Answer :
To determine if [tex]\(x-2\)[/tex] is a factor of the polynomial [tex]\(3x^4 - 6x^3 - 5x + 10\)[/tex] using synthetic division, follow these steps:
1. Write down the coefficients of the polynomial [tex]\(3x^4 - 6x^3 - 5x + 10\)[/tex]. Note that for any missing powers of [tex]\(x\)[/tex], include a coefficient of 0.
- The coefficients are: [tex]\(3, -6, 0, -5, 10\)[/tex]
2. The root from the first expression [tex]\(x-2\)[/tex] is [tex]\(2\)[/tex].
3. Perform synthetic division using these coefficients and the root [tex]\(2\)[/tex].
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & -10 \\ \hline & 3 & 0 & 0 & -5 & 0 \\ \end{array} \][/tex]
4. Bring down the first coefficient [tex]\(3\)[/tex].
- Multiply [tex]\(3\)[/tex] by [tex]\(2\)[/tex], writing the result ([tex]\(6\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & & & \\ \hline & 3 & 0 & & & \\ \end{array} \][/tex]
5. Add the second coefficient [tex]\(-6\)[/tex] and [tex]\(6\)[/tex] to get [tex]\(0\)[/tex].
- Multiply [tex]\(0\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(0\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & & \\ \hline & 3 & 0 & 0 & & \\ \end{array} \][/tex]
6. Add the third coefficient [tex]\(0\)[/tex] and [tex]\(0\)[/tex] to get [tex]\(0\)[/tex].
- Multiply [tex]\(0\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(0\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & \\ \hline & 3 & 0 & 0 & -5 & \\ \end{array} \][/tex]
7. Add the fourth coefficient [tex]\(-5\)[/tex] and [tex]\(0\)[/tex] to get [tex]\(-5\)[/tex].
- Multiply [tex]\(-5\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(-10\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & -10 \\ \hline & 3 & 0 & 0 & -5 & 0 \\ \end{array} \][/tex]
8. Add the fifth coefficient [tex]\(10\)[/tex] and [tex]\(-10\)[/tex] to get [tex]\(0\)[/tex].
The final row of numbers [tex]\(3, 0, 0, -5, 0\)[/tex] represents the coefficients of the quotient polynomial [tex]\(3x^3 + 0x^2 + 0x - 5\)[/tex], and the last number (0) is the remainder.
Since the remainder is zero, the divisor [tex]\(x-2\)[/tex] is indeed a factor of the polynomial [tex]\(3x^4 - 6x^3 - 5x + 10\)[/tex]. The quotient polynomial is:
[tex]\[3x^3 + 0x^2 + 0x - 5\][/tex]
Therefore, the answer is:
[tex]\[3x^3-5\][/tex]
1. Write down the coefficients of the polynomial [tex]\(3x^4 - 6x^3 - 5x + 10\)[/tex]. Note that for any missing powers of [tex]\(x\)[/tex], include a coefficient of 0.
- The coefficients are: [tex]\(3, -6, 0, -5, 10\)[/tex]
2. The root from the first expression [tex]\(x-2\)[/tex] is [tex]\(2\)[/tex].
3. Perform synthetic division using these coefficients and the root [tex]\(2\)[/tex].
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & -10 \\ \hline & 3 & 0 & 0 & -5 & 0 \\ \end{array} \][/tex]
4. Bring down the first coefficient [tex]\(3\)[/tex].
- Multiply [tex]\(3\)[/tex] by [tex]\(2\)[/tex], writing the result ([tex]\(6\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & & & \\ \hline & 3 & 0 & & & \\ \end{array} \][/tex]
5. Add the second coefficient [tex]\(-6\)[/tex] and [tex]\(6\)[/tex] to get [tex]\(0\)[/tex].
- Multiply [tex]\(0\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(0\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & & \\ \hline & 3 & 0 & 0 & & \\ \end{array} \][/tex]
6. Add the third coefficient [tex]\(0\)[/tex] and [tex]\(0\)[/tex] to get [tex]\(0\)[/tex].
- Multiply [tex]\(0\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(0\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & \\ \hline & 3 & 0 & 0 & -5 & \\ \end{array} \][/tex]
7. Add the fourth coefficient [tex]\(-5\)[/tex] and [tex]\(0\)[/tex] to get [tex]\(-5\)[/tex].
- Multiply [tex]\(-5\)[/tex] by [tex]\(2\)[/tex], the root, writing the result ([tex]\(-10\)[/tex]) under the next coefficient.
[tex]\[ \begin{array}{r|rrrrr} 2 & 3 & -6 & 0 & -5 & 10 \\ & & 6 & 0 & 0 & -10 \\ \hline & 3 & 0 & 0 & -5 & 0 \\ \end{array} \][/tex]
8. Add the fifth coefficient [tex]\(10\)[/tex] and [tex]\(-10\)[/tex] to get [tex]\(0\)[/tex].
The final row of numbers [tex]\(3, 0, 0, -5, 0\)[/tex] represents the coefficients of the quotient polynomial [tex]\(3x^3 + 0x^2 + 0x - 5\)[/tex], and the last number (0) is the remainder.
Since the remainder is zero, the divisor [tex]\(x-2\)[/tex] is indeed a factor of the polynomial [tex]\(3x^4 - 6x^3 - 5x + 10\)[/tex]. The quotient polynomial is:
[tex]\[3x^3 + 0x^2 + 0x - 5\][/tex]
Therefore, the answer is:
[tex]\[3x^3-5\][/tex]