Question

question
use synthetic division to find the result when (4x^{3}+13x^{2}+6x + 9) is divided by (x + 3).

Explanation:

Step1: Identify root of divisor

For \(x + 3\), set \(x+3 = 0\), so \(x=-3\). The coefficients of the dividend \(4x^{3}+13x^{2}+6x + 9\) are \(4\), \(13\), \(6\), \(9\).

Step2: Set up synthetic division

Write the root \(-3\) on the left and the coefficients \(4\), \(13\), \(6\), \(9\) in a row:
\[

$$\begin{array}{r|rrrr} -3 & 4 & 13 & 6 & 9 \\ & & -12 & -3 & -9 \\ \hline & 4 & 1 & 3 & 0 \\ \end{array}$$

\]
Bring down the first coefficient \(4\). Multiply \(-3\) by \(4\) to get \(-12\), add to \(13\) ( \(13-12 = 1\) ). Multiply \(-3\) by \(1\) to get \(-3\), add to \(6\) ( \(6 - 3=3\) ). Multiply \(-3\) by \(3\) to get \(-9\), add to \(9\) ( \(9-9 = 0\) ).

Step3: Interpret the result

The last number is the remainder (which is \(0\)), and the other numbers are the coefficients of the quotient polynomial. The quotient is a quadratic polynomial with coefficients \(4\), \(1\), \(3\), so the quotient is \(4x^{2}+x + 3\) and the remainder is \(0\). So \(\frac{4x^{3}+13x^{2}+6x + 9}{x + 3}=4x^{2}+x + 3\) (since remainder is \(0\)).

Answer:

\(4x^{2}+x + 3\) (with a remainder of \(0\))