wrote an equation for the line. passing through (-5,7) and parallel to the line whise equation is 5x-3y-4=0

Step-by-step explanation:

5x-3y-4=0(1)

Slope = - coefficient of X/ coefficient of y

= -5/(-3)

= 5/3

(x1,y1)= (-5,7)

Equation of the line,

Y-Y1= slope(X-X1)

Y-7= (5/3)×(X-(-5))

Y-7= (5x+25)/3

3y-7=5x+25

5x-3y+32=0 is the required eqn

f(x) = x^3 -2x^2 -3x

x( x^2 -2x-3)

x( x^2 +x -3x -3)

x( x(x+1) -3(x+1))

x( x-3)( x+1)

so zeroes are 0, 3,-1

its (2) option