Answer :
To find the remaining solutions of the polynomial equation [tex]\(x^4-11x^3+21x^2+59x-70 = 0\)[/tex] after two known solutions ([tex]\(-2\)[/tex] and [tex]\(1\)[/tex]), we will use synthetic division.
1. First Synthetic Division with root [tex]\(-2\)[/tex]:
Set up the synthetic division using the polynomial coefficients: [tex]\(1, -11, 21, 59, -70\)[/tex].
- Bring down the leading coefficient: [tex]\(1\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(1 \times -2 = -2\)[/tex].
- Add to the next coefficient: [tex]\(-11 + (-2) = -13\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(-13 \times -2 = 26\)[/tex].
- Add to the next coefficient: [tex]\(21 + 26 = 47\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(47 \times -2 = -94\)[/tex].
- Add to the next coefficient: [tex]\(59 + (-94) = -35\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(-35 \times -2 = 70\)[/tex].
- Add to the last coefficient: [tex]\(-70 + 70 = 0\)[/tex].
The result is a new polynomial: [tex]\(1x^3 - 13x^2 + 47x - 35\)[/tex].
2. Second Synthetic Division with root [tex]\(1\)[/tex]:
Use the coefficients of the new cubic polynomial: [tex]\(1, -13, 47, -35\)[/tex].
- Bring down the leading coefficient: [tex]\(1\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(1 \times 1 = 1\)[/tex].
- Add to the next coefficient: [tex]\(-13 + 1 = -12\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(-12 \times 1 = -12\)[/tex].
- Add to the next coefficient: [tex]\(47 + (-12) = 35\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(35 \times 1 = 35\)[/tex].
- Add to the last coefficient: [tex]\(-35 + 35 = 0\)[/tex].
The result is a new quadratic polynomial: [tex]\(1x^2 - 12x + 35\)[/tex].
3. Solve the Quadratic Equation:
The resulting quadratic equation is [tex]\(x^2 - 12x + 35 = 0\)[/tex].
Factor this equation to find the remaining solutions:
- This factors into [tex]\((x - 7)(x - 5) = 0\)[/tex].
- Setting each factor to zero gives the solutions:
[tex]\[
x - 7 = 0 \quad \Rightarrow \quad x = 7
\][/tex]
[tex]\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\][/tex]
Thus, the remaining solutions of the polynomial are [tex]\(x = 7\)[/tex] and [tex]\(x = 5\)[/tex]. Together with the known solutions, the complete set of solutions is [tex]\(-2, 1, 5, \)[/tex] and [tex]\(7\)[/tex].
1. First Synthetic Division with root [tex]\(-2\)[/tex]:
Set up the synthetic division using the polynomial coefficients: [tex]\(1, -11, 21, 59, -70\)[/tex].
- Bring down the leading coefficient: [tex]\(1\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(1 \times -2 = -2\)[/tex].
- Add to the next coefficient: [tex]\(-11 + (-2) = -13\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(-13 \times -2 = 26\)[/tex].
- Add to the next coefficient: [tex]\(21 + 26 = 47\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(47 \times -2 = -94\)[/tex].
- Add to the next coefficient: [tex]\(59 + (-94) = -35\)[/tex].
- Multiply by [tex]\(-2\)[/tex]: [tex]\(-35 \times -2 = 70\)[/tex].
- Add to the last coefficient: [tex]\(-70 + 70 = 0\)[/tex].
The result is a new polynomial: [tex]\(1x^3 - 13x^2 + 47x - 35\)[/tex].
2. Second Synthetic Division with root [tex]\(1\)[/tex]:
Use the coefficients of the new cubic polynomial: [tex]\(1, -13, 47, -35\)[/tex].
- Bring down the leading coefficient: [tex]\(1\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(1 \times 1 = 1\)[/tex].
- Add to the next coefficient: [tex]\(-13 + 1 = -12\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(-12 \times 1 = -12\)[/tex].
- Add to the next coefficient: [tex]\(47 + (-12) = 35\)[/tex].
- Multiply by [tex]\(1\)[/tex]: [tex]\(35 \times 1 = 35\)[/tex].
- Add to the last coefficient: [tex]\(-35 + 35 = 0\)[/tex].
The result is a new quadratic polynomial: [tex]\(1x^2 - 12x + 35\)[/tex].
3. Solve the Quadratic Equation:
The resulting quadratic equation is [tex]\(x^2 - 12x + 35 = 0\)[/tex].
Factor this equation to find the remaining solutions:
- This factors into [tex]\((x - 7)(x - 5) = 0\)[/tex].
- Setting each factor to zero gives the solutions:
[tex]\[
x - 7 = 0 \quad \Rightarrow \quad x = 7
\][/tex]
[tex]\[
x - 5 = 0 \quad \Rightarrow \quad x = 5
\][/tex]
Thus, the remaining solutions of the polynomial are [tex]\(x = 7\)[/tex] and [tex]\(x = 5\)[/tex]. Together with the known solutions, the complete set of solutions is [tex]\(-2, 1, 5, \)[/tex] and [tex]\(7\)[/tex].
