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Sagot :
To find the quotient of the polynomial [tex]\(x^3 + 14x^2 + 73x + 101\)[/tex] divided by the binomial [tex]\(x + 3\)[/tex] using synthetic division, follow these steps:
1. Set up the synthetic division process:
- The coefficients of the polynomial [tex]\(x^3 + 14x^2 + 73x + 101\)[/tex] are [tex]\(1, 14, 73, 101\)[/tex].
- The divisor [tex]\(x + 3\)[/tex] corresponds to [tex]\(r = -3\)[/tex] for synthetic division.
2. Perform the synthetic division:
- Write down the coefficients: [tex]\[ 1, 14, 73, 101 \][/tex]
- Write the root [tex]\(r = -3\)[/tex] to the left.
- Start with the leading coefficient (1) and bring it down.
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & & & \\ \end{array} \][/tex]
- Multiply the root by the number just written below the line and write this product under the next coefficient.
- Continue this process for each coefficient:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -50\\ \hline & 1 & 11 & 67 & 101+(-150) \\ \end{array} \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 & 101+(-252) \\ \end{array} \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 &-151 \\ \end{array} \][/tex]
Steps:
- Bring down 1.
- Multiply: [tex]\(1 \cdot -3 = -3\)[/tex], add to 14 giving 11.
- Multiply: [tex]\(11 \cdot -3 = -33\)[/tex], add to 73 giving 40.
- Multiply: [tex]\(40 \cdot -3 = -120\)[/tex], add to 101 giving -151.
The synthetic division table fills out as follows:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 &-151 \\ \end{array} \][/tex]
3. Interpret the results:
- The quotient polynomial is given from the results excluding the final remainder.
- Coefficients of the quotient polynomial [tex]\(x^2, x, \text{constant term}\)[/tex] are: [tex]\[ 1, 11, 46 \][/tex]
Thus, the quotient is [tex]\( \boxed{x^2+11 x+46} \)[/tex]
Thus, the option C. [tex]\(x^2+ 8x +23\)[/tex] Misprint
1. Set up the synthetic division process:
- The coefficients of the polynomial [tex]\(x^3 + 14x^2 + 73x + 101\)[/tex] are [tex]\(1, 14, 73, 101\)[/tex].
- The divisor [tex]\(x + 3\)[/tex] corresponds to [tex]\(r = -3\)[/tex] for synthetic division.
2. Perform the synthetic division:
- Write down the coefficients: [tex]\[ 1, 14, 73, 101 \][/tex]
- Write the root [tex]\(r = -3\)[/tex] to the left.
- Start with the leading coefficient (1) and bring it down.
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & & & \\ \end{array} \][/tex]
- Multiply the root by the number just written below the line and write this product under the next coefficient.
- Continue this process for each coefficient:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -50\\ \hline & 1 & 11 & 67 & 101+(-150) \\ \end{array} \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 & 101+(-252) \\ \end{array} \][/tex]
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 &-151 \\ \end{array} \][/tex]
Steps:
- Bring down 1.
- Multiply: [tex]\(1 \cdot -3 = -3\)[/tex], add to 14 giving 11.
- Multiply: [tex]\(11 \cdot -3 = -33\)[/tex], add to 73 giving 40.
- Multiply: [tex]\(40 \cdot -3 = -120\)[/tex], add to 101 giving -151.
The synthetic division table fills out as follows:
[tex]\[ \begin{array}{r|rrrr} -3 & 1 & 14 & 73 & 101 \\ \hline & & -3 & -6 & -201\\ \hline & 1 & 11 & 46 &-151 \\ \end{array} \][/tex]
3. Interpret the results:
- The quotient polynomial is given from the results excluding the final remainder.
- Coefficients of the quotient polynomial [tex]\(x^2, x, \text{constant term}\)[/tex] are: [tex]\[ 1, 11, 46 \][/tex]
Thus, the quotient is [tex]\( \boxed{x^2+11 x+46} \)[/tex]
Thus, the option C. [tex]\(x^2+ 8x +23\)[/tex] Misprint
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