Westonci.ca offers quick and accurate answers to your questions. Join our community and get the insights you need today. Our Q&A platform provides quick and trustworthy answers to your questions from experienced professionals in different areas of expertise. Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform.
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
We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Find reliable answers at Westonci.ca. Visit us again for the latest updates and expert advice.