the left and the coefficients in a row:

$$
LATEXBLOCK0
$$

Step3: Bring down the leading coefficient

Bring down the \(3\):

$$
LATEXBLOCK1
$$

Step4: Multiply and add

Multiply the brought - down number (\(3\)) by the root (\(-2\)): \(3\times(-2)=-6\). Add this to the next coefficient (\(14\)): \(14+(-6) = 8\).

$$
LATEXBLOCK2
$$

Multiply the new number (\(8\)) by the root (\(-2\)): \(8\times(-2)=-16\). Add this to the next coefficient (\(9\)): \(9+(-16)=-7\).

$$
LATEXBLOCK3
$$

Multiply the new number (\(-7\)) by the root (\(-2\)): \(-7\times(-2) = 14\). Add this to the next coefficient (\(-12\)): \(-12 + 14=2\).

$$
LATEXBLOCK4
$$

Multiply the new number (\(2\)) by the root (\(-2\)): \(2\times(-2)=-4\). Add this to the last coefficient (\(4\)): \(4+(-4)=0\).

$$
LATEXBLOCK5
$$

The numbers in the bottom row (except the last one, which is the remainder) are the coefficients of the quotient polynomial. Since the original polynomial is of degree \(4\), the quotient polynomial is of degree \(3\). The coefficients are \(3,8,-7,2\), so the quotient is \(3x^{3}+8x^{2}-7x + 2\) and the remainder is \(0\). So \(\frac{3x^{4}+14x^{3}+9x^{2}-12x + 4}{x + 2}=3x^{3}+8x^{2}-7x + 2\)

Answer:

\(3x^{3}+8x^{2}-7x + 2\)