To use synthetic division to divide \(2x^3 + 3x^2 - 21x + 20\) by \(x - 2\), we first note that we will be using \(2\) (the root of \(x - 2 = 0\)) in our synthetic division process.
Here are the steps:
Step 1: Set Up Synthetic Division
Write down the coefficients of the polynomial \(2x^3 + 3x^2 - 21x + 20\):
- Coefficients are: \(2, 3, -21, 20\)
Now we represent this as:
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & & & & \ \end{array} \]
Step 2: Perform Synthetic Division
- Bring down the leading coefficient (the 2).
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & 2 & & & \ \end{array} \]
- Multiply \(2\) (the value we are dividing by) with the value just written below the line, and write the result under the next coefficient:
- Multiply: \(2 \times 2 = 4\)
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & 2 & 4 & & \ \end{array} \]
- Now add the column:
- \(3 + 4 = 7\)
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & 2 & 7 & & \ \end{array} \]
- Repeat the process:
- Multiply \(2\) by \(7\) (the value we just calculated): \(2 \times 7 = 14\)
- Add to the next coefficient: \(-21 + 14 = -7\)
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & 2 & 7 & -7 & \ \end{array} \]
- Multiply \(2\) by \(-7\): \(2 \times -7 = -14\)
- Add to the last coefficient: \(20 - 14 = 6\)
\[ \begin{array}{r|rrrr} 2 & 2 & 3 & -21 & 20 \ \hline & 2 & 7 & -7 & 6 \ \end{array} \]
Step 3: Write the Result
At the end of the synthetic division, the bottom row gives us the coefficients of the quotient polynomial. The last number (6) is the remainder.
The quotient polynomial is \(2x^2 + 7x - 7\) and the remainder is \(6\).
Final Result
So, when \(2x^3 + 3x^2 - 21x + 20\) is divided by \(x - 2\), we can express the result as:
\[ q(x) + \frac{r(x)}{b(x)} = 2x^2 + 7x - 7 + \frac{6}{x - 2} \]
Thus, the final answer is:
\[ 2x^2 + 7x - 7 + \frac{6}{x - 2} \]