Take out x as the common factor:
x(2x^2-5x-3)=0;
From here, you can just factorise normally or complete the square.
x(2x^2+x-6x-3)=0;
x(x(2x+1)-3(2x+1))=0;
x(2x+1)(x-3)=0;
x=-1/2,0,3
SOLVE BY FACTORING...THANK YOU
2X^3 - 5X^2 - 3X=0
2 answers
2 x³ - 5 x² - 3 x = 0
x ( 2 x² - 5 x - 3 ) = 0
Factor of 2 x² - 5 x - 3:
2 x² - 5 x - 3 = ( 2 x² + x ) - 6 x - 3 = x ( 2 x + 1 ) - ( 6 x + 3 ) =
x ( 2 x + 1 ) - 3 ( 2 x + 1 ) =
( 2 x + 1 ) ∙ x - ( 2 x + 1 ) ∙ 3 = ( 2 x + 1 ) ( x - 3 )
2 x³ - 5 x² - 3 x = x ( 2 x + 1 ) ( x - 3 )
Using the Zero Factor Principle: If a b = 0 then a = 0 or b = 0 or both a = 0 and b = 0
In this case x = 0 and ( 2 x + 1 ) ( x - 3 ) = 0
So one of solution is x = 0
Now solve 2 x + 1 = 0
2 x = - 1
x = - 1 / 2
Solve x - 3 = 0
x = 3
The solutions are:
x = - 1 / 2 , x = 0 and x = 3
x ( 2 x² - 5 x - 3 ) = 0
Factor of 2 x² - 5 x - 3:
2 x² - 5 x - 3 = ( 2 x² + x ) - 6 x - 3 = x ( 2 x + 1 ) - ( 6 x + 3 ) =
x ( 2 x + 1 ) - 3 ( 2 x + 1 ) =
( 2 x + 1 ) ∙ x - ( 2 x + 1 ) ∙ 3 = ( 2 x + 1 ) ( x - 3 )
2 x³ - 5 x² - 3 x = x ( 2 x + 1 ) ( x - 3 )
Using the Zero Factor Principle: If a b = 0 then a = 0 or b = 0 or both a = 0 and b = 0
In this case x = 0 and ( 2 x + 1 ) ( x - 3 ) = 0
So one of solution is x = 0
Now solve 2 x + 1 = 0
2 x = - 1
x = - 1 / 2
Solve x - 3 = 0
x = 3
The solutions are:
x = - 1 / 2 , x = 0 and x = 3