Solve $18 x^{3} + 81 x^{2} + 121 x + 60 = 0$ give that the root is equal to the sum of the remaining roots?

Open in App

Solution

$18 x^{3} + 81 x^{2} + 121 x + 60 = 0$

$18 x^{3} + 81 x^{2} + 121 x + 60$

$= (2 x + 3) \frac{18 x^{3} + 81 x^{2} + 121 x + 60}{2 x + 3}$

$= 9 x^{2} + 27 x + 20$

$= (2 x + 3) (3 x + 4) (3 x + 5)$

$(2 x + 3) (3 x + 4) (3 x + 5) = 0$

$x = - \frac{3}{2}, x = - \frac{4}{3}, x = - \frac{5}{3}$

Suggest Corrections

Similar questions

18 x^{3} + 81 x^{2} + 121 x + 60 = 0

18 x^{3} + 81 x^{2} + 121 x + 60 = 0

24 x^{3} - 14 x^{2} - 63 x + λ = 0

λ

18 x^{3} + 81 x^{2} + λ x + 60 = 0

λ

x^{4} - 18 x^{3} + k x^{2} + 200 x - 1984 = 0

(- 32)

k

x^{3} - 12 x^{2} + k x - 18 = 0

k =

Join BYJU'S Learning Program