To perform synthetic division of the polynomial \( 5x^4 + 12x^3 - 21x^2 - 40x - 12 \) by \( x + 3 \), we will use the root of \( x + 3 \), which is \( -3 \).
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Set up the synthetic division:
- Write the coefficients of the polynomial: \( 5, 12, -21, -40, -12 \).
- Use \( -3 \) as the divisor.
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Synthetic division steps:
- Write down the coefficients:
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & & & & \ \end{array} \]
- Bring down the leading coefficient (5):
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & & & & \ & 5 & & & & \ \end{array} \]
- Multiply \( 5 \) by \( -3 \) and write the result under the next coefficient:
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & -15 & & & \ & 5 & & & & \ \end{array} \]
- Add the second column:
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & -15 & & & \ & 5 & -3 & & & \ \end{array} \]
- Continue this process:
\[ -3 \times -3 = 9 \quad \text{(add to -21:)} \quad -21 + 9 = -12 \]
- Continue this:
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & -15 & 9 & & \ & 5 & -3 & -12 & & \ \end{array} \]
- Multiply \( -12 \) by \( -3 \):
\[ -12 \times -3 = 36 \]
- Add to -40:
\[ -40 + 36 = -4 \]
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & -15 & 9 & 36 & \ & 5 & -3 & -12 & -4 & \ \end{array} \]
- Finally, multiply \( -4 \) by \( -3 \):
\[ -4 \times -3 = 12 \]
- Add to -12:
\[ -12 + 12 = 0 \]
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Complete synthetic division:
The synthetic division is complete:
\[ \begin{array}{r|rrrrr} -3 & 5 & 12 & -21 & -40 & -12 \ & & -15 & 9 & 36 & 12 \ & 5 & -3 & -12 & -4 & 0 \ \end{array} \]
- The result of the synthetic division is \( 5x^3 - 3x^2 - 12x - 4 \) (with a remainder of 0).
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Write the answer in standard form:
Thus, the solution to the division of \( 5x^4 + 12x^3 - 21x^2 - 40x - 12 \) by \( x + 3 \) is:
\[ 5x^3 - 3x^2 - 12x - 4 \]