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Sagot :
To solve the polynomial division [tex]\( (2x^3 + 4x^2 - 35x + 15) \div (x-3) \)[/tex] using synthetic division, we proceed as follows:
1. Identify the coefficients of the polynomial:
The polynomial [tex]\( 2x^3 + 4x^2 - 35x + 15 \)[/tex] has coefficients [tex]\([2, 4, -35, 15]\)[/tex].
2. Determine the root for synthetic division:
For the divisor [tex]\( (x-3) \)[/tex], the root is [tex]\( x = 3 \)[/tex].
3. Set up the synthetic division table:
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & & 6 & 30 & -15 \\ \hline & 2 & 10 & -5 & 0 \\ \end{array} \][/tex]
4. Start with the leading coefficient:
The first coefficient is [tex]\(2\)[/tex], which goes directly into the row below the line.
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & & & \\ \hline & 2 & & & \\ \end{array} \][/tex]
5. Multiply and add:
- Multiply the leading coefficient [tex]\(2\)[/tex] by the root [tex]\(3\)[/tex] and place the result under the next coefficient [tex]\(4\)[/tex]:
[tex]\[2 \times 3 = 6\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & & \\ \hline & 2 & 10& & \\ \end{array} \][/tex]
- Add this result ([tex]\(6\)[/tex]) to the next coefficient ([tex]\(4\)[/tex]):
[tex]\[4 + 6 = 10\][/tex]
6. Repeat the process:
- Multiply the new result ([tex]\(10\)[/tex]) by [tex]\(3\)[/tex] and place it under the next coefficient ([tex]\(-35\)[/tex]):
[tex]\[10 \times 3 = 30\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & 30 & \\ \hline & 2 & 10& -5 & \\ \end{array} \][/tex]
- Add the new result ([tex]\(30\)[/tex]) to the next coefficient ([tex]\(-35\)[/tex]):
[tex]\[-35 + 30 = -5\][/tex]
- Multiply the new result ([tex]\(-5\)[/tex]) by [tex]\(3\)[/tex] and place it under the next coefficient ([tex]\(15\)[/tex]):
[tex]\[-5 \times 3 = -15\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & 30 & -15 \\ \hline & 2 & 10& -5 & 0 \\ \end{array} \][/tex]
- Add this result ([tex]\(-15\)[/tex]) to the last coefficient ([tex]\(15\)[/tex]):
[tex]\[15 + (-15) = 0\][/tex]
7. Result of synthetic division:
The quotient is given by the numbers we have in the last row except the last term, and the remainder is the last term in the last row.
- Quotient: [tex]\[ 2x^2 + 10x - 5 \][/tex]
- Remainder: [tex]\[ 0 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{2x^2 + 10x - 5} \][/tex]
1. Identify the coefficients of the polynomial:
The polynomial [tex]\( 2x^3 + 4x^2 - 35x + 15 \)[/tex] has coefficients [tex]\([2, 4, -35, 15]\)[/tex].
2. Determine the root for synthetic division:
For the divisor [tex]\( (x-3) \)[/tex], the root is [tex]\( x = 3 \)[/tex].
3. Set up the synthetic division table:
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & & 6 & 30 & -15 \\ \hline & 2 & 10 & -5 & 0 \\ \end{array} \][/tex]
4. Start with the leading coefficient:
The first coefficient is [tex]\(2\)[/tex], which goes directly into the row below the line.
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & & & \\ \hline & 2 & & & \\ \end{array} \][/tex]
5. Multiply and add:
- Multiply the leading coefficient [tex]\(2\)[/tex] by the root [tex]\(3\)[/tex] and place the result under the next coefficient [tex]\(4\)[/tex]:
[tex]\[2 \times 3 = 6\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & & \\ \hline & 2 & 10& & \\ \end{array} \][/tex]
- Add this result ([tex]\(6\)[/tex]) to the next coefficient ([tex]\(4\)[/tex]):
[tex]\[4 + 6 = 10\][/tex]
6. Repeat the process:
- Multiply the new result ([tex]\(10\)[/tex]) by [tex]\(3\)[/tex] and place it under the next coefficient ([tex]\(-35\)[/tex]):
[tex]\[10 \times 3 = 30\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & 30 & \\ \hline & 2 & 10& -5 & \\ \end{array} \][/tex]
- Add the new result ([tex]\(30\)[/tex]) to the next coefficient ([tex]\(-35\)[/tex]):
[tex]\[-35 + 30 = -5\][/tex]
- Multiply the new result ([tex]\(-5\)[/tex]) by [tex]\(3\)[/tex] and place it under the next coefficient ([tex]\(15\)[/tex]):
[tex]\[-5 \times 3 = -15\][/tex]
[tex]\[ \begin{array}{c|cccc} 3 & 2 & 4 & -35 & 15 \\ \hline & 2 & 6 & 30 & -15 \\ \hline & 2 & 10& -5 & 0 \\ \end{array} \][/tex]
- Add this result ([tex]\(-15\)[/tex]) to the last coefficient ([tex]\(15\)[/tex]):
[tex]\[15 + (-15) = 0\][/tex]
7. Result of synthetic division:
The quotient is given by the numbers we have in the last row except the last term, and the remainder is the last term in the last row.
- Quotient: [tex]\[ 2x^2 + 10x - 5 \][/tex]
- Remainder: [tex]\[ 0 \][/tex]
Therefore, the correct answer is:
[tex]\[ \boxed{2x^2 + 10x - 5} \][/tex]
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