Solve the equation 3 x 3 - 4 x 2 + x + 88 = 0 which has 2 - - 7 as a root.

Given that, 2 - 7 i is one root. ⇒ 2 + 7 i is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± 7 i is x 2 - 2 - 7 i + 2 + 7 i x + 2 - 7 i 2 + 7 i = 0 ⇒ x 2 - 4 x + 11 = 0 . ∴ x 2 - 4 x + 11 = 0 is a factor of 3 x 3 - 4 x 2 + x + 88 = 0 . Applying synthetic division. 4 - 11 3 0 0 - 4 12 0 1 32 - 33 88 0 - 88 3 8 0 0 ⇒ 3 x + 8 = 0 ⇒ x - 8 3 Therefore, the roots of the given equation are 2 ± 7 i , - 8 3 .

Solve x 4 - 4 x 2 + 8 x + 35 = 0 , given that 2 + i 3 is a root.

Given that, 2 + i 3 is one root. ⇒ 2 - i 3 is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± i 3 is x 2 - 2 + i 3 + 2 - i 3 x + 2 + i 3 2 - i 3 = 0 ⇒ x 2 - 4 x + 7 = 0 . ∴ x 2 - 4 x + 7 = 0 is a factor of x 4 - 4 x 2 + 8 x + 35 = 0 . 4 - 7 1 0 0 0 4 0 - 4 16 - 7 8 20 - 28 35 0 - 35 1 4 5 0 0 ⇒ x 2 + 4 x + 5 = 0 Now, applying quadratic formula. ⇒ x = - 4 ± 16 - 20 2 x = - 4 ± 2 i 2 = - 2 ± i Therefore, the roots of the given equation are 2 ± i 3 , - 2 ± i .

Solve the equation x 4 - 6 x 3 + 11 x 2 - 10 x + 2 = 0 , given that 2 + 3 is a root of the equation.

Given that, 2 - 3 i is one root. ⇒ 2 + 3 i is another root. We know that, the equation having roots α , β is x 2 - α + β x + α β = 0 The equation having roots 2 ± i 3 is x 2 - 2 + 3 i + 2 - 3 i x + 2 + 3 i 2 + 3 i = 0 ⇒ x 2 - 4 x + 1 = 0 . ∴ x 2 - 4 x + 1 = 0 is a factor of x 4 - 6 x 3 + 11 x 2 - 10 x + 2 = 0 . 4 - 1 1 0 0 - 6 4 0 11 - 8 - 1 - 10 8 2 2 0 - 2 1 - 2 2 0 0 ⇒ x 2 - 2 x + 2 = 0 Now, applying quadratic formula. ⇒ x = 2 ± 4 - 8 2 = - 2 ± 2 i 2 ⇒ x = - 1 ± i Therefore, the roots of the given equation are 2 ± 3 , 1 ± i .

Given that - 2 + - 7 is a root of the equation x 4 + 2 x 2 - 16 x + 77 = 0 , solve it completely.

Given that, - 2 - 7 i is one root. ⇒ - 2 + 7 i is another root. The equation having roots - 2 ± 7 i is x 2 - - 2 - 7 i + - 2 + 7 i x + - 2 - 7 i - 2 + 7 i = 0 ⇒ x 2 + 4 x + 11 = 0 . ∴ x 2 + 4 x + 11 = 0 is a factor of x 4 + 2 x 2 - 16 x + 77 = 0 . Now, applying synthetic division - 4 - 11 1 0 0 0 - 4 0 2 16 - 11 - 16 - 28 44 77 0 - 77 1 - 4 7 0 0 ⇒ x 2 - 4 x + 7 = 0 Now, applying quadratic formula ⇒ x = 4 ± 16 - 28 2 = 4 ± 2 3 i 2 = 2 ± 3 i Therefore, the roots of the given equation are - 2 ± i 7 , 2 ± 3 i .

Solve the equation x 4 + 2 x 3 - 16 x 2 - 22 x + 7 = 0 , given that 2 - 3 is one of its roots.

Let 2 - 3 is one root. If a - i b is a root then a + i b is also a root. ⇒ 2 + 3 is another root. The equation having roots α , β is given by x 2 - α + β x + α β = 0 The equation having roots 2 ± 3 is x 2 - 2 - 3 + 2 + 3 x + 2 - 3 2 + 3 = 0 ⇒ x 2 - 4 x + 1 = 0 . ∴ x 2 - 4 x + 1 = 0 is a factor of x 4 + 2 x 3 - 16 x 2 - 22 x + 7 = 0 . Now, applying synthetic division 4 - 1 1 0 0 2 4 0 - 16 24 - 1 - 22 28 - 6 7 0 - 7 1 6 7 0 0 ⇒ x 2 + 6 x + 7 = 0 On applying quadratic formula ⇒ x = - 6 ± 36 - 28 2 x = - 6 ± 8 2 = - 3 ± 2 Therefore, the roots of the given equation are 2 ± 3 , - 3 ± 2 .

Solve the equation 3 x 5 - 4 x 4 - 42 x 3 + 56 x 2 + 27 x - 36 = 0 given that 2 + 5 is one of its roots.

Let 2 + 5 is one root. ⇒ 2 - 5 is another root. The equation having roots 2 ± 5 is x 2 - ( 2 + 5 + 2 - 5 ) x + ( 2 + 5 ) ( 2 - 5 ) = 0 ⇒ x 2 - 2 2 x - 3 = 0 The equation having roots ± 2 ± 5 is x 2 + 2 2 x - 3 x 2 - 2 2 x - 3 = 0 ⇒ x 2 - 3 + 2 2 x 2 - 3 - 2 2 = 0 ⇒ x 2 - 3 2 - ( 2 2 x ) 2 = 0 ⇒ x 4 - 6 x 2 + 9 - 8 x 2 = 0 ⇒ x 4 - 14 x 2 + 9 = 0 Given 3 x 5 - 4 x 4 - 42 x 3 + 56 x 2 + 27 x - 36 = 0 ⇒ 3 x x 4 - 14 x 2 + 9 - 4 x 4 - 14 x 2 + 9 = 0 ⇒ x 4 - 14 x 2 + 9 ( 3 x - 4 ) = 0 ⇒ x = ± 2 ± 5 or x = 4 3 Therefore, the roots of the given equation are ± 2 ± 5 , 4 3 .