Welcome to Westonci.ca, your one-stop destination for finding answers to all your questions. Join our expert community now! Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly 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 hope this information was helpful. Feel free to return anytime for more answers to your questions and concerns. We appreciate your time. Please come back anytime for the latest information and answers to your questions. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.
