Solve each of the following cubic equations.
1) x^{3} + 2x^{2} - x - 2 = 0
2) 2x^{3} - x^{2} - 4x + 2 = 0
3) x^{3} - 2x^{2} - 5x + 6 = 0
4) 4x^{3} - 7x + 3 = 0
5) x^{3 }- 23x^{2 }+ 142x - 120 = 0
6) 4x^{3 }- 5x^{2 }+ 7x - 6 = 0
7) x^{3} - 7x + 6 = 0
8) x^{3} + 13x^{2 }+ 32x + 20 = 0
1. Answer :
x^{3} + 2x^{2} - x - 2 = 0
The given cubic equation can be factored by grouping.
Factor and solve.
x^{3} + 2x^{2} - x - 2 = 0
x^{2}(x + 2) - 1(x + 2) = 0
(x + 2)(x^{2} - 1) = 0
x + 2 = 0 x = -2 |
x^{2} - 1 = 0 x^{2} - 1^{2} = 0 (x + 1)(x - 1) = 0 x = -1 or x = 1 |
So, the solution is {-2, -1, 1}.
2. Answer :
2x^{3} - x^{2} - 4x + 2 = 0
The given cubic equation can be factored by grouping.
Factor and solve.
2x^{3} - x^{2} - 4x + 2 = 0
x^{2}(2x - 1) - 2(2x - 1) = 0
(2x - 1)(x^{2} - 2) = 0
2x - 1 = 0 x = 1/2 |
x^{2} - 2 = 0 x^{2} = 2 x = ±√2 |
So, the solution is {1/2, √2, -√2}.
3. Answer :
x^{3} - 2x^{2} - 5x + 6 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
1 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2 }- x - 6 = 0
Factor and solve.
(x - 3)(x + 2) = 0
x = 3 or x = -2
So, the solution is {1, 3, -2}.
4. Answer :
4x^{3} - 7x + 3 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
1 is one of the roots. The other roots can be determined by solving the quadratic equation
4x^{2} + 4x - 3 = 0
Factor and solve.
4x^{2} + 6x - 2x - 3 = 0
2x(2x + 3) - 1(2x + 3) = 0
(2x + 3)(2x - 1) = 0
x = -3/2 or x = 1/2
So, the solution is {1, -3/2, 1/2}.
5. Answer :
x^{3 }- 23x^{2 }+ 142x - 120 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
1 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2} - 22x + 120 = 0
Factor and solve.
(x - 10)(x - 12) = 0
x = 10 or x = 12
So, the solution is {1, 10, 12}.
6. Answer :
4x^{3 }- 5x^{2 }+ 7x - 6 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
1 is one of the roots. The other roots can be determined by solving the quadratic equation
4x^{2 }- x + 6 = 0
This quadratic equation can not be solved by factoring. So use quadratic formula and solve.
x = [-b ± √(b^{2} - 4ac)]/2a
a = 4, b = -1 and c = 6,
x = (1 ± √-95)/8
For the given cubic equation, there is only one real root, that is 1. The remaining two roots are imaginary.
7. Answer :
x^{3} - 7x + 6 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
1 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2 }+ x - 6 = 0
Factor and solve.
(x + 3)(x - 2) = 0
x = -3 or x = 2
So, the solution is {1, -3, 2}.
8. Answer :
x^{3} + 13x^{2 }+ 32x + 20 = 0
Check whether the values 1, -1, 2, -2, ...... could be one of the solutions using synthetic division.
-1 is one of the roots. The other roots can be determined by solving the quadratic equation
x^{2 }+ 12x + 20 = 0
Factor and solve.
(x + 2)(x + 10) = 0
x = -2 or x = -10
So, the solution is {-1, -2, -10}.
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