Solve Cubic Equations - Total Sum Method

[khaizzmoko]

sir where did u get the 6, 3, 2 that multiply those #?

[vinayagankuppusamy]

You must show another example for cases where the sum of constants not equal to zero. Thank you sir

[rezwan]

YOUR EXPLANATIONS AND METHODS ARE GREAT. THANKS A LOT.

[benjaminkarazi]

Hello,
6*x²+1*x-15=0 ⇒ (2*x-3)*(3*x+5)=0, how would you determine -3 and +5 for +1 and -15?
Thank you,
Regards,

[rockstar]

If the roots are xyz:
x+y+z=-(B/A),
xyz=-(D/A),
xy+yz+zx=C/A.

[vinayagankuppusamy]

What if the sum of the four constants is not equal to zero?

[aberzinci]

Can you solve it step by step and easily please x³+5x²-7x+9=0

[nalinikantamohapatra]

If the sum of A B C D will not be zero

[riteshsinghkinwar]

If there is not Sum of A, B, C&D equal to 0

[devondevon]

Answer x = 1 , x= - 5/3 and x= 3/2
6x^3 -5x^2 -16x +15=0
1 6 -5 -16 + 15
6 1 -15
6 1 -15 0 using synthetic division
(x-1)( 6x^2 + x -15)
Therefore 6x^3 -5x^2-16 + 15= (x-1)(6x^2 + x -15)=0
6x^2+ x-15 = (3x+5)(2x-3)
Therefore (x-1)(3x+5)(2x-3)=0 when x is +1 or 3x is -5 or 2x is +3, the left side will = 0 thus the same as the right side
x = 1
3x+5=0, 3x=-5 x= -5/3
2x-3=0 2x=3 x=3/2

[theophonchana]

6x^(2) + x - 15 = 0
(2x - 3) × (3x + 5) = 0

[reinamaeda]

I plugged the x=-3/5 into the original equation, but I'm confused as to why I didn't get0 no matter how I calculated it. X=3/2 turned out to be 100% correct.