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Sagot :
To solve the synthetic division problem for [tex]\( \frac{3x^3 - 25x^2 + 12x - 32}{x - 8} \)[/tex], let's carry out the step-by-step process. The divisor root is 8 and the polynomial's coefficients are 3, -25, 12, and -32.
1. Set up the synthetic division:
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & & & & \\ \end{array} \][/tex]
2. Bring down the first coefficient (3):
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & & & \\ \end{array} \][/tex]
3. Multiply 8 (the root) by 3 (the first value in the bottom row) and place the result under -25:
[tex]\[ 8 \times 3 = 24 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & 24 & & \\ \end{array} \][/tex]
4. Add -25 and 24 (combine them) to get -1:
[tex]\[ -25 + 24 = -1 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & & \\ \end{array} \][/tex]
5. Multiply 8 (divisor root) by -1 (the latest value in the bottom row) and place the result under 12:
[tex]\[ 8 \times -1 = -8 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & -8 & \\ \end{array} \][/tex]
6. Add 12 and -8 (combine them) to get 4:
[tex]\[ 12 + -8 = 4 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & \\ \end{array} \][/tex]
7. Multiply 8 (divisor root) by 4 (the latest value in the bottom row) and place the result under -32:
[tex]\[ 8 \times 4 = 32 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 32 \\ \end{array} \][/tex]
8. Add -32 and 32 (combine them) to get 0, which is the remainder:
[tex]\[ -32 + 32 = 0 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 0 \\ \end{array} \][/tex]
The quotient is [tex]\(3x^2 - x + 4\)[/tex] and the remainder is 0. Here is the completed synthetic division:
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 0 \\ \end{array} \][/tex]
Therefore, we've successfully found that
[tex]\[ \frac{3x^3 - 25x^2 + 12x - 32}{x - 8} = 3x^2 - x + 4 \quad \text{with a remainder of} \quad 0. \][/tex]
1. Set up the synthetic division:
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & & & & \\ \end{array} \][/tex]
2. Bring down the first coefficient (3):
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & & & \\ \end{array} \][/tex]
3. Multiply 8 (the root) by 3 (the first value in the bottom row) and place the result under -25:
[tex]\[ 8 \times 3 = 24 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & 24 & & \\ \end{array} \][/tex]
4. Add -25 and 24 (combine them) to get -1:
[tex]\[ -25 + 24 = -1 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & & \\ \end{array} \][/tex]
5. Multiply 8 (divisor root) by -1 (the latest value in the bottom row) and place the result under 12:
[tex]\[ 8 \times -1 = -8 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & -8 & \\ \end{array} \][/tex]
6. Add 12 and -8 (combine them) to get 4:
[tex]\[ 12 + -8 = 4 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & \\ \end{array} \][/tex]
7. Multiply 8 (divisor root) by 4 (the latest value in the bottom row) and place the result under -32:
[tex]\[ 8 \times 4 = 32 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 32 \\ \end{array} \][/tex]
8. Add -32 and 32 (combine them) to get 0, which is the remainder:
[tex]\[ -32 + 32 = 0 \][/tex]
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 0 \\ \end{array} \][/tex]
The quotient is [tex]\(3x^2 - x + 4\)[/tex] and the remainder is 0. Here is the completed synthetic division:
[tex]\[ \begin{array}{r|cccc} 8 & 3 & -25 & 12 & -32 \\ \hline & 3 & -1 & 4 & 0 \\ \end{array} \][/tex]
Therefore, we've successfully found that
[tex]\[ \frac{3x^3 - 25x^2 + 12x - 32}{x - 8} = 3x^2 - x + 4 \quad \text{with a remainder of} \quad 0. \][/tex]
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