Question

X^3+x^2-20x=0.solve this cubic equation.

Answer 1

Answer:

${x}^{3} + {x}^{2} - 20x = 0 \\ \\ x( {x}^{2} + x - 20) = 0 \\ \\ x( {x}^{2} + 5x - 4x - 20) = 0 \\ \\ x(x + 5)(x - 4) = 0 \\ \\1. \: x = 0 \\ 2. \: x = 4 \\ 3. \: x = - 5$

Here's The Answer My Friend....

Answer 2

Answer:

Given: x^3+x^2-20x=0

Step-by-step explanation:

x(x^2+x-20)=0

x(x^2+5x-4x-20)=0

x(x+5)-4(x-5)=0

(x+5)(x-4)=0

x= -5, x= 4

values are,

1. X=0

2. X=4

3. X= -5