Question 251549
v[1] = v^3

v[2] = v^3 + 2v^2 - 3v


let a = amount height was decreased.
let b = amount width was increased.


v * (v-a) * (v+b) = v^3 + 2v^2 - 3v.


find the roots of v^3 + 2v^2 - 3v.


set the equation equal to 0 to get:


v^3 + 2v^2 - 3v = 0


factor out a v to get:


v * (v^2 + 2v - 3) = 0


factor v^2 + 2v - 3 to get:


(v-1) * (v+3) * v = 0


you get:


a = 1
b = 3


confirm by substituting in original equation to get:


v * (v-1) * (v+3) = (v^2 -v) * (v+3) = v^3 + 3v^2 -v^2 - 3v.


combine like terms to get:


v * (v-1) * (v+3) = v^3 + 2v^2 - 3v


since this is identical to the original equation of v^3 + 2v^2 - 3v, then the values for a and b are good.


v*v*v = v^3 = original equation for the cube.


v*(v-1)*(v+3) = v^3 + 2v^2 - 3v = original equation after taking 1 unit off the length and adding 3 units to the width.


answer is:


height was decreased by 1 unit.
width was increased by 3 units.